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# Rare stickers

In an earlier post we looked at the number of packets of Panini stickers you’d be likely to need in order to complete an album. In that post I made the throwaway and obvious remark that if some stickers were rarer than others, you’d be likely to need even more packs than the standard calculations suggest. But is there any evidence that some stickers are rarer than others?

The official Panini response is given here. In reply to the question “Does Panini deliberately print limited edition or rare stickers?”, the company answer is:

No. Every collection consists of a number of stickers printed on one or two print sheets, which in turn contain one of each sticker and are printed in the same quantity as the albums.

So that seems clear enough. But the experience of collectors is rather different, with many suggesting that some stickers are much harder to find than others. So, what’s the data and what type of statistical methods can be used to judge the evidence?

The first thing to say is that, as we saw in the previous post, the nature of collecting this way can be a frustrating experience. So that even if the average number of packs needed was less than 1000, we saw that some collectors are likely to need more than 2000, even assuming all stickers are equally frequent. To put this into perspective, suppose you’ve already collected 662 of the 682 stickers, and need just another 20 stickers to complete the album. By the same type of calculation that we made before, the expected number of further packs needed to complete the album is 491, which is well over half the expected total number of packs needed (which was 969). You might like to try that calculation yourself, using the same method as in the previous post.

In other words, once you reach the point of needing just 20 more stickers, you’re not even half-way through your collection task in terms of the number of packs you’re likely to need. This can make it seem like certain stickers – the 20 you are missing – are rarer than others, even if they’re not.

But that’s not to say that rare stickers don’t exist – it’s just that the fact you’re like to have to wait so long to get the remaining few stickers might make you feel like they are rare. So again: do rare stickers really exist?

Well, here’s a site that tried to conduct the experiment by buying 1000 packs from the 2018 world cup sticker series, listing every single sticker they got and the number of times they got it. Since each pack contains 5 stickers, this means that they collected a total of 5000 stickers. They then used the proportion of times they got each sticker as an estimate of the probability of getting each sticker. For example, they were fortunate enough to get the Belgium team photo sticker 18 times, so they estimated the probability of getting such a sticker as

$18/5000 \approx 1/278$

Using this method, they were able to calculate which team collectively were the easiest and most difficult to complete. The results are summarised in the following figure:

On this basis, stickers of players from Senegal and Columbia were easy to obtain, while those of players from Belgium and Saudi Arabia were much harder. So although the Belgium team photo was one of the most frequent stickers in their collection, the individual players from Belgium were among the least frequent.

Now, you won’t need me to tell you that this is pretty much a waste of time. With just 5000 stickers collected, it’s bound to be that some stickers occur more often than others, and we don’t learn anything this way about whether stickers of certain teams or players are genuinely more difficult to find than others. One could try to ascertain whether the pattern of results here is consistent with an equal distribution of stickers, but there’s no mention of such an analysis being done, and a sample of just 5000 stickers would probably be too small to reach definitive conclusions anyway.

One fun thing though: with the 5000 stickers collected, these guys managed to complete the album except for one missing sticker: Radja Nainggolan of Belgium.  But he ended up not being selected for the World Cup anyway 😀.

This doesn’t really bring us any closer to the question of whether rare stickers exist or not. One interesting suggestion is to look at frequencies of requests in sites that handle markets for second-hand stickers. And indeed, this site finds jumps in frequencies of requests for certain types of stickers, suggesting such types are rarer than others. However, it’s not necessarily the case that the rare stickers are the ones with the most requests: Lionel Messi might be a popular trade, because… Messi… and not because his sticker is rare. Still, the post is a fun read, with complete details about how you might approach this type of analysis.

Finally, far be it for me to promote the exploits of one of our competitors, but some of the guys at ATASS pooled their world cup collections to address exactly this issue. A complete description of their findings can be found here. In summary, from an analysis of nearly 11,000 stickers, they found evidence that shiny stickers – of which there were 50 in the 2018 World Cup series – are much rarer than standard stickers.

Moreover, the strength of the evidence is completely overwhelming. This is partly because the number of stickers collected is large – 11,000 rather than 5,000 in the study I mentioned above – but also because there are 50 shiny stickers, and all of them occurred with a lower frequency than an average sticker. In fact, overall, shiny stickers occurred at a rate that’s around half that of a normal sticker. Now, if a single sticker isn’t found, it’s reasonable to put that down to chance; but it’s beyond the realms of plausibility that 50 stickers of a certain kind all occurred at below the average rate.

