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## Thursday, March 12, 2015

### ATLAS: a 3-sigma supersymmetric excess

Three standard deviations mean that the null hypothesis doesn't seem "great". The deviation from the predictions isn't enough for a discovery in a hard scientific discipline such as particle physics. However, I am convinced that a 3-sigma deviation – formally equal to a 99.7% certainty of a new effect – simply has to be publicized because it's interesting enough. We've covered lots of 2-sigma or 2.5-sigma deviations so it would be unfair to be silent about a 3-sigma one.

The new ATLAS preprint with this potentially interesting finding is called
Search for supersymmetry in events containing a same-flavour opposite-sign dilepton pair, jets, and large missing transverse momentum in $$\sqrt{s}=8\TeV$$ $$pp$$-collisions with the ATLAS detector
So look at it: we want final states that contain $$\ell^+\ell^-$$, a lepton pair – it's the leptons that apparently follow from a decay of the Z-bosons that will matter; jets; and large missing transverse momentum (assumed to be composed of the lightest superpartners such as the lightest neutralinos – possibly particles of dark matter). The Standard Model didn't do too well.

One shouldn't overstate the deviations from the Standard Model. Most of the channels are OK. For example, a particular channel in the ATLAS data refused to confirm a recently announced CMS' 2.6-sigma excess – although the kinematic constraints were slightly different in the two experiments.

However, while rejecting this CMS excess, ATLAS saw an even larger one. It was in the events in which the lepton pair looked like one from a Z-boson of the correct mass (no neutrinos are produced in that). The key table is table 7 which describes SR-Z (the signal region with a Z-boson):$\begin{array}{|c|c|c|c|} \hline {\rm Channel} & e^+e^- & \mu^+\mu^- & \ell^+\ell^- combo\\ \hline {\rm expect} &4.2\pm 1.6& 6.4\pm 2.2 & 10.6\pm 3.2\\ \hline {\rm observed} & 16&13&29 \\ \hline \end{array}$ The excess is mostly in the electron pair channel, 3.0 sigma again, but even in the dimuon channel, there is some excess, 1.7 sigma. When combined, the dilepton channels are back to 3 sigma. One expects $$10\pm 3$$ and gets $$29$$ events, not bad! I vaguely remember some other recent excess that – surprisingly or disappointingly – appeared mostly in the electron channel only but I can't figure out where it was. Do you know where we saw it?

As far as I understand, CMS hasn't released its own data on the on-Z signal region where ATLAS is seeing the excess yet.

If you want to believe that this excess is from new physics, it may be a gluino or a squark that is decaying. One exclusion plot – Figure 13b – weakly suggests that the exclusion curve is being repelled from a point with a $$1\TeV$$ gluino and a $$600\GeV$$ lightest neutralino which may be "the truth" for those who want to be real optimists.

(On Monday, March 16th, this paper by Barenboim et al. will appear that will make pretty much the same conclusion as I did 4 days earlier. A SUSY scenario of "general gauge mediation" is the simplest thing that is needed. The gravitino is light and it's the LSP. Fundamental higher-energy scales are at hundreds of TeV. Only this paragraph was added after March 12th.)

Your humble correspondent doesn't really understand whether they assume these strongly interacting superpartners to be pair-produced and if they do, why the other member of the pair doesn't destroy the identity of the final state. But maybe it's giving the jets and the missing transverse energy is only required to "be there" and its magnitude isn't important. Can you help me?

New Higgs CMS excess

CMS will release a paper tomorrow looking for heavier friends of the now familiar $$125\GeV$$ Higgs boson. There is a 2.56-2.64 sigma excess in the mass range $$700$$-$$800\GeV$$. The newer papers released recently seem to have many more excesses than 1+ year ago – which supports the conjecture that whenever they see something "not quite mundane", they (both ATLAS and CMS) are delaying the publication, an approach that may be reasonable but also a potentially dangerous bias we should be aware of (and force us to deduce lessons from "comparative literature" – like the statistical evaluation "how many excesses have been seen anywhere" – much more carefully than otherwise).

Incidentally, a few hours ago, there was a CERN webcast kickstarting the new season, the 2015 run. You may watch the recorded press conference here. Be ready for lots of French and German accents. I believe that the content isn't important enough to be analyzed.

1. Assume that there is betting and if she guess correctly she wins some money. If you are using a betting argument, the total stake when coin lands heads or tails should be same. This way you one easily sees that the correct answer should be 1/2. If you bet same money each time she awakens, in effect you bet more money when it lands tails. You should bet 1/3 $each time she awakens when coin lands tails, if you bet 1$ when she awakens when it lands heads.

2. Hi again Lubos,
fine post, with such nonsense out there I am starting to understand why you need to be so "tough".

I don't understand how some people who even are hired scientists cannot get that you are right on this. Maybe this explanation will help.

Imagine a hundred sleeping beauties, and a hundred beds. I paint half of them red, and leave half of them white. The hundred beauties are now put to sleep, distributed on the beds at random, and the ones on the white beds are woken up once and asked which color bed they think they lie on, while the ones on the red beds are woken up twice (or a million or a billion times, it doesn't matter for this part of the question) and asked about the probability of the color of their bed. Because there are 50 beds of each kind it is pretty clear that every beauty will say that there are a 50 percent chance they are lying on a red bed, and 50 percent that they are lying on a white bed.

3. If you regard probability as a subjective evaluation of the probability of certainty, then the withholding of information is a condition that must be taken into account, not a cheat. The correct probability under this definition is then 1/3. And whining about the cheating is like whining about what things in QM are really like, instead of just accepting the probabilities of the observables are all we can know.

Of course, by a frequentist interpretation, you are correct. Modern mathematics is apparently based on rigorous proofs but the conclusions still seem to depend upon what one chooses to assume. This has left me unsure about what kind of knowledge mathematics is.

PS There is a typo in Bayes' theorem above I think. Also, it is definitely wrong to imply there's anything Marxist about DeLong, who is a committed anti-Marxist.

4. This sleeping beauty thing also reminds me of the Monty Hall problem. 3 doors, behind two of them you find a sheep and behind the last you find a prize. You are asked to pick one at the beginning, then the game host will open another door revealing a sheep (he always chooses a door that reveals a sheep) and the participant can now keep the first door he chose or switch to the other option. What should he do? Assume the prize was really behind the first door, and play out the three different scenarios where he chooses number one, number two or number three at first. So if he chooses to keep his first choice he only wins when his first choice was door number 1, that is 1/3 of the time. If however he chose door 2 or 3 at first and switches (in which case door 3 and 2 respectively is revealed with a sheep behind) he wins 2/3 times, so therefore it is more sensible to switch.
Another way to make it easier to see is to choose a door among 1 million, and then the game host opens all the other doors (always revealing sheep, he knows where the prize is) and now it is obvious that it is better to switch.

In the tv-game deal or no deal, the participant is asked to choose a suitcase among 20 containing different money amounts from almost nothing up to a million. After this first choice he is asked to gradually choose one at a time from the remaining ones revealing their content. Say he continues the game until he only has his first choice and one remaining suitcase among the remaining ones, and the amounts one dollar and one million dollars are left. Should he switch?
In this case it doesn't matter because he chose all the suitcases himself and choosing which suitcases to eliminate is no different from choosing the ones you want to keep(as he did with the first one). So it makes no difference. However for unknown psychological reasons I find it much cooler to keep the first choice ;-).

5. Yes. I think the 1/3-er crowd find this confusing because they sense intuitively (and correctly) that if this was a game where the win condition was to get >50% correct guesses over repeated runs of the experiment, then obviously a winning strategy for beauty is to predetermine to always guess tails.

But this is due simply to the fact that a correct guess of tails is effectively rewarded with "2 points", whereas a correct guess of heads is only awarded "1 point". We can imagine an isomorphic game without the distraction of the repeated awakenings + memory erasure where you repeatedly toss a coin and I simply guess tails all the time and collect my 2/3 (points-weighted) overall score.

Obviously on a single run of the experiment with only one coin toss, whether she consistently guesses heads or tails beauty has a 50% chance to win and there is no winning strategy, and over repeated large-N runs half the coin tosses will have come up heads and the other half tails.

The one thirders seem almost to imagine that were beauty to be woken 99 times in the case of tails, that this would somehow retrocausally bias the actual coin tosses to be 99% tails!

6. Exactly, Michael, you are using a similar emphasis as Bob Walters, see the Markov text

http://motls.blogspot.com/2014/08/sleeping-beauty-thirders-rudimentary.html?m=1

The key in your analogous description of the situation is that the total number of beds - the denominator used for the probabilities - is fixed to constant at the beginning.

Only with this choice, one may interpret the fractions as probabilities. In effect, this is equivalent to saying that we are computing - and we must be computing - probabilities of evolution to a certain state during the week but *from* the known initial state.

The possibilities are mutually exclusive only if the "tree of evolution" branches. But if one follows a single branch, like beauty with tails on Monday and Tuesday, those really share the same number of "beds" to start with, the same probability, and if one wants to decide about the probability of coin_state+day, the probabilities are "pouring from one day to another" but just in between the points on the same branch, not to other branches, as long as something on the branch is alive.

7. Dear Steven, the bets that the bookmaker offers on Monday and on Tuesday are *different* bets with *different* odds for heads and tails (50-50 vs 0-100), and if he pretends that they're the same while he can influence whether the client is woken up on one day or another, he may manipulate the betting to his advantage.

If she is offered these bets in the "thirder" way, she can of course figure out that she should demand 3 times the price of the lottery ticket by which she bets on "heads". But this sentence is not equivalent to the statement that the probability of "heads" at the point when she is woken up is 1/3 because the probabilities are only linked to the corresponding ratios from the betting if the bookmaker can't manipulate with the asymmetric knowledge.

8. I don't see how there's any room for confusion about this problem except that people *think* they have a precise understanding of some terms but actually don't. There's no need to dodge the bookmaker scenario. Simply describe in advance what betting tickets will be offered (a neutral constable will enforce this contract). If we're making odds for a betting ticket to be offered either once in event of H or twice for T, this should lead naturally to Thirder indifferent-odds (H ticket paying off 2:1). It's up to you if you choose to call that contractually-agreed payoff ratio the 'proper subjective credence'. I do like your argument that even if you're going to be woken (w/ memory erasure) thousands of times if you're in Guantanamo you shouldn't be subjectively near-certain that you're being tortured in Guantanamo :)

9. Do you agree, Michael, that the same logic that gives 1/2 for the sleeping beauty is the logic that gives 2/3 for the Monty Hall problem? This numerical difference may be confusing - because the results are kind of reverted in the two problems.