On this basis, Panini’s claim that all stickers are equally likely is completely implausible. The only way it could really hold true, assuming the ATASS analysis to be correct, is if there were variations in the physical distribution of stickers, either geographically or temporally. So, although Panini might produce all stickers in equal numbers, variations in distribution might mean that some stickers were harder to get at certain times in certain places.

That seems unlikely though, and the evidence does seem to point to the fact that shiny stickers in the World Cup 2018 series were indeed harder to find than the others.

In summary: yes, rare stickers exist.

# Here’s to all the money at the end of the world

I made the point in last week’s Valentine’s Day post, that although the emphasis of this blog is about the methodology of using Statistics to understand the world through the analysis of data, it’s often the case that statistics in themselves tell their own story. In this way we learnt that a good proportion of the population of the UK buy their pets presents for Valentine’s Day.

As if that wasn’t bad enough, I now have to report to you the statistical evidence for the fact that nature itself is dying. Or as the Guardian puts it:

Plummeting insect numbers `threaten collapse of nature’

The statistical and scientific evidence now points to the fact that, at current rates of decline, all insects could be extinct by the end of the century. Admittedly, it’s probably not great science or statistics to extrapolate the current annual loss of 2.5% in that way, but nevertheless it gives you a picture of the way things are going. This projected elimination of insects would be, by some definitions, the sixth mass extinction event on earth. (Earlier versions wiped out dinosaurs and so on).

And before you go all Donald Trump, and say ‘bring it on: mosquito-free holidays’, you need to remember that life on earth is a complex ecological system in which the big things (including humans) are indirectly dependent on the little things (including insects) via complex bio-mechanisms for mutual survival. So if all the insects go, all the humans go too. And this is by the end of the century, remember.

Here’s First Dog on the Moon’s take on it:

So, yeah, let’s do our best to make money for our clients. But let’s also not forget that money only has value if we have a world to spend it in, and use Statistics and all other means at our disposal to fight for the survival of our planet and all the species that live on it.

# Happy Valentine’s Day

Happy Valentine’s Day. In case you didn’t get any cards or gifts today, please know that Smartodds loves Statistics loves you.

Anyway, I thought it might be interesting to research some statistics about Valentine’s day, and found this article, from which I learned much more about the population of Britain than I was expecting to.

Here are some of the highlights:

1. A significant number of people spend money for Valentine’s day on their pets. This number varies per generation, and is as high as 8.7% for millennials.
2. A slightly smaller, but still significant,  number of people spend money on themselves for Valentine’s. Again, this trend is most prevalent among millennials, and also more common for women than men.
3. 36.2% of people get unwanted gifts most years.
4. 19% of people plan to celebrate Valentine’s late in order to save money by buying cards and gifts once the prices have dropped.

I’m not sure which of these statistics I found to be the more shocking.

Most of the posts in this blog are about the way Statistics as a science can be used to investigate problems and interpret data. But sometimes, the statistics are fascinating in themselves, and don’t require any kind of mathematical sophistication to reveal the secrets that they contain.

Anyway, I have to run now to buy myself my girlfriend a gift

Happy Valentine’s…

# The bean machine

Take a look at the following video…

It shows the operation of a mechanical device that is variously known as a bean machine, a quincunx or a Galton board. When the machine is flipped, a large number of small balls or beans fall through a funnel at the top of the device. Below the funnel is a layered grid of pegs. As each bean hits a peg it can fall left or right – with equal probability if the board is carefully made – down to the next layer, where it hits another peg and can again go left or right. This repeats for a number of layers, and the beans are then collected in groups, according to the position they fall in the final layer. At the end you get a kind of physical histogram, where the height of the column of beans corresponds to the frequency with which the beans have fallen in that slot.

Remarkably, every time this experiment is repeated, the pattern of beans at the bottom is pretty much the same: it’s symmetric, high in the middle, low at the edges and has a kind of general bell-shape. In fact, the shape of this histogram will be a good approximation to the well-known normal distribution curve:

As you probably know, it turns out that the relative frequencies of many naturally occurring phenomena look exactly like this normal curve: heights of plants, people’s IQ, brightness of stars…. and indeed (with some slight imperfections) the differences in team points in sports like basketball.

Anyway, if you look at the bottom of the bean machine at the end of the video, you’ll see that the heights of the columns of beans – which in itself represents the frequency of beans falling in each position – resembles this same bell-shaped curve. And this will happen – with different small irregularities – every time the bean machine is re-started.

Obviously, just replaying the video will always lead to identical results, so you’ll have to take my word for it that the results are similar every time the machine is operated. There are some simulators available, but my feeling is you lose something by not seeing the actual physics of real-world beans falling into place. Take a look here if you’re interested, though I suggest you crank the size and speed buttons up to their maximum values first.

But why should it be that the bean machine, like many naturally occurring phenomena, leads to frequencies that closely match the normal curve?

Well, the final position of each bean is the result of several random steps in which the bean could go left or right. If we count +1 every time the bean goes right and -1 every time the bean goes left, then the final position is the sum of these random +/-1 outcomes. And it turns out, that under fairly general conditions, that whenever you have a process that is the sum of several random experiments, the final distribution is bound to look like this bell-shaped normal curve.

This is a remarkable phenomenon. The trajectory of any individual bean is unpredictable. It could go way to the left, or way to the right, though it’s more likely that it will stay fairly central. Anything is possible, though some outcomes are more likely than others. However, while the trajectory of individual beans is unpredictable, the collective behaviour of several thousand beans is entirely predictable to a very high degree of accuracy: the frequencies within any individual range will match very closely the values predicted by the normal distribution curve. This is really what makes statistics tick. We can predict very well how a population will behave, even if we can’t predict how individuals will behave.

Even more remarkably, if the bean machine has enough layers of pegs, the eventual physical histogram of beans will still look like the normal distribution curve, even if the machine has some sort of bias. For example, suppose the beans were released, but that the machine wasn’t quite vertical, so that the beans had a higher tendency to go left, rather than right, when they hit a peg. In this case, as long as there were sufficiently many layers of pegs, the final spread of beans would still resemble the normal curve, albeit no longer centred at the middle of the board. You can try this in the simulator by moving the left/right button away from 50%.

Technically, the bean machine is a physical illustration of a mathematical result generally termed the Central Limit Theorem. This states that in situations like those illustrated by the bean machine, where a phenomenon can be regarded as a sum of random experiments, then under general conditions the distribution of final results will look very much like the well-known bell-shaped normal curve.

It’s difficult to overstate the importance of this result – which is fundamental to almost all areas of statistical theory and practice – since it lets us handle probabilities in populations, even when we don’t know how individuals behave. And the beauty of the bean machine is that it demonstrates that the Central Limit Theorem is meaningful in the real physical world, and not just a mathematical artefact.

Can’t live without your own desktop bean machine? I have good news for you…

# Groundhog day

Fed up of the cold, snow and rain? Don’t worry, spring is forecast to be here earlier than usual. Two caveats though:

1. ‘Here’ is some unspecified region of the United States, and might not extend as far as the UK;
2. This prediction was made by a rodent.

Yes, Saturday (February 2nd) was Groundhog Day in the US. And since Punxsutawney Phil failed to see his shadow, spring is forecast to arrive early.

You probably know about Groundhog Day from the Bill Murray movie

… but it’s actually a real event. It’s celebrated in many locations of the US and Canada, though it’s the event in Punxsutawney, Pennsylvania, which has become the most famous, and around which the movie was based. As Wikipedia says:

The Groundhog Day ceremony held at Punxsutawney in western Pennsylvania, centering around a semi-mythical groundhog named Punxsutawney Phil, has become the most attended.

Semi-mythical, no less. If you’d like to know more about Punxsutawney Phil, there’s plenty of information at The Punxsutawney Groundhog Club website, including a dataset of his predictions. These include the entry from 1937 when Phil had an ‘unfortunate meeting with a skunk’. (And whoever said data analysis was boring?)