10. Thanks for the answer Lubos.
I will look at the other article :-)

11. The same logic, hmm, I agree that the answer is 1/2 for the sleeping beauty as I also mentioned in my comment below about the colored beds. I am the same Michael as him. But maybe you know that and mean something else?

12. There were a few more but the Markov one was a bit more focused on formalism - and based on a different man who carefully studied it - so it may be more interesting than to read my other, similar texts.

13. Apologies, I don't understand why you seem anxious here.

Did I count Michael at one moment and Michael at another moment as two different people?

That would be the fallacy that the "thirders" are doing to the sleeping beauty - they effectively pretend that the tails-sleeping beauty when woken up on Tuesday is a different lady than the tails madam who woke up a day earlier.

14. Exactly - to interpret the fractions of winning tickets as probabilities, one has to reward each successful guess with 1 point, while the correct "tails guess" is rewarded by 2 points in the counting of the thirders.

A way to fix it is to guarantee that e.g. when it's tails, she is sold a half-price ticket on Monday and half-price ticket on Tuesday so that in total, she buys 1 full ticket in the case of tails just like she does in the case of heads.

With this fix, she may assume that the probabilities for heads and tails are the correct 50% and she will get her money back if the winning tickets pay 2 times what she bet. In effect, the only deviation from the simple head-tails bet is that in the tails week, the same ticket is divided to two halves.

Your 99-day extremization of the wrong calculation ad absurdum is almost equivalent to my previous blog post about the sleeping beauty in Guantanamo Bay:

http://motls.blogspot.com/2014/08/sleeping-beauty-in-guantanamo-bay.html?m=1

15. Okay, sorry.
The probabilities change when the beauty is given some new information about the day, as you explained, and the probabilities also gets updated in a similar way in the Monty Hall problem when the participant learns something new.

16. I agree with Luboš that this is a good different example. The problem I have is easily illustrated with it:
The sleeping beauties all know that they will not be woken in 1/4 of the cases. So only 75 beauties are effectively asked whether they are among the 50 on one side or the twenty five on the other side of bed. far as I can tell the beauties are given the information that they are making a selection among three equally likely outcomes because they know 25 of 100 beauties are never asked.

17. Hey awesome twist and a much more dramatic and clear explanation! :-)

This kind of confusion is similar to the kinds that arise with anthropic reasoning about multiverse "populations" and "democratically selected" typical-observers etc, though I'll grant that it's harder to keep a clear head when infinite measures are involved.

In finite cases like with sleeping beauty, I can't see that there's much excuse for a "professional" getting it wrong though!

Here's an example of an infinite case of a probability measure "paradox" (countably infinite prisoners with black and white hats) which is fun:

https://cornellmath.wordpress.com/2007/09/13/the-axiom-of-choice-is-wrong/

*Really* stupid blog-post title, sorry, (well it's not my blog!). But there are a couple of great posts from Terry Tao in the comments section about the breakdown of probability additivity assumptions when infinite (and especially non-measurable) sets are in play.

Don't show it to DeLong and the 1/3-ers though, it's liable to blow their minds/cause smoke to come out their ears... ;-)

18. You only update the probabilities when you get new information. Imagine that *you* are lying in the bed, remember nothing about if you already have been woken up or not, and you know that half of the beds are red and half white. You will answer 50 percent chance for the color of your bed without additional information.
But if you are told the day, like Monday, its different. First calculate the probabilities for this case to happen in the two cases. For one color of the bed you get 50 percent for the color multiplied by the probability that it is this day in the subspace you are in. What do you get? You get (as Lubos said) 0,5(the color red)*(the day, monday for this case)1 , and the probability for the other event is 0,5(the color white)*(the day, monday)0,5. So 50 percent and 25 percent. Now you know that it is Monday so you must "renormalize" to find the probability for the color. So 50 percent out of 50+25 percent is 2/3, which is your probability for the red bed when told Monday.

19. Michael KolczynskiMar 12, 2015, 7:12:00 PM

Luboš, I posted a comment just under you giving him a basic course in Bayesian inference. He deleted my comment then sent me an email saying "I don't think you read the problem correctly..." I sure did.

20. LOL, he also has an ego. I posted a comment mentioning that this new blog entry exists and it's still there. Some avoidance of anything controversial - if I avoid the term sycophancy - may help. ;-)

21. Michael KolczynskiMar 12, 2015, 8:10:00 PM

That was exactly what my reply to him was about. He said it's unsupported so I used a matrix and then just the plain old Bayes Rule. He merely said I don't understand the problem.

Turning a Bayesian question into a game theoretic one might be an interesting exercise in AI, but it's not what it is now.

22. Oops I claimed this to be so simple and now I have succesfully confused myself. The multiplication of the probabilities and the 50 percent for Monday or Tuesday in the case where she is woken up both days, that are multiplied on, are treated as exclusive "events" in the above comment, either one happens or the other. But they don't exclude one another, both Monday and Tuesday will happen as time flows. So now I think telling her Monday won't change anything, because Tuesday wasn't an option yet, it doesn't "live" with Monday. Therefore Monday and Tuesday cannot be clumped together, so the probability remains 50-50 after being told Monday. Being told Tuesday will of course still complete reveal the color of the bed. Sorry, that I either am confused now or was confused a moment ago. How embarrassing.

23. The sleeping beauty knows she is being asked. The information that she is awake and asked tells her she is not in the 1/4 of not awake beauties who are not asked. Among the 75 beauties who may be asked these are evenly the other three outcomes of day and flips. The outcome resulting in the question being asked eliminates the fourth outcome.

24. Nice :-)!

25. A lot of naysayers will regret their naysaying if SSM is measured at the upgraded LHC.

26. As I tried to write a reply to Rehbock, and repeating the calculation that the probability change when she is told its Monday, made me think that there is a mistake in the conditional probabilities concerning learning about it being Monday. The reason is that Tuesday and Monday don't "live together as possibilities at the same moment". If she was thrown into a Monday box always in the case of tails and a Tuesday box half the time and a Monday box the other half if it was heads, the calculation of the 2/3 for tails upon learning about it being Monday would be true. But because Monday and Tuesday are separated in time and Monday always happens and she is always woken up on Mondays I don't think telling her its Monday changes anything, so even this additional information keeps the probability for heads or tails as 50-50. Do you agree that the separation in time makes this difference?

27. Speaking of three-sigmas (I usually use a max. of two for AMS dating, hopefully aided by Bayesian if they come-out in order in stratigraphic sequence), how about this for a three-Sigma outlier event, a recent political suicide (there was also a 'defenestration 10 days ago), by an ex-Yank ally:

http://www.bbc.com/news/world-europe-31855700

28. I'm sure Jeffrey Epstein or Rajendra Pachauri would be willing to do the experiments if someone can supply the babes.

29. DeLong is not a big admirer of Marx's. He makes only three positive comments on Marx's economic work. First, he approves Marx's recognition there is a business cycle, not just inexplicable crises. Big but, though, he believes the business cycle has been corrected (yes, really, events of the last decade are just a mistake.) so this is purely of historical interest. Second, he approves Marx's advocacy of industrial civilization, which strikes me as approving someone for liking Mom and loving apple pie, whereas disapproving seems at best churlish. And the third he approves is some economic history. For economists, economic history is not even "of historical interest," so this is not high praise. The rest is negative.

As for Sleeping Beauty? Again, the probabilities can be viewed as measures of the subjective certainty of outcome. As such, she can meaningfully calculate probabilities, that is, how certain her knowledge of possible outcomes is, even if she doesn't know the day. If she wants to know the magnitude of her uncertainty, she has to calculate. What she can't do, which I think is what you're saying, is calculate the real frequency from the information she has. The thing is, from a subjective probability standpoint, frequency is not probability, and "real" is moot.

30. I am sure that the whole thing lies in the way one understands the question. This is how I understand it: she is woken and asked to bet her money on whether it was tail or heads at the beginning of the particular week (or the previous Friday or whatever). She knows she is equally likely to have been woken heads/mo, tails/mo and tails/tu. If she always bets against heads, she will be winning 2/3 of the time. In other words, in tails weeks, she is given the chance to win twice, while in head weeks she only loses once... Am I wrong in the interpretation of the question?

31. I think that all the confusion is caused by the fact that the question for the sleeping beauty is a bit ambiguous. Probability of exactly what are we asking about?

1) The whole experiment is the coin toss and nothing else. All the sleeping and waking stuff after that is irrelevant. The probability is 1/2.

2) The experiment consists of the coin toss PLUS some follow-up actions that depend on the toss outcome. It all ends with a "measurement" of the result. The scheme is such that only 1/3 of measurements will come out as heads, the remaining 2/3 will be tails. If you ask "what is the probability that any specific measurement will give a heads result", then the answer is 1/3. You're asking not just about the "coin+measure" system, but about the "coin+sleep/wake+measure" system.

It's telling that both authors formulate the question differently:

L: calculate the probability that the coin landed "heads" on Sunday.
B: what is your confidence that the coin flip winds up heads?

The first formulation is much more precise than the sloppy second one.

32. Sorry I was reading this "ℓ+ℓ−, a lepton pair – " and try to figure out what the hell you meant when I was distracted by an ad for womens underwear. Then I came to this page where I was distracted by an ad for womens underwear. For some reason I find woman in underwear distracting. If you want me to focus on physics you might want to have fewer ads.

33. DeLong is a ginormous asshole. His primary intention to discredit Lewis is obvious. He used you as a pawn expecting that quoting your usual non-PC style and global warming opinions will win him an argument with little effort on his side, since his entire post is otherwise an incoherent babble and confused self-gloating.

That this elementary sophist strategy worked is clear from some responses.

34. Dear Jaroslave, accuracy is great and important but I don't see in what interpretation your "L" sentence and your "B" sentence could possibly differ, and I don't even see why you claim that one of them is formulated more accurately than the other. And I don't see what is the exercise that gave you the result of 1/3.

As I see it, all your formulations of the problem are equivalent and accurate enough to be unambiguous. It is some of your reasoning that is totally sloppy and wrong!

35. Well, his is just another hard left blog so it attracts lots of people who are living with the left-wing ideology such as global warming.