Anyway, the theory is that if, at 7.30 a.m. on the second of February, Phil the groundhog sees his shadow, there will be six more weeks of winter; if not, spring will arrive early. Now, it seems a little unlikely that a groundhog will have powers of meteorological prediction, but since the legend has persisted, and there is other evidence of animal behaviour serving as a weather predictor,  it seems reasonable to assess the evidence.

Disappointingly, Phil’s success rate is rather low. This article gives it as 39%. I’m not sure if it’s obvious or not, but the article also states (correctly) that if you were to guess randomly, by tossing a coin, say, then your expected chance of guessing correctly is 50%. The reason I say it might not be obvious, is because the chance of spring arriving early is unlikely to be 50%. It might be 40%, say. Yet, randomly guessing with a coin will still have a 50% expected success rate. As such, Phil is doing worse than someone who guesses at random, or via coin tossing.

However, if Phil’s 39% success rate is a genuine measure of his predictive powers – rather than a reflection of the fact that his guesses are also random, and he’s just been a bit unlucky over the years – then he’s still a very useful companion for predictive purposes. You just need to take his predictions, and predict the opposite. That way you’ll have a 61% success rate – rather better than random guessing. Unfortunately, this means you will have to put up with another 6 weeks of winter.

Meantime, if you simply want more Groundhog Day statistics, you can fill your boots here.

And finally, if you think I’m wasting my time on this stuff, check out the Washington Post who have done a geo-spatial analysis of the whole of the United States to colour-map the regions in which Phil has been respectively more and less successful with his predictions over the years.

🤣

# Groundhog day

Fed up of the cold, snow and rain? Don’t worry, spring is forecast to be here earlier than usual. Two caveats though:

1. ‘Here’ is some unspecified region of the United States, and might not extend as far as the UK;
2. This prediction was made by a rodent.

Yes, Saturday (February 2nd) was Groundhog Day in the US. And since Punxsutawney Phil failed to see his shadow, spring is forecast to arrive early.

You probably know about Groundhog Day from the Bill Murray movie

… but it’s actually a real event. It’s celebrated in many locations of the US and Canada, though it’s the event in Punxsutawney, Pennsylvania, which has become the most famous, and around which the movie was based. As Wikipedia says:

The Groundhog Day ceremony held at Punxsutawney in western Pennsylvania, centering around a semi-mythical groundhog named Punxsutawney Phil, has become the most attended.

Semi-mythical, no less. If you’d like to know more about Punxsutawney Phil, there’s plenty of information at The Punxsutawney Groundhog Club website, including a dataset of his predictions. These include the entry from 1937 when Phil had an ‘unfortunate meeting with a skunk’. (And whoever said data analysis was boring?)

Anyway, the theory is that if, at 7.30 a.m. on the second of February, Phil the groundhog sees his shadow, there will be six more weeks of winter; if not, spring will arrive early. Now, it seems a little unlikely that a groundhog will have powers of meteorological prediction, but since the legend has persisted, and there is other evidence of animal behaviour serving as a weather predictor,  it seems reasonable to assess the evidence.

Disappointingly, Phil’s success rate is rather low. This article gives it as 39%. I’m not sure if it’s obvious or not, but the article also states (correctly) that if you were to guess randomly, by tossing a coin, say, then your expected chance of guessing correctly is 50%. The reason I say it might not be obvious, is because the chance of spring arriving early is unlikely to be 50%. It might be 40%, say. Yet, randomly guessing with a coin will still have a 50% expected success rate. As such, Phil is doing worse than someone who guesses at random, or via coin tossing.

However, if Phil’s 39% success rate is a genuine measure of his predictive powers – rather than a reflection of the fact that his guesses are also random, and he’s just been a bit unlucky over the years – then he’s still a very useful companion for predictive purposes. You just need to take his predictions, and predict the opposite. That way you’ll have a 61% success rate – rather better than random guessing. Unfortunately, this means you will have to put up with another 6 weeks of winter.

Meantime, if you simply want more Groundhog Day statistics, you can fill your boots here.

And finally, if you think I’m wasting my time on this stuff, check out the Washington Post who have done a geo-spatial analysis of the whole of the United States to colour-map the regions in which Phil has been respectively more and less successful with his predictions over the years.

# Stickers

Last year’s Fifa© world cup Panini sticker album had spaces for 682 stickers. Stickers were sold in packs of 5, at a cost of 80 pence per pack. How much was it likely to cost to fill the whole album? Maybe have a guess at this before moving on.