But at the end, I think that these blog posts have nothing to do with the climate or politics and I don't really care what those chronic politically defective commenters are saying - I had to care when I was living in an environment polluted by them from the floor to the roof but now I can freely despise and ignore these scumbags (and there are many over there, indeed!).

At the end, it's primarily blog posts about the statistical reasoning and it just seems that my explanations are totally getting through and dominate even among DeLong's readers despite his vigorous attempt to suggest that the correct solution is so stupid that it must only be an April Fools Day prank.

36. The whole thing lies in accepting the strict definition of probability. That is:

- the number of events that realize a particular outcome (you winning, for example) divided by the number of all possible, *exclusive* events.

However, you waking up on Tuesday and winning (by betting on tails) is not an exclusive event w.r.t. the other waking events. It is clearly dependent on the event of your waking up also on Monday.

The tail event is what caused these two *different* but obviously dependent waking events to occur.

That is, the only thing that was really exclusive was that it was a tail since it obviously can't be both tail and head at the same time.

You won but you didn't win twice in the sense of two independent events. It was 100% sure that you would win on Tuesday once you won on Monday. That is really one event in the sense of the definition, even if physically it was two events. The definition requires to measure the events in the same units.

Otherwise, you could toss a coin once, then take several pictures of the result and claim that the number of your wins is equal to the number of pictures you have taken of the coin in its final state. Well, once the first picture shows the win, clearly all the subsequent pictures will show a win too. It still is a single win.

37. Yeah, but I still somehow think that the creators of the puzzle had in mind the subjective outlook of the Beauty: "given you were woken up, what coin toss would you bet your money on?" The Beauty needs not know more about the world than "given I was woken up, it is twice more likely for the toss to have been tails"... and why should she care that she might be double counting one particular toss? She is not asked about the chance of heads in one unbiased toss, she is I believe asked a different question. She is asked about her personal perspective of the situation.

38. I think the whole confusion is in language and meaning of the question. This is quote from BDL page: "When awakened and asked "what is your confidence that the coin flip winds up heads?", what answer should he give? If he is then told "it is Monday; what is your confidence that the coin flip winds up heads?", what answer should he give?" I believe that the question means "what answer should he give to be right as often as possible". This of course leads to the tails answer.

39. Note that the word "probability" is nowhere to be seen in the question.

40. Dear Mažňáku, "credence", "confidence", and "[subjective] probability" are synonymous, so your assertion that the word "probability" may be true in some literal sense, but it is false morally.

41. I don't see what the change in terminology really changes in essence. If I am confident about something that means I attribute it a high probability. One can also say I consider that event more likely.

However, by redefining what it means to win, or what it means for a certain event to occur in relation to a bunch of other, somehow related events that may also occur, one can surely arrive at many different ways to 'calculate the probability'. I just wouldn't really call it a calculus of probability.

42. LOL, Tony, did you write your comment independently of my reply to the same maznak's comment below? Because we wrote almost the same thing. ;-)

43. Yes, completely independently. My mobile browser was showing note that there is one new comment as I was about to press send.

44. Good - your comment had some extra content. But good to see this synergy. ;-)

Being careful and distinguish similar words is great and often important - and people often err.

On the other hand, this objection is also often used in a wrong way when people insist on distinguishing things, concepts, ideas, and words that are actually the same. In those cases, their focus on accuracy and exact terminology is spurious and only increases chaos.

45. Many of them will probably regret silently, pretending that they have never naysayed anything. I sincerely hope that it won't be the case of Adam Falkowski who will owe me $10,000 at that point! ;-) My$100 for him in the case that SUSY isn't seen soon (when 30/fb of luminosity is surpassed and all data analyzed) is ready.

46. Dear Tom, she was provided by Google to make the point sharper. The opposites attract and the two leptons have the opposite charge, like you and her.

47. Ok, so you say that if the question was interpreted as "what should she say to be right as often as possible", it would be a wrong interpretation? Somehow it still seems to me to be the most natural one...

48. Tony, I am not jealous about the question you got. What to do with it?

The question is "what is your confidence that the coin landed heads?" Why should this totally clear sentence be "interpreted" as the incomprehensible sentence "what should she say to be right as often as possible"?

This "interpretation", Maznak, is much more confusing than anything else that was said before. You may think that you are making things better or more accurate but you are doing exactly the opposite.

The word "often" in your version of the question is also enough to see that you completely misunderstand the probability calculus. We are talking about the *confidence*, i.e. the subjective probability, and with this interpretation, the situation is really occurring only *once* in a lifetime. So the word "often" is nonsense. It is *you* who is completely distorting the discussion and making it chaotic and misleading.

49. So are you saying that it makes a difference whether it is a one-off situation or a repeated experiment? Not trying to be difficult, just trying to learn something, honestly.

50. Tom, the ads here are tailored to each viewer differently based on their Google profile. Thanks for letting us know you better :)

51. I got one for home insurance and one for car insurance. Shows what an interesting life I lead.

52. You mean thirders are wrong that she is different in the information she has? She is different because she knows she is awake and is therefore to choose among the three tosses that result in her being awake.

53. Jaroslav ŠnajdrMar 13, 2015, 2:13:00 PM

The exercise giving me a 1/3 result is for example this one:

Every day I knock on your door and give you a sealed envelope. Inside is a piece of paper that says either HEADS or TAILS. What is the probability that the content is HEADS, assuming the following rules?

1) On Day 0: I toss a coin and write the result into the envelope.

2) On Day n + 1: If I tossed a coin on Day n and the result was TAILS, then I perform no toss today and write TAILS. Otherwise, I do a new toss, like on Day 0, and write the result.

3) You never remember any of the previous results.

The ratio of HEADS:TAILS in this sequence is 1:2, so the probability of HEADS on any day n > 0 is 1/3, isn't it? Can this 1/3 be interpreted as a probability, even though the values in the sequence are not independent? I don't see clearly why not.

Of course, this exercise is not exactly the original sleeping beauty problem, but rather an "iterated sleeping beauty", where I repeat the experiment over and over and wake her up once or twice every week.

54. Right, 1/3 of the words in a long sequence of yours will be "heads" because your rule may be simplified: toss contains essentially once a day but if you get "tails", write the result twice.

If one watches the paper on a "random" day, then the fraction 1/3 may be interpreted as the probability of "heads".

In the case of the sleeping beauty, which is a different problem, the whole tension about the right result is about the counterpart of the concept of the "random day".

Among "random days", the halfers say that 1/2 are the days when the coin shows "heads" and 1/2 are the days when it shows "tails", so that "tails days" have to be included with 1/2 weights of the heads' weights, while the "thirders" naively insert all the awakening days from different histories with the same weight, and that's why they end with the wrong result 1/3 for the probability.

This mistake is equivalent to saying that the possibility that one gets "tails" occurs with a doubled probability just because a woman is awaken twice during the possibility. But it doesn't. The probability doesn't get modified as a function of awakenings of any woman - or any man, either.

55. Exactly, Maznak. One may learn quite something. People with similar complaints like Tom should realize that it is not the blog owner or the readers who are choosing the ads for him. It's a rather sophisticated automatic program.

56. Dear Maznaku, you may repeat the experiment and if everything will be the same, the subjective probability will always be the same.

But the subjective probability will *not* be equal to some frequentist ratio of cases that go one way or another if that ratio is calculated sloppily.

To calculate the ratio properly needs to assign the correct weights to different scenarios when the experiment is repeated.

So to get the right probabilities from the frequentist ratios, the weeks following the "heads" toss must be repeated twice, so that the number of awakenings remains the same in each week. In other words, the "tails" weeks have to be given the weight 1/2 times the weight of the "heads" - exactly to cancel the fact that she is awaken twice on that week.

When you do repeat these experiments with these right weights, you will get the correct probability of heads after awakening, P=1/2, even from the ratio of awakenings that show heads after many weeks.

If you get P=1/3, it's because you used wrong weights.

57. Not sure whether I understand your comment, Rehbock - almost certainly No.

I am saying something very simple. They treat the three combinations - heads-Monday, tails-Monday, tails-Tuesday - as if they were three equally likely places in space and if the woman were choosing at which of these places she is located.

But this Z_3-like symmetry doesn't exist. In reality, the tails-Monday and tails-Tuesday awakenings describe the same woman in the same history - in the same "parallel world", if someone loves to think about the many worlds interpretation of probability both in classical and quantum mechanics - so they must always be viewed as following from the same probability associated with the past events (that decided that this history takes place), and if one also wants to decide "when" she woke up, the different times when it could have happened (Monday or Tuesday) must divide this fixed probability (the number of beds) - they can't "steal" it from completely different histories.

This has been said so many times in so many creatively different ways (like the beds) that I don't believe that it makes any sense to continue this discussion. Who hasn't gotten it by now will probably never get it.

58. All I am trying to understand flaw in this: - sleeping beauty is awake in only three of four coin outcomes. She knows and we know that she is awake because she answers question. So the possible outcomes drops to three when she is asked omitting the coin day combination in which she is not awake.

59. Right, there are 3 coin_state-day arrangements, but they are *not* equally likely. WTF?

60. SteveBrooklineMAMar 13, 2015, 3:41:00 PM

I thought maybe generalizing the problem might help explain it. Consider the 8 mutually exclusive events:

E000: tails, sleep monday, sleep tuesday,
E001: tails, sleep monday, awake tuesday,
E010: tails, awake monday, sleep tuesday,
E011: tails, awake monday, awake tuesday,
E100: heads, sleep monday, sleep tuesday,
E101: heads, sleep monday, awake tuesday,
E110: heads, awake monday, sleep tuesday,
E111: heads, awake monday, awake tuesday,

with the sum of the corresponding probabilities equal to 1. Given that an event includes at least one awakening, what is the probability that it is a heads event? Clearly, it's

P=(pE101+pE110+pE111)/(1-pE000-pE100)

where the p-prefix means probability. Now suppose we know a priori that pE000=pE100=0. If we also know the coin is fair, it follows that pE101+pE110+pE111=1/2=P.

The conclusion is reached even without knowing exactly what the probabilities of six of the events are.

61. Here is my last try. I am sorry that I am unable to explain my point. Perhaps I am misunderstanding what we can know. Here was why and what I thought they are equally likely:
The coin has two choices and the question(s) may be on either day. That means only four combinations of day and coin HM TM and HT TT are equally possible. Equally possible it is M or T and equally H or T. Until she is awake she knows only that it is one of those four equal possibilities two for each side. When she is awake she knows it is not the one of four combinations in which she is left asleep. That new information allows her to choose among the remaining three equally likely outcomes of the combination of coin and day when answering.
I am not trying to be obstinate or otherwise perturb you. I still miss why we don't have three equal possibilities based on her learning she is awake.

62. Sorry Lubos, I can't resist either ;)

What should she say to be right as often as possible?

Well, the best thing then is to say what would make her 100% correct each and every time! For example, she can become a lover with one of the guys who are waking her up. He would secretly signal her the actual result of the coin toss and she would say the right thing.

After 10 coin tosses which happen to be, say exactly 5 heads and 5 tails, she would be right 15 times, a whooping 150%!

I'm joking of course, just to demonstrate how ridiculous conclusions can be reached if one mixes and/or shifts the interpretations, definitions and conditions.

63. If one has to determine the probabilities of the heads-Monday, tails-Monday, and tails-Tuesday combinations (possible arrangements of the state of the coin and the day when she is woken up), the two tails possibilities have to share the 50%, so if Monday and Tuesday are equally likely for tails, the most sensible arrangement is 25% for tails-Monday and 25% for tails-Tuesday.

The probabilities are 50-25-25 for the three arrangements which also allows you to say that the probability is 50/(50+25) = 2/3 that it's "heads" if she's told that it's Monday, 25/(50+25) = 1/3 for "tails" if she's told it's Monday

This is wrong. If she's told it's Monday, the probabilities are (still) 50/50. And the 50-25-25 "arrangement" doesn't correspond to anything meaningful.

64. It may be counterintuitive or subtle but if she is told it's Monday, the probability of heads jumps to 2/3 because in that case, she *does* learn new information she didn't know on Sunday.

It was possible that the wakeup was on Tuesday, which would mean that the coin landed tails, but because she learned that it's not Tuesday, tails became less likely, and their probability dropped to 1/3.

At any rate, be aware that both Elga and Lewis agree with me that the probability jumps by 1/6 if she learns it's Monday, search e.g. for E6 and L7 in Lewis' paper

65. Did you see my comment where I updated my stance on this? I am now also claiming that the probability remains 50-50 upon learning Monday (and of course 0-100 on being told Tuesday). The reason being that Monday and Tuesday do not exist as potential options at the same moment, (and this separation in time is the crucial thing) so being told Monday eliminates Tuesday from the thinking. One is now just lying in a bed knowing that 50 are red and 50 white, and also knowing that it is Monday. What were to be done to one in the future doesn't matter because its known where one is in time. My other comment was this:

As I tried to write a reply to Rehbock, and repeating the calculation that the probability change when she is told its Monday, made me think that there is a mistake in the conditional probabilities concerning learning about it being Monday. The reason is that Tuesday and Monday don't "live together as possibilities at the same moment". If she was thrown into a Monday box always in the case of tails and a Tuesday box half the time and a Monday box the other half if it was heads, the calculation of the 2/3 for tails upon learning about it being Monday would be true. But because Monday and Tuesday are separated in time and Monday always happens and she is always woken up on Mondays I don't think telling her its Monday changes anything, so even this additional information keeps the probability for heads or tails as 50-50. Do you agree that the separation in time makes this difference?

Drawn:

---------White bed------Monday-------Tuesday

/

---------Red bed----Monday

50-50 for learning monday.

vs:
-----------Monday
--------White bed
-----------Tuesday

------Red bed ------------Monday

2/3-1/3 for learning Monday

waking Tuesday is never possible on a Monday, and being told Monday makes the place in time sharp and destroys the clumping together of days in the probabilities.

66. My last drawing was deformed a bit by posting, sorry, it was better before

67. There *must* be a way to divide 100% to these possibilities for her, right?

I don't think so. I mean you can do it the way that the thirders do, which is to sample across the space of all awakenings, but then you get their answer, which has nothing to do with the likelihood that the coin is heads. But otherwise the whole idea of trying to assign consistent probabilities to each scenario is a trap as far as I can tell.

68. Lubos - I think dreamfeed is correct. Bob Walters says much the same here: http://rfcwalters.blogspot.it/2014/08/the-sleeping-beauty-problem-how-some.html - 'There is a complication I didn't mention explicitly in the last post. In the case of heads the beauty knows that only one step has occurred. Hence she should calculate the probability of the heads case using the probabilities in one step, not one or two steps. Hence she should estimate the probability of Monday as 1/2 after heads.'
By the way, what is going on with upvote/downvote by guests? Today it's not possible. A few days ago it was possible. Prior to that only upvoting was possible. I don't know what probabilities to assign. Am I the unwitting subject of some experiment?
Cheers!!!

69. SteveBrooklineMAMar 13, 2015, 9:01:00 PM

I think this also makes leads to a 50/50 "double half-er" conclusion as well. Given that an event includes a Monday awakening, what is the probability that it is a heads event? It's P2=(pE110+pE111)/(pE010+pE011+pE110+pE111). Now, turning to the specific Beauty problem, we have pE100=pE101=pE111=0, pE110=1/2, pE000=pE001=pE010=0, pE011=1/2, giving P2=1/2.

70. I wonder if Brad Delong answered 50% to the Monty Hall problem.

71. The problem said 'sometimes' on Monday.

72. dreamfeed - I agree with the spirit of your remarks here, but I think that the trap lies elsewhere.
Of course one can assign consistent probabilities to the awakenings! The mistake lies in assuming, or requiring, them to be mutually exclusive. The weighted average used by Lubos and Walters is an unneccessary 'fix' to a non-existent problem and allows them to address the 'thirders' on the 'thirders' terms. It's only virtue is that it yields the correct answer to the *original* question.
The probabilty of (mon,heads) = probabilty of (mon,tails) = probabilty of (tue,tails) = 1/2. This is perfectly consistent because the probablity of ((mon,tails) or (tue,tails)) = 1/2 as well. No weighted average is needed, just an elementary calculation.
To illustrate, I refer you to the Markov chain diagram in this post: http://motls.blogspot.co.uk/2014/08/sleeping-beauty-thirders-rudimentary.html?m=1. The probabilities of the awakenings, or nodes, given the initial state, are the product of successive transition probabilities. (for more general diagrams it would be a sum over all possible ways of arriving at a particular node from the initial state - much like quantum mechanics) When a transition probability is equal to 1 the conditional probability of the second w.r.t. the first is 1; this is part of the elementary calculation mentioned above.
I view this as a pseudo-problem, with most of the arguments being about whose misconceptions should prevail.

73. I'm also leaning towards the opinion that dreamfeed/Walters are correct here, Perhaps Lewis made a minor error in his paper stemming from giving too much credence to the other side's "ensemble of potential consciousness" argument...

I think it can be seen from "double guantanamo-izing" the problem.

Instead of tossing a fair coin, the experimenter will roll a fair 1000-sided dice. If it comes up 1-999 beauty will be woken once on monday, if it comes up 1000 she will be woken a million times over the coming days with induced amnesia etc. She will be asked to guess whether or not the dice came up 1000.

Additionally the experimenter informs her that he has arbitrarily predetermined that he will inform her what day it is on either none or all of any subsequent awakenings in the experiment.

Clearly if she is woken and told that it is monday she should still guess that the dice came up in the range 1-999 and not 1000, regardless of the putative existence of a million other awakening events that are (most probably) counterfactual in any case.

I think the subtlety here is that although she has gained new information on being informed that it's monday, (she now knows herself to be at the first, and probably last, step of the process), the extra information is redundant when it comes to assigning probabilities to which of the two branches she might be on.

The thirder wants to answer a different question - namely, what if a large ensemble of runs of the experiment exists, and a "random consciousness injection genie" will inject his consciousness into one of the beauties at a random awakening event - in which case "the dice came up 1000" is overwhelmingly likely to be the correct guess.

But this is changing the definition of what are the "independent events" whose likelyhood we are considering, in way that may be logically consistent, but that has nothing to do with any physical events or correlations.

(Normally we would consider the tails-monday, tails-teusday awakenings of a given beauty to be maximally correlated and not independent events!)

The random consciousness injection genie may be fun to think about, but morally he corresponds to an (admittedly interesting) "unphysical solution", and I wish physicists would leave him alone!

I wonder... if a truly committed thirder was playing the double-guantanamo version and was informed that he would be killed in the event of making a wrong guess, would he still have the balls to guess that the dice came up 1000 - even if he was informed it was monday? :D

74. I agree and I used frequentist definition because it is more familiar to people in simple cases like coin toss. Weight is of course correct and better term than saying that events must be "in the same units". I think that Maznak got that part anyway, but isn't willing to accept, or is not knowledgeable enough to see the merit of well defined, consistent set of rules.

75. I've been giving this some thought, and I'm also leaning towards the opinion that dreamfeed/Walters are correct here, Perhaps Lewis made a minor error in his paper stemming from giving too much credence to the other side's "ensemble of potential consciousness events" argument...

I think it can be seen from "double Guantanamo-izing" the problem.

Instead of tossing a fair coin, the experimenter will roll a fair 1000-sided dice. If it comes up 1000 beauty will be woken once on monday, if it comes up 1-999 she will be woken a million times over the coming days with induced amnesia etc. She will be asked to guess whether or not the dice came up 1000.

Additionally the experimenter tells her that he has arbitrarily predetermined that he will inform her what day it is, on either none or all of any subsequent awakenings in the experiment.

Clearly if she is woken and told that it is monday, she should still guess that the dice came up in the range 1-999 and not 1000, regardless of the highly likely existence of a million (minus one) other awakening events that she can rule out being on, but that are as yet still unobserved/counterfactual.

Knowing that it's monday and that moday awakenings will be a small proportion of all awakenings in a typical run of the experiment, is not an indicator that the dice came up 1000 with probablity approaching 1.

To sharpen this let's add that beauty will be killed if at any point in the experiment she makes a wrong guess, (Her goal is to survive).

Clearly if she adopts a predetermined strategy that if awoken and informed that it is monday, she will guess that the dice came up 1000, she's going to be in big trouble (99.9% probability to die), regardless of the fact that she in that case knows it's not tuesday or any of the subsequent days when she will (most probably) be woken!

I think the subtlety here is that although she has gained new information on being informed that it's monday, (she now knows herself to be at the first, and probably not last, step of the process), the extra information is redundant when it comes to assigning probabilities to which of the two branches she might be on.