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Well, to get 682 stickers you need 137 packs, so the obvious (but wrong) answer is 137 times 80 pence, which is ￡109.60. It’s wrong, of course, because it doesn’t take into account duplicate stickers: as the album fills up, when you buy a new pack, it’s likely that at least some of the new stickers will be stickers that you’ve already collected. And the more stickers you’ve already collected, the more likely it is that a new pack will contain stickers that you’ve already got. So, you’re likely to need many more than 137 packs and spend much more than ￡109.60. But how much more?

It turns out (see below) that on average the number of packs needed can be calculated as

$(682/682 + 682/681 + 682/680 + \ldots + 682/1) /5 \approx 969$

where the “…” means “plus all the terms in-between”. So the next term in the sequence you have to add is 682/679 and then 682/678 and so on, all the way down to the final term in the sequence which is given as 682/1.

So the average cost of filling the album  is around $969 \times 80$ pence, or £775. You can probably also guess how this calculation changes if the number of spaces in the album were different from 682 or if the number of stickers per pack were different from 5.

Well, actually, there’s a small mistake in this calculation. Strictly speaking, when you buy packs of 5 stickers, none of the stickers in a pack will be duplicates among themselves. The above calculation ignores this fact, and assumes that duplicates could occur within packs. However, it turns out that doing the mathematics more carefully – which is quite a bit more complicated – leads to a not-very-different answer of £773. So, we might have simplified things in our calculation of £775, but we didn’t lose much in terms of accuracy.

Anyway, a question that’s just as interesting as the accuracy of the answer is what the value of £775 means in practice. Though it’s the average value that would be spent by many collectors in filling the album, the actual experience of any individual collector might be quite different from this. The mathematics is more complicated again in this case, but we can avoid the complexity by simulating the process. The figure below shows a histogram of the number of packs needed to fill the album in a simulation of 10,000 albums.

So, for example,  I needed roughly 800 packs to complete the album in around 1500 of the simulated albums. Of course, the average number of packs needed turns out to be close to the theoretical average of 969. But although sometimes fewer than this number were needed, the asymmetry of the histogram means that on many occasions far more than the average number was needed. For example, on a significant number of  occasions more than 1000 packs were needed; on several occasions more than 1500 packs were needed; and on a few occasions more than 2000 packs were needed (at a cost of over £1600!). By contrast, there were no occasions on which 500 packs were sufficient to complete the album. So, even though an average spend of £775 probably sounded like a lot of money to fill the album, any individual collector might need to spend as much as £2000 or more, while all collectors would have need to spend at least £400.

This illustrates an important point about Statistics in general – an average is exactly that: an average. And individual experiences might differ considerably from that average value. Moreover, asymmetry in the underlying probability distribution – as seen in the histogram above – will imply that variations from the average are likely to be bigger in one direction than the other. In the case of Panini sticker albums, you might end up paying a lot more than the average of £775, but are unlikely to spend very much less.

To be fair to Panini, it’s common for collectors to swap duplicate stickers with those of other collectors. This obviously has the effect of reducing the number of packs needed to complete the album. Furthermore, Panini now provide an option for collectors to order up to 50 specific stickers, enabling collectors who have nearly finished the album to do so without buying further packs when the chance of duplication is at its highest. So for both these reasons, the expected costs of completing the album as calculated above are over-estimates. On the other hand, if certain stickers are made deliberately rarer than others, the expected number of packs will increase! Would Panini do that? We’ll discuss that in a future post.

Meantime, for maths enthusiasts, and just in case you’re interested, let’s see where the formula

$(682/682 + 682/681 + 682/680 + \ldots + 682/1) /5 \approx 969$

comes from. You might remember from an earlier post, that if I repeat an experiment that has probability p of success until I get my first success,  I will have to repeat the experiment an average of 1/p times. Well, buying new stickers until I get one that’s different from those I’ve already collected is an experiment of exactly this type, so I can use this result. But as the number of stickers I’ve already collected changes, so does the probability of obtaining a different sticker.