If we invert the setup so that 1-999 results in 1 awakening, and 1000 results in a million awakenings, it also sheds light on what's going on in the mind of the convinced "thirder".

The thirder wants to answer a different question to what was the likelyhood of the dice coming up 1000 in a given experiment - namely, what if a sufficiently large ensemble of runs of the experiment exists, and a "random consciousness injection genie" will inject the thirder's consciousness into one of the beauties' streams of consciousness at a random awakening event - in which case "the dice came up 1000" is of course overwhelmingly likely to be the correct guess,

But this is changing the definition of what are the "independent events" whose likelyhood we are considering, in a way that may be logically consistent, but that has nothing to do with any physical events or correlations.

(Normally we would consider the tails-monday, tails-teusday awakenings of a given beauty to be maximally correlated and not in any reasonable sense independent events!)

The random consciousness injection genie may be fun to think about, but morally he corresponds to an (admittedly interesting) "unphysical solution", and I wish physicists would leave him alone!

I wonder... if a truly committed thirder was playing the inverted double-guantanamo version and was informed that he would be killed in the event of making a wrong guess, would he still have the balls to guess that the dice came up 1000 - even if he was informed it was monday? :D

76. I have a problem with some statements, which may be due to a typo or my minterpreting (or both;)).

First, all awakenings are exclusive in the sense that you are either awaken on Tuesday or on Monday, when toss resulted in either head or tail. Pairs (day, toss result) are also exclusive. However, some pairs are dependent on each other. This is a typo that I had in one of my posts as well.

Second, the total of all probabilities of all exclusive events should be 1. We need to use weights, or renormalize, to get 1.

77. Well, initially so did Erdős Pál, one of the greatest mathematicians who ever lived :D :D

(He changed his mind when it was explained to him of course)

Probability is hard...

78. Tony - Given the statement of the problem, and the Markov chain model, awakenings are not exclusive. If she is awakened on Monday after a throw of tails then she is certain to be awakened on Tuesday. Of course she doesn't know the day or the throw, so it doesn't matter to her.
'Monday' and 'Tuesday' are labels, or indices, indicating the number of steps taken in the process, i.e. n=1 and n=2. Along with the result of the toss they can also label the nodes. Of course they must be distinct, but they do not serve any other purpose. One might say they are parameters and not observables. The state of the system at any given time, or day, or value of n. after the initial state is a pair of nodes (one from either trajectory), each with a probability of 1/2 and these nodes are mutually exclusive (as are the trajectories themselves, and any pair of nodes - one from each trajectory, or even a node from one trajectory and a pair of nodes from the other). However, all pairs are not exclusive - (mon, tails) and (tue, tails) are not exclusive, while (mon,heads) and ((mon, tails) or (tue, tails)) are exclusive. Nodes on the same trajectory are not exclusive and weighting them does not make them exclusive.
Practically the first thing one learns about probability is the addition rule, and this is all that is required here. The weighting simply panders to the 'thirders' misconceptions about the importance of awakenings.

79. Let's first agree on what we mean by exclusive, shall we?

You seem to use it in the sense that one event (awakening on Monday after tail) doesn't *prevent* another (Tuesday with tail) from happening later on.

I use it in the sense that if it is Monday then it is not Tuesday. Unless, of course, we start considering other reference frames, etc. which is probably not a kind of complexity that we want to include in the analysis of this problem.

We are going in quite different directions. You are seeing paths and chains while I am looking at the probability space if total measure 1 and seeing two delta functions sitting on head and tail respectively, then tail splitting into two, one with label Monday another with label Tuesday.

If I look at awakenings label (which is a pair) I see three delta functions (M,h), (M,t) and (T,t), if I look at coin toss results label I see two, where I added (M,t) and (T,t) into one, the second being (M, h).

The set of three or the set of two, all exclusive elements of the corresponding set, given the criteria which makes them distinct and non-overlapping (they are delta functions after all)

80. Yes, I am going back to the axioms:

http://en.wikipedia.org/wiki/Probability_axioms

- First axiom

The probability of an event is a non-negative real number: P(E) where E is event in event space Omega which is always finite.

- Second axiom

P(Omega) = 1

This is the assumption of unit measure: that the probability that some elementary event in the entire sample space will occur is 1. More specifically, there are no elementary events outside the sample space.

This is often overlooked in some mistaken probability calculations; if you cannot precisely define the whole sample space, then the probability of any subset cannot be defined either.

- Third axiom

This is the assumption of additivity:

Any countable sequence of disjoint (synonymous with mutually exclusive) events P(E1), P(E2) ...satisfies the formula : probability of the union is the sum of probabilities of each element of the union.

81. The light just came on. Sorry. Of course the three waking are no longer equal for same reason - i was just locked into that four box

82. The assumption that N choices are equally likely i.e. the probability of each is 1/N is rather tempting.

One funny example was this Walter Wagner on Comedy Central who claimed that the LHC will destroy the world with 50% probability - because it will either happen, or not, so it's 50-50.

But somewhat less dramatically, I remember how my classmate at the basic school, a great historian - and today a professional one - vigorously fought with our mathematics teacher and he claimed that if you toss 2 identically looking dice, there are 15 choices what you can get, so the probability of 5-6 is the same as 6-6 and it is 1/15.

If the believer in such a thing is stubborn, and he was, it is not easy to convince him that 5-6 is actually 2 times more likely than 6-6 because 5-6 includes 5-6 and 6-5 with the dice which are in principle distinguishable but in practice not, and 5-6 and 6-5 and 6-6 in this sense are equally likely, with P = 1/36 each.

More generally, when people get used to the fact that the probabilities are not this uniform and always equal to 1/N, they want some "easy universal physical visualization" of probabilities.

So they imagine that probabilities are always directly proportional to some volume, or duration, or some mass, etc. But in general, probabilities are given by numbers and formulae that are much more complex and general than any of these "extensive" things. The laws of physics e.g. quantum physics are hard and they produce the probability or probability amplitude from complicated unitary operators integrated in complicated ways etc. None of the simple rules for probabilities is right universally, for all problems in the whole Universe.

A priori, we don't know what the probability of something is, and sometimes we can calculate it if we are lucky. The "naive" believers think that probabilities are something trivial, given by counting or weights or timing etc. But probabilities are in no way trivial: everything that is complicated about the world may be interpreted as complexity of some probability distributions - everything that quantum mechanics predicts, in particular, are probabilities that may be very difficult and structured.

83. Hi Tony/RAF III/dreamfeed.

I'm now sure that this is indeed correct.

The upshot of my "wordy" post above is that if clarifies the situation to consider deformations of the probability of Heads versus Tails away from 50-50.

If I set P(Heads) = 1/n with n arbitrarily large, then choose a number of tail-awakenings N such that N >> n, then by the Elga/Lewis counting this gives me P(Heads|Monday) -> 1 as N -> inf.

So i can makes heads arbitrarily unlikely, but according to the Elgin/Lewis

on waking and being told it's monday I can be certain the the coin came up heads!

It's useful to consider deforming probabilities away from 50-50 because you can no longer get away with sloppy arguments based on counting branches or nodes, (like the QM MWI crowd who can't cleanly interpret wavefunction amplitudes).

I agree that the error here with 50-25-25 is that Monday and Tuesday are not independent events... P(Tuesday|Monday) = 1 not zero! The Markov chain interpretation of this as the transition probability along the branch makes the situation perfectly clear, and the canonical labeling of the nodes is 50-50-50 not 50-25-25. This doesn't violate normalisation as the second two are perfectly correlated events not independent events!

50-25-25 can be useful as post-hoc weighting to restore even betting or give nice weights for the thirders' "consciousness injection genie" to respect, but thats adding to the model.

Walters says it best:

"In a Markov process one cannot in general talk about the probability of being in a state. One must talk about the probability of passing from state x to state y in n steps. (If the process is ergodic then in a limiting distribution there
is some sense of talking about the probability of being in a state, but
that is not the case here.)"

I guess physicists can get confused because they're used to considering thermal ensembles in an ergodic phase space where all states are "equally likely".

Thirders can of course ask the question "whats the probability of heads given a random node on the Markov chain" and it's well defined, but not physically meaningful/interesting.

To show a thirder the error of his ways, have him consider deforming so that P(Tails) is small and N awakenings is large and see what happens... ;)

84. I think the told it is Monday probability is 1/2.
To see this, consider the other question, but with the coin tossed after she was awakened on Monday to determine if she will be awakened on Tuesday. I think the probabilities are the same in both versions, and makes it clear that the awakening probability should be 1/2.
However, for being told it is Monday, you are saying she should argue that this means the probability of an upcoming coin toss will be 2/3.

85. Tony - Please consider this a reply to both of yours above.
I would first like to say that I envy the clarity with which you express your thoughts and reasoning. Even though (I think) I somehow knew all along how you and others were thinking about this, the clarity of your remarks makes it much easier for me to respond.


text:
# Why were the antennas on the spherical surface of some early satellites spiral-shaped?

In the question What's the spiral pattern on this satellite? I show this image of a spiral-shaped pattern on the spherical surface of a (presumably) vintage satellite or model thereof. @OrganicMarble's answer links to an article which identifies this as a logarithmic spiral antenna.

In this answer I show this image of a project ELINT satellite aka GRAB-1 (for an interesting story, read further there) which @OrganicMarble pointed out is the little one on top, and the one with the spiral pattern is another satellite of the same type as that above.

I'm wondering why these spiral-shaped antennas were actually necessary. Does it have to do with limitations of the electronics, or the nature of the RF signals, or the spherical shape of the spacecraft, or something else like "packaging"?

Compare to images of earlier spherical satellites in these questions and their answers, which all used several straight rod antennas sticking out from the sphere

Question: Why were the antennas on the spherical surface of some early satellites spiral-shaped?

• Let's not forget Lunokhod's "Unicorn horn". – SF. Sep 11 '18 at 10:11
• @SF. I love it! It's not a spherical unicorn, but it's nice nonetheless. – uhoh Sep 11 '18 at 10:14

Those are spiral antennas.

Spiral antennas belong to the class of frequency independent antennas which operate over a wide range of frequencies. Polarization, radiation pattern and impedance of such antennas remain unchanged over large bandwidth.[3] ... Spiral antennas are reduced size antennas with its windings making it an extremely small structure.

So, an antenna that can operate at a large frequency range with a nice predictable performance, and with a small footprint. All desirable qualities for an antenna to be used in a spacecraft.

For the Transit navigation satellites, I'll have to make a guess. The early Transits were launched on Scout rockets which meant severe constraints on size and weight. That makes an antenna that can be painted on the sat's outside surface an attractive option. Their function doesn't seem to dictate a wide frequency range: it broadcast on two fixed frequencies.

• Then is it possible to understand why such a large frequency range was needed? Or were they used because of the compactness instead? I'm looking for the "root cause" if possible. The satellites shown in the four bulleted linked questions seem to have gotten by without them. – uhoh Sep 8 '18 at 15:29
• ieeexplore.ieee.org/document/4066067 I found that reference in Table C-1 in dtic.mil/dtic/tr/fulltext/u2/a066299.pdf which I found in this answer – uhoh Sep 9 '18 at 11:03

Radio waves -- just like light -- are electromagnetic waves. Because they are transverse waves of the electric and magnetic fields, they can be polarized. There are two ways (*1) to polarize E-M waves: linear polarization and circular polarization.

Linear polarization

• The electric field vibrates in one direction perpendicular to the direction of propagation, but with an amplitude that fluctuates periodically.
• Light example: polarized sunglasses

Circular polarization

• The electric field maintains constant amplitude, but the direction changes as the wave propagates.
• Light example: polarized 3-D movie glasses
• Radio examples: spiral antennas, helical antennas, Apollo scimitar antenna

When E-M waves reflect off smooth surfaces, they can become linearly polarized. The best example of this is glare, which is light that reflects off a road, snow, or ground. In this case, the electric field perpendicular to the surface is reduced by destructive interference, and so we say the reflected light is polarized parallel to the surface. By wearing polarized sunglasses that are vertically polarized, you can block the glare, yet still see the ambient light.

The same destructive interference by reflection can also happen with radio waves. Television, radio, and WiFi are transmitted by dipole antennas that create a radio wave polarized in the direction of the antenna (vertical in the case of TV and radio). However, when the signal reflects off certain geographic features, you can get cancellation of the linearly-polarized signal. This is why some locations have a hard time getting TV or radio signals. However, this problem doesn't happen with circularly polarized signals.

Another issue with linear polarization is that when you rotate the transmitter (or receiver), the amplitude of the signal changes. If you tilt your head while wearing polarized sunglasses, the glare will start to appear. You could make a 3-D movie with linearly polarized glasses (one eye horizontally- and the other eye vertically-polarized), but you would see double-images if you tilt your head. It's also why the antenna on your WiFi router can be rotated into various orientations. This is another problem that doesn't happen with circularly polarized signals. Thus, 3-D movie glasses use circular polarization.

The disadvantage of spiral antennas is that they are inefficient. They tend to spread their signal in a wide beam, which means much of the energy of the radio transmission is being broadcast in the wrong direction. As @Hobbes notes, they can transmit on a wide bandwidth; you don't want to do this as it requires more transmitter power.

Putting a spiral or helical antenna on orbiting spacecraft is a smart idea. It creates circularly-polarized radio waves, which means the orientation of the antennas don't matter. That's important for an orbiting spacecraft, which is constantly changing its orientation relative to positions on Earth. There is also the advantage of circularly-polarized wave being less susceptible to interference by reflection. You can see a picture of a helical antenna on a ground-tracking station in my answer here.

In contrast, spacecraft beyond orbit -- think Apollo or Voyager -- have fairly stable orientations, and so they use more efficient linearly-polarized antennas. That's why Apollo had so many dang antennas: some for when the orientation was stable (*2), and others for when the orientation was changing (*3).

(*1) Elliptical polarization is a third way to polarize E-M waves, but they add nothing to this discussion.

(*2) CSM to Earth (trip between Earth and moon); LEM to Earth (on the moon); the S-IVB ("third stage") to Earth; and the ALSEP experiments left behind.

(*3) CSM to Earth (Earth or lunar orbit); LEM to CSM; LEM to Earth (ascending/descending); lunar rover to CSM.

• I don't believe circularly polarized beams "spread faster" than linearly polarized beams. Every antenna design has its own radiation pattern, but spreading is not really caused by circular polarization. Same with the bandwidth; it's a function of the antenna design, not the circular polarization. – uhoh Sep 11 '18 at 10:10
• @uhoh: True, helical antennas are circularly-polarized but directional. I've fixed the answer. – DrSheldon Sep 11 '18 at 10:21
• Thirdly, let's confirm that this particular, unusual, unique configuration, where a sphere is completely covered in spiral, really does have circular polarization to begin with. I'm not convinced that this particular design falls neatly into the category of helical antennas. – uhoh Sep 11 '18 at 10:35

See A Broad-Band Spherical Satellite Antenna (paywalled)

tl;dr:

• Broad range of frequencies by being an angles-only shape without characteristic length
• isotropic with no strong nulls, worst case is -10 dB.

Some of the spherical satellites with spiral patterns, from: Artificial Earth Satellites, Designed and Fabricated by The Johns Hopkins University Applied Physics Laboratory

Satellite                      Transmitters
1-A     17-Sep-1959      54/108 MHz, 162/216 MHz, 108 MHz (TM).
1-B     13-Apr-1960     162/216 MHz,  54/324 MHz
2-A     22-Jun-1960     162/216 MHz,  54/324 MHz, 108 MHz (TM), and NOTS infrared scanner on 107.9 MHz
3-A     30-Nov-1960     162/216 MHz,  54/324 MHz, 108 MHz (TM)
3-B     21-Feb-1961     162/216 MHz,  54/324 MHz, 136 MHz (TM), and 224/421/448 MHz (SECOR)

After these, the spacecraft changed shape and the spirals were not used. According to Spacecraft Design Innovations in the APL Space Department (Johns Hopkins APL Technical Digesl. Volume 13. Number 1 (1992)):

The Space Department needed to develop several novel antennas to radiate the Doppler signal to the ground. The first Transits used an APL-invented spherical projection of a logarithmic spiral (Fig. 3). When later spacecraft departed from spherical shape, the design switched to the aptly-named "lamp shade" antenna. That antenna, however, suffered from poor polarization and a null in its gain pattern at nadir. A new antenna was clearly needed-one that could provide broad beam coverage with good circular polarization yet be small compared with the 2-m wavelength and the spacecraft body itself. Charles C. Kilgus answered this need by inventing the resonant quadrifilar helix, first flown on Triad (and visible in Fig. 7)8

8Kilgus, C. C., "Resonant Quadrifilar Helix Design," Microwave 1. l3 (12), 49-54 (1970).

From the US patent US3,034,121, May 8, 1962:

The requirement for a broadband antenna, covering a frequency range of at least four to one and providing radiation characteristics which would produce no nulls greater than l0 db with respect to the maximum in any plane, and at any angle with respect to the antenna within a spherical boundary, was generated by the development of a special spherical aerial vehicle. This requirement was so imposed that regardless of the attitude of the vehicle in space, the received signal variation would not be greater than 10 db due to antenna pattern changes.

The design of the antenna constituting the present invention is based on the established properties of the class of broad band antennas using equiangular spiral slots or conductors. Such an antenna has the property that its shape is specified entirely by angles, so that its performance is independent of wavelength. A familiar example of such an antenna is the infinite biconical horn.

The logarithmic spiral has the property that an angle between the tangent tangent to the curve and its radius vector is a constant and greater than 90 degrees. Such a spiral is shown in FIG. 1. A spiral slot antenna can be constructed by plotting two spirals slipped in angle with respect to. each other, as shown, in FIG. 2. Similarly, as shown in FIG. 3, a double spiral slot can be produced by plotting two spirals (FIG. 2) 180 degrees with respect to each other. In FIG. 3 the feed points are shown at X-X.

The present invention contemplates the application of a logarithmic spiral antenna to the surface of a hemisphere rather than to a flat surface or sheet as in conventional practice. More specifically, and as shown in FIG. 4 of the drawings, the present invention utilizes a pair of concentric logarithmic spiral antennas applied back to back The sphere is shown at 10 and is mounted on a suitable support 12 which also carries the feed line, to be described hereinafter. passes through the equator of the sphere, indicated at 14, and is "thus positioned normal to the polar axis of the sphere. Suitably secured to or formed on each half of the sphere is a pair of concentric spiral logarithmic antenna's elements 16 and 18 separated by dielectric filled in FIG. 3.

It continues, and later says:

When the two spirals which are fed independently at the poles are fed in phase, by equal length lines, from a common point and the radiation components are in phase all around the equator, the radiation pattern is a figure of revolution about the polar axis.

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0
RESEARCH PAPERS

# Resonance of Natural Convection Inside a Bidisperse Porous Medium Enclosure

[+] Author Affiliations
Arunn Narasimhan1

Department of Mechanical Engineering, Heat Transfer and Thermal Power Laboratory, Indian Institute of Technology, Madras, Chennai 600036, India, arunn@iitm.ac.in

B. V. K. Reddy

Department of Mechanical Engineering, Heat Transfer and Thermal Power Laboratory, Indian Institute of Technology, Madras, Chennai 600036, India, bvkreddy680@gmail.com

1

Corresponding author.

J. Heat Transfer 133(4), 042601 (Jan 06, 2011) (9 pages) doi:10.1115/1.4001316 History: Received September 22, 2009; Revised January 27, 2010; Published January 06, 2011; Online January 06, 2011

## Abstract

An enclosure filled with blocks made of a microscale porous medium, separated by macrosize pores, can be treated as a bidisperse porous medium (BDPM) enclosure. This study investigates natural convection resonance inside such a BDPM enclosure subjected to time-periodic heat flux at a side wall, with the opposite wall kept isothermal while the top and bottom walls are adiabatic. The BDPM enclosure is made from uniformly spaced, disconnected, square, porous blocks that form the microporous medium saturated with a fluid of $Pr=0.7$. For Rayleigh numbers, $Ra=108$ and $107$, the pulsating wall heat flux is varied over a frequency range of $0.01≤f≤0.1$, and the bidispersion effect is induced by varying both the internal (micropore) Darcy number $(DaI)$ and external (macropore) Darcy number $DaE$. Natural convection resonance is observed in the BDPM enclosure and the resonance heating frequency $fr$ increases with Ra, $DaI$, and $DaE$. However, $fr$ of the BDPM enclosure is always less than that of the corresponding clear fluid enclosure limit (at $DaI→∞$). Predictions of $fr$ using a modified scale analysis incorporating BDPM effects agree well with that arrived by numerical methods.