• At the start, I have 0 stickers, so the probability the next sticker will be a new sticker is 682/682, and the expected number of stickers I’ll need till the next new sticker is 682/682. (No surprises there.)
• I will then have 1 sticker, and the probability the next sticker will be a new sticker is 681/682.  So the expected number of stickers I’ll need till the next new sticker is 682/681.
• I will then have 2 different stickers, and the probability the next sticker will be a new sticker is 680/682.  So the expected number of stickers I’ll need till the next new sticker is 682/680.
• This goes on and on till I have 681 stickers and the probability the next sticker will be a new sticker is 1/682.  So the expected number of stickers I’ll need till the next new sticker is 682/1.

At that point I’ll have a complete collection. Adding together all these expected numbers of stickers gives

$(682/682 + 682/681 + 682/680 + \ldots + 682/1)$

But each pack contains 5 stickers, so the expected number of packs I’ll need  is

$(682/682 + 682/681 + 682/680 + \ldots + 682/1) /5 \approx 969$

Just in case anyone attempted the Royal Statistical Society Christmas quiz that I included in an earlier post, the solutions are now available here. I managed to match my personal all-time record of zero. Funny though, looking at the solutions, how obvious everything should have been. Bit like looking at last weekend’s results and spotting all the obvious bets!

# Who wants to win £194,375?

In an earlier post I included a link to Oscar predictions by film critic Mark Kermode over the years, which included 100% success rate across all of the main categories in a couple of years. I also recounted his story of how he failed to make a fortune in 1992 by not knowing about accumulator bets.

Well, it’s almost Oscar season, and fabien.mauroy@smartodds.co.uk pointed me to this article, which includes Mark’s personal shortlist for the coming awards. Now, these aren’t the same as predictions: in some year’s, Mark has listed his own personal favourites as well as what he believes to be the likely winners, and there’s often very little in common. On the other hand, these lists have been produced prior to the nominations, so you’re likely to get better prices on bets now, rather than later. You’ll have to be quick though, as the nominations are announced in a couple of hours.

Anyway, maybe you’d like to sift through Mark’s recommendations, look for hints as to who he thinks the winner is likely to be, and make a bet accordingly. But if you do make a bet based on these lists, here are a few things to take into account:

1. Please remember the difference between an accumulator bet and single bets;
3. Please don’t blame me if you lose.

If Mark subsequently publishes actual predictions for the Oscars, I’ll include a link to those as well.

Update: the nominations have now been announced and are listed here. Comparing the nominations with Mark Kermode’s own list, the number of nominations which appear in Mark’s personal list for each category are as follows:

Best Picture: 1

Best Director: 2

Best Actor: 1

Best Actress: 2

Best supporting Actor: 3

Best supporting Actress: 1

Best Score: 2

In each case except Best Picture, there are 5 nominations and Mark’s list also comprised 5 contenders. For Best Picture, there are 8 nominations, though Mark only provided 5 suggestions.

So, not much overlap. But again, these weren’t intended to be Mark’s predictions. They were his own choices. I’ll aim to update with Mark’s actual predictions if he publishes them.

# Pulp Fiction (Our Esteemed Leader’s cut)

The previous post had a cinematic theme. That got me remembering an offsite a while back where Matthew.Benham@smartbapps.co.uk gave a talk that I think he called ‘Do the Right Thing’, which is the title of a 1989 Spike Lee film. Midway through his talk Matthew gave a premiere screening of his own version of a scene from Pulp Fiction. Unfortunately, I’ve been unable to get hold of a copy of Matthew’s cut, so we’ll just have to make do with the inferior original….

The theme of Matthew’s talk was the importance of always acting in relation to best knowledge, even if it contradicts previous actions taken when different information was available. So, given the knowledge and information you had at the start of a game, you might have bet on team A. But if the game evolves in such a way that a bet on team B becomes positive value, you should do that. Always do the right thing. And the point of the scene from Pulp Fiction? Don’t let pride get in the way of that principle.

These issues will make a great topic for this blog sometime. But this post is about something else…

Dependence is a big issue in Statistics, and we’re likely to return to it in different ways in future posts. Loosely speaking, two events are said to be independent if knowing the outcome of one, doesn’t affect the probabilities of the outcomes of the other. For example, it’s usually reasonable to treat the outcomes of two different football matches taking place on the same day as independent. If we know one match finished 3-0, that information is unlikely to affect any judgements we might have about the possible outcomes of a later match. Events that are not independent are said to be dependent: in this case, knowing the outcome of one will affect the outcome of the other.  In tennis matches, for example, the outcome of one set tends to affect the chances of who will win a subsequent set, so set winners are dependent events.