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## Figures

Figure 1

Schematic of enclosure with 4×4 porous block array (a) geometry, (b) mesh, and (c) pulsating heat flux

Figure 2

The response of instantaneous average Nuc, Num, and θ¯h for a BDPM enclosure at DaI→∞ (the clear fluid limit) and Ra=108 with different heat input frequencies: (a) 0.01, (b) 0.029, and (c) 0.04

Figure 3

The response of instantaneous average Nuc, Num, and θ¯h for a BDPM enclosure at N2=4×4, DaI=10−3, and Ra=108 with different heat input frequencies: (a) 0.01, (b) 0.02, and (c) 0.04

Figure 4

The response of instantaneous average Nuc, Num, and θ¯h for a BDPM enclosure at N2=4×4, DaI→0 (monodisperse limit), and Ra=108 with different heat input frequencies: (a) 0.01, (b) 0.02, and (c) 0.04

Figure 5

Instantaneous volume averaged velocity variation with average hot wall temperature at Ra=108 and f=fr for enclosure with (a) clear fluid, (b) porous blocks (4×4, DaI=10−3), and (c) solid blocks (4×4)

Figure 6

Instantaneous streamlines and isotherms for a BDPM enclosure for DaI→∞ (clear fluid limit) at Ra=108 and resonance frequency fr=0.029

Figure 7

Instantaneous streamlines and isotherms for a BDPM enclosure at N2=4×4, DaI=10−3, ϕE=0.64, and ϕI=0.5 at Ra=108 and resonance frequency fr=0.02

Figure 8

Instantaneous streamlines and isotherms for a BDPM enclosure at N2=4×4, DaI→0 (monodisperse limit), ϕE=0.64 at Ra=108, and resonance frequency fr=0.01

Figure 9

BDPM enclosure Num,max variation with pulsating heat input frequency in the steady oscillatory state for several DaI values at N2=4×4, ϕE=0.64, and Pr=0.7: (a) Ra=108 and (b) Ra=107

Figure 10

Effect of bidispersion (DaI) on the resonance (heating) frequency (fr) for Pr=0.7, N2=4×4, ϕE=0.64, and ϕI=0.5. Comparison of predictions from scale analysis and numerical method.

Figure 11

Effect of bidispersion (by varying ϕE) on the BDPM enclosure Num,max in the steady oscillatory state at N2=4×4, Ra=108, and Pr=0.7

## Discussions

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# Which mutated enzymes act as better drugs than their non-mutated form?

Is there an enzyme which is used as a drug (let it be called version $A$) and a mutated version of it ($M$), where the activity of $M$ was found to be higher than $A$?

Any references on such a case would be greatly acknowledged.

• A possible option: "Site-saturation mutagenesis of Tyr-105 reveals its importance in substrate stabilization and discrimination in TEM-1 β-lactamase" Feb 28, 2013 at 11:45
• Another option: "Most Efficient Cocaine Hydrolase Designed by Virtual Screening of Transition States" Mar 4, 2013 at 15:00
• A third option: "Why Does the G117H Mutation Considerably Improve the Activity of Human Butyrylcholinesterase against Sarin? Insights from Quantum Mechanical/Molecular Mechanical Free Energy Calculations" Mar 4, 2013 at 15:00
• Design of High-Activity Mutants of Human Butyrylcholinesterase against (−)-Cocaine: Structural and Energetic Factors Affecting the Catalytic Efficiency Mar 4, 2013 at 20:17

## 1 Answer

There are mutants of clotting factor VIII that have longer half lives and better activity than natural Factor VIII. Factor VIII is one of the missing proteins commonly involved in hemophilia. Hemophiliacs usually use injections of Factor VIII to prevent bleeding. The mutant form should last longer in the blood, with better activity, resulting in fewer injections.

Since this is a 2007 patent, I don't know if this product is actually being used to treat patients, drug approval takes a long time.


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# Testing production scenarios for (anti-)(hyper-)nuclei and exotica at energies available at the CERN Large Hadron Collider

Bellini, Francesca (European Organisation for Nuclear Research (CERN), 1211 Geneva, Switzerland) ; Kalweit, Alexander P. (European Organisation for Nuclear Research (CERN), 1211 Geneva, Switzerland)

28 May 2019

Abstract: We present a detailed comparison of coalescence and thermal-statistical models for the production of (anti-) (hyper-)nuclei in high-energy collisions. For the first time, such a study is carried out as a function of the size of the object relative to the size of the particle emitting source. Our study reveals large differences between the two scenarios for the production of objects with extended wave functions. While both models give similar predictions and show similar agreement with experimental data for (anti-)deuterons and (anti-)${}^{3}\mathrm{He}$ nuclei, they largely differ in their description of (anti-)hypertriton production. We propose to address experimentally the comparison of the production models by measuring the coalescence parameter systematically for different (anti-)(hyper-)nuclei in different collision systems and differentially in multiplicity. Such measurements are feasible with the current and upgraded Large Hadron Collider experiments. Our findings highlight the unique potential of ultrarelativistic heavy-ion collisions as a laboratory to clarify the internal structure of exotic QCD objects and can serve as a basis for more refined calculations in the future.

Published in: Physical Review C 99 (2019)


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Question

# Ratio of 200 ml to 2 litre is:(a) 20:1(b) 1:10(c) 2:200(d) 1:20

Verified
114.3K+ Views
Hint: Start by converting both the parts of the ratio to the same unit either to litre or millilitre.

Let us first know what a ratio is.

A ratio, in basic words, is a quantity used to define a comparison between two quantities. A bit toward the advanced side, it is the quantity that defines how many times of one quantity is that of others.

At our level, apart from the definition, we will treat it as a simple fraction that defines a relation between two given quantities.

Now, starting with the solution to the above question, first moving to the conversion of 2 litre to ml.

We know that 1 litre is equal to 1000 ml. Then we can say that 2 litre is equal to $1000\times 2=2000ml$

We will now try to find the ratio of 200 ml to 2 litre.

$\therefore ratio=\dfrac{200}{2000}=\dfrac{1}{10}$

The above ratio can also be written as 1:10.

So, the ratio of 200 ml to 2 litre is 1:10, i.e. the correct answer is option (b) 1:10

Note: Read the question carefully as in the question, including ratio, there is always a chance that the question might have a twist hidden in the words of the question. Also, it is preferred to convert the quantities to the smaller unit of the two units as this ensures that none of the numbers are in decimal form, simplifying the calculations.


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1. Latex

Latex text doesn't get converted into the the normal font in the tapatalk app

3. Re: Latex

Last night, the site was upgraded with a new and improved Latex system. Does this display for you:

$$1 ~,~ \sqrt 2 ~,\&~ \sqrt {\frac{3}{2}}$$


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# Prove $|f(z)|\leq M|z|$ inside unit disk if $f(0)=0$ and $|f(z)|\leq M$

The problem goes as follows:

Assume $f(z)$ is analytic inside the unit disk $D=\{z:|z|<1\}$. Also, $f(0)=0$ and $|f(z)|\leq M$ in $D$.

Prove that $|f(z)|\leq M|z|$ in $D$. When does equality occur?

This seems like an easy problem, but I can't seem to come up with a solution.

I have tried using Cauchy's integral formula and the ML-inequality, but I am stuck here:

$|f(z)|=|\frac{1}{2\pi i}\oint_{C_r} \frac{f(\zeta)}{\zeta-z}d\zeta|\leq\frac{1}{2\pi}\frac{M}{r-|z|}2\pi r=\frac{Mr}{r-|z|}$

where $C_r$ is a circle of radius $r$ inside $D$.

Any help would be appreciated.

• Try to factor $f$ as $f(z)=zg(z)$. – Yai0Phah Oct 22 '15 at 18:55
• I've considered that, but don't know how to continue from there.. – Anton Lundberg Oct 22 '15 at 19:16
• You can determine that the maximum of the function g is M. Then you are done. You could use maximum modulus to get the bound. – random123 Oct 22 '15 at 20:39

As suggested, write $f(z) = zg(z)$ (where $g$ is also holomorphic on $|z|<1$). Then for $|z|\le r$ and $r < 1$, $$|g(z)| = \frac{|f(z)|}{|z]} \le \frac{M}{r}.$$ Letting $r \to 1^-$, it follows that $$|g(z)| \le M$$ for all $z \in D$. Hence $|f(z)| \le M |z|$ for $z \in D$. (This is a variant of Schwarz' lemma.)

Equality can only happen (at $z=0$ or) if $|g(z)| = M$ for some $z \in D$. But then, the maxmimum modulus principle forces $g(z)$ to be constant on $D$, $g(z) = Me^{i\theta}$ for some real $\theta$, and consequently $f(z) = Mze^{i\theta}$.

• Really, the question to have asked the OP is: Do you know the Schwarz Lemma? If not, are they asking the OP to prove the SL? That would be a little weird. But if so, why is eveyone reproving the SL? – zhw. Oct 22 '15 at 23:48

HINT: define $$g(z):=\frac{f(z)}M$$ and apply Schwarz Lemma.


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# Tag Info

1

It's all because of the wavelength of light. In most bands the radio telescope is about the size of the wavelength it's observing - so it can only see a single point at once anyway. It would be like having an optical telescope that was a tiny microscopic pinhole - there wouldn't be much point in having a megapixel camera behind it. Radio telescopes that ...

5

While I agree with Alfred Centauri's answer, I am not sure it gives a direct answer to your specific question for the following reason. If there is a receiving antenna somewhere, there is always some reflection, however minute. If the antenna is connected to a circuit, the conditions of reflection will change, and the transmitter can "notice" that. In ...

4

As an intermediate step, consider a sinusoidal source driving an infinite transmission line with some characteristic impedance $Z_0 = 50\Omega$. The source "sees" a real impedance of $50\Omega$ and so, power is delivered to the line and, since the TL is infinitely long, the power is transported down the line, via an electromagnetic wave, without reflection. ...

-1

I like thought experiments ! No need to replace those expensive melted bits !! And that is relevant to this experiment. We will wind up the RF output of the transmitter until the aerial itself is glowing a nice cherry red . Still intact (just !) as a functional aerial but emitting light (you can see its red). So assuming we have still got a good SWR and all ...

Top 50 recent answers are included


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# Different types of communication in MPI

In this subsection we will learn how the types of communication in MPI are divided and we will learn about the Point to Point communication paradigms.

There are two criteria by which we can divide the types of communication in MPI.

First way to define types of communication is to divide it according to the number of processes that are involved. So, if there are only two processes involved in the communication, meaning only one sender and one receiver, then it is referred to as point to point communication. This is the simplest form of message passing where one process sends a message to another. The other type of communication in this category is the collective communication in which we have multiple processes that are involved. This implicates communication between either one processes with many other processes or even many processes communicating with several other processes. So, there are different ways that this can be executed. This is one criteria for distinguishing the types of communication, i.e., distinction by the number of processes involved.

The second criteria and perhaps more complex is by defining the type of communication into blocking and non blocking types. A blocking routine returns only when the operation has completed. This means blocking implies that if we send a message, we can’t proceed to the next steps until the receiver actually returns us information that it has received the message.

Image courtesy: Rolf Rabenseifner (HLRS)

The non blocking communication is more complicated than the simpler blocking counterpart. In this case it returns immediately and allows the sub-program to perform other work. It differs from the blocking communication in a way that if we send something to the receiver, we can execute some other tasks in between and after some time, we can check if the receiver has actually returned the information, i.e., it has received the message, or everything is OK. Many real applications usually employ this type of communication.

Image courtesy: Rolf Rabenseifner (HLRS)

### Point-to-Point Communication

As we already saw in the previous section, Point-to-Point Communication is the simplest communication as it involves only two processes. One of the processes acts as a sender and the other one as the receiver. Here, the source or the sender sends a message via the communicator to the receiver. In order to do so, the environment has to know who is the sender and who is the receiver and this is taken care of by specifying the ranks of the processes.

Image courtesy: Rolf Rabenseifner (HLRS)

In order to send a message with Point-to-Point Communication, we use the function

MPI_Send(void *buf, int count, MPI_Datatype datatype, int dest, int tag, MPI_Comm comm);

We will now see what arguments this routine actually needs.

• buf is the pointer to the data with count elements and each of them is described with datatype. So first of all, we have to specify the address of the data that we would like to send. It’s a pointer to the data of the sender and then the second argument is actually the number of elements we send. For example, if we just send 1 integer, the size will be 1. If you send an array with 100 integers, this will be 100 and so on. The third argument this function would like to have is the datatype. So, if we are sending an integer, we will have to specify here MPI_INT and so on.
• dest is the rank of the destination process. In this argument we specify the rank of the receiver. So, for instance, in the previous example, this will be 5.
• tag is an additional integer information sent with the message. tag is basically a number by which we identify the message. So, usually if we just send only one message, we can just put the 0 tag there or maybe any number that we would like.
• The last argument is the communicator and as we already discussed we usually use MPI_COMM_WORLD.

These are the arguments that are the most important information the MPI environment needs in order to know what data is sent and to whom.

As we saw the sender needs a function to send a message and obviously the receiver has to call the receive function. This means that for the communication to work, we need two processes and two ranks. So, one will call the MPI_Send function and the other will similarly call MPI_Recv to receive. To be able to receive, we use the function

MPI_Recv(void *buf, int count, MPI_Datatype datatype, int source, int tag, MPI_Comm comm, MPI_Status *status);

The arguments needed by this function are similar to the MPI_Send function.

• The buf/count/datatype describe the received data. In the first argument we specify the address of the data, i.e., the address where we would like to receive the data and similarly, we put the data type here.
• source is the rank of the sender process. We have to have the rank of the sender process. In the previous example we would specify number 3, because the rank with number three is trying to send us a message.
• Similar to MPI_Send, we also have a tag here. It is really important to make sure that we match this number with the sender. So, if the sender specifies that the message has tag 0, the receiver also has to specify the same number here. Otherwise, this will be an infinite loop as we will not receive anything because we would not match our receive function call with the send function call.
• The next argument is the communicator and again we would just use the MPI_COMM_WORLD.
• The last argument is not so important for us at the moment as this is something that we will be learning in the next exercise. For now we will just use
MPI_STATUS_IGNORE

So, let’s go through an example to understand again the prerequisites for this communication to work efficiently and how this would actually work in a code.

Image courtesy: Rolf Rabenseifner (HLRS)

Here, the left is the sender and the right is the receiver. Let’s suppose that the sender would like to send this buffer array that has n floats over to some other process. For this, it calls the MPI_Send routine function. As we already know, the first one is the pointer to the data. So, this is the send buffer. Then it needs to specify the number of data. In this case, it is n. The next argument is MPI_FLOAT, and we need to make sure that this data matches with the one mentioned earlier. As we previously discussed, this is the MPI data type that the environment defines, but it has to match with this one, otherwise the communication will not work. Another thing we need to keep in mind is that this data type has to match with the receiver. So, we have to be careful when we write these functions that all of these have to match. Now, the receiver has to call the receiver function with the same data type. Here it has to first define an array where it would like to receive this data, the receive buffer. The communicator, of course, has to be the same because they are bound in the same program. But we usually use the MPI_COMM_WORLD communicator. The next important part is that the tag has to match. And finally, the type of the message or type of the data has to match.


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## Warm-up

In Ab-surd!, we explored rationalising the denominators of fractions with square roots.

What would happen if instead of square roots, we had cube roots?

Can you rationalise the denominator of $\dfrac{1}{\sqrt[3]{2}}$?


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_Summary and Peer-review_ ### A comprehensive evaluation of normalization methods for Illumina high-throughput RNA sequencing data analysis ( [Dillies et al. 2013](http://bib.oxfordjournals.org/content/14/6/671)) _Trang Tran_

### 7 normalization methods were evaluated

\begin{align} \DeclareMathOperator*{\quantile}{quantile} \DeclareMathOperator*{\median}{median} \end{align}
\begin{align} \text{Total count (TC) } a_{ij} &= \frac{c_{ij}}{\sum\limits_{j=1}^{N} c_j} \end{align}
\begin{align} \text{Upper Quartile (UQ) } a_{ij} &= \frac{c_{ij}}{\quantile\limits_{c_j > 0}^{0.75} c_j} \end{align}
\begin{align} \text{Median (M) } a_{ij} &= \frac{c_{ij}}{\median\limits_{c_j > 0} c_j} \end{align}
\begin{align} \text{DESeq } a_{ij} &= \frac{c_{ij}}{\median\limits_{j = 1}^m \frac{c_{ij}}{\frac{1}{m} \sum\limits_{j = 1}^m c_{ij}}} \end{align}
\begin{align} \text{TMM } a_{ij} &= \end{align}
\begin{align} \text{Quantile: Not explained in the paper} \end{align}
\begin{align} \text{RPKM } a_{ij} &= \frac{c_{ij}}{\ell_i \cdot M \cdot 10^{-9}} \end{align}
\begin{align} c_{ij} &: \text{read count of gene } i \text{ in sample } j \\ N &: \text{number of genes} \\ m &: \text{number of samples} \\ M &: \text{number of mapped reads} \end{align}

Real data

Simulation data

### Real data sets

Table 1:
D = directional, ND = non-directional

Organism Type Number of genes Replicates per condition Minimum library size Maximum library size Correlation between replicates Correlation between conditions % Reads associated with the most expressed gene Library type Sequencing machine
h. sapiens rna 26 437 {3, 3} 2.0 × 107 2.8 × 107 (0.98, 0.99) (0.93, 0.96) ≈1% sr 54, nd gaiix
a. fumigatus rna 9248 {2, 2} 8.6 × 106 2.9 × 107 (0.92, 0.94) (0.88, 0.94) ≈1% sr 50, d hiseq2000
e. histolytica rna 5277 {3, 3} 2.1 × 107 3.3 × 107 (0.85, 0.92) (0.81, 0.98) 6.4–16.2% pe 100, nd hiseq2000
m. musculus mirna 669 {3, 2, 2} 2.0 × 106 5.9 × 106 (0.95, 0.99) (0.09, 0.75) 17.4–51.1% sr 36, d gaiix

### Normalization on real data are compared in terms of

A. Distribution of normalized expression

B. Intra-group variance

C. Differential expression analysis

z. House-keeping gene variance

### A. Distribution of normalized expression

Each box corresponds to one sample.

Organism Type Number of genes
h. sapiens rna 26 437
a. fumigatus rna 9248
e. histolytica rna 5277
m. musculus mirna 669

### C. Effects on Differential Analysis

Log-transformed of the ratio of expression levels between 2 conditions (y-axis) vs their Average (x-axis). A value well above/below zero region is indicative of differentially expressed genes

Genes are identified as differentially expressed by all methods are colored in GREY.

Both DESeq and TSPM require a table of count data as input.

The outcome of a DEG analysis also depends on the compatibility of the normalization method used for input and the DE method.

### z. Housekeeping genes variance

30 Housekeeping genes are selected from the list identified by Eisenberg & Levanon (2003)

However, the authors have rightfully noted about the caveats of using housekeeping genes as internal reference in RNA-seq normalization

#### Housekeeping genes are inappropriate for use as internal reference in RNA-seq

• They are often affected by various factors that are not controlled
• They are usually highly expressed thus not representing genes of low intensities
• HKG are usually a very small subset, so ﬂuctuations in their intensities are highly affected by random or systematic errors
• This use requires the a priori knowledge of the housekeeping genes

### Performance on simulation data were based on

False positive rate

Power (Recall)

### Simulation settings

1. Equal library size, no dominant gene
2. Non-equal library size, no dominant gene
3. Equal library size, some dominant genes

### Consensus and discordance

Log-transformed of the ratio of expression levels between 2 conditions (y-axis) vs their Average (x-axis). A value well above/below zero region is indicative of differentially expressed genes

Genes are identified as differentially expressed by all methods are colored in GREY.

Both DESeq and TSPM require a table of count data as input.

The outcome of a DEG analysis also depends on the compatibility of the normalization method used for input and the DE method.

### Other note: GC-bias

Log-transformed counts vs GC-content.

Each line is one sample, with color-coded conditions (A,B,C).

It's not clear how this plot was generated. A guess would be that each data point is a transcript, and the x-value is the GC-composition of that transcript, while the y-value is the log-transformed of accumulated read counts. Thus each line of the plot might have been fitted over the raw data, and the choice of regression method might have biased toward an a priori model.

Assuming the fitting of those lines are not biased towards any pre-defined model, it seems there is siginificant correlation between GC composition and read counts, in contrast to the authors' claim that no such bias was observed in the data. Although this relation might not be linear nor monotonic, it seems to have the same pattern within a species, but sample-specific parameters.