With this in mind, let’s follow-up the discussion in the previous 2 posts (here and here) about accumulator bets. By multiplying prices from separate bets together, bookmakers are assuming that the events are independent. But if there were dependence between the events, it’s possible that an accumulator offers a value bet, even if the individual bets are of negative value. This might be part of the reason why Mark Kermode has been successful in several accumulator bets over the years (or would have been if he’d taken his predictions to the bookmaker and actually placed an accumulator bet).

Let me illustrate this with some entirely made-up numbers. Let’s suppose ‘Pulp Fiction (Our Esteemed Leader’s cut)’, is up for a best movie award, and its upstart director, Matthew Benham, has also been nominated for best director. The numbers for single bets on PF and MB are given in the following table. We’ll suppose the bookmakers are accurate in their evaluation of the probabilities, and that they guarantee themselves an expected profit by offering prices that are below the fair prices (see the earlier post).

True Probability Fair Price Bookmaker Price
Best Movie: PF 0.4 2.5 2
Best Director: MB 0.25 4 3.5

Because the available prices are lower than the fair prices and the probabilities are correct, both individual bets have negative value (-0.2 and -0.125 respectively for a unit stake). The overall price for a PF/MB accumulator bet is 7, which assuming independence is an even poorer value bet, since the expected winnings from a unit stake are

$0.4 \times 0.25 \times 7 -1 = -0.3$

However, suppose voters for the awards tend to have similar preferences across categories, so that if they like a particular movie, there’s an increased chance they’ll also like the director of that movie. In that case, although the table above might be correct, the probability of MB winning the director award if PF (MB cut) is the movie winner is likely to be greater than 0.25. For argument’s sake, let’s suppose it’s 0.5. Then, the expected winnings from a unit stake accumulator bet become

$0.4 \times 0.5 \times 7 -1 = 0.4$

That’s to say, although the individual bets are still both negative value, the accumulator bet is extremely good value. This situation arises because of the implicit assumption of independence in the calculation of accumulator prices. When the assumption is wrong, the true expected winnings will be different from those implied by the bookmaker prices, potentially generating a positive value bet.

Obviously with most accumulator bets – like multiple football results – independence is more realistic, and this discussion is unhelpful. But for speciality bets like the Oscars, or perhaps some political bets where late swings in votes are likely to affect more that one region, there may be considerable value in accumulator bets if available.

If anyone has a copy of Our Esteemed Leader’s cut of the Pulp Fiction scene on a pen-drive somewhere, and would kindly pass it to me, I will happily update this post to include it.


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# How can I define a Step-Wise function in Mathematica (Not using Heaviside Step Function)? [closed]

I need to define a function, which has very different behaviour in different regions. There are about 13 different regions. A sample of my function is the foloowing table:

I want to define it as a function of x, and I demand that for example for 0<=x<=1221.5, f[x_]:=13.088-8.84*x^2, and for 1221.5<=x<=3480, f[x_]:=12.58-1.26*x-3.64*x^2-5.53*x^3 and .... .

How can I do it in a wise way in Mathematica?

-
Take a look at Piecewise. –  VLC Mar 19 '13 at 18:44
@VLC, Thanks for your fast reply. You meen for example f[x_]:=Piecewise[{{x^2, x < 0}, {x, x > 0}}] ? –  ZKT Mar 19 '13 at 18:46
Indeed, that's how you can use it. –  VLC Mar 19 '13 at 18:48
@VLC Thanks a lot again. I love this forum, my problem got solved just 1 minute after posting my question in here! :) –  ZKT Mar 19 '13 at 18:51

## closed as too localized by Artes, m_goldberg, Jens, Oleksandr R., rm -rf♦Mar 21 '13 at 20:23

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Your function is double valued at x=1221.5 but if you are happy to chose one of the limits to be strictly greater than (or less than), then you can define your function as:
f[x_]=Piecewise[{{13.088-8.84x^2,0<=x<1221.5},{12.58-1.26x-3.64x^2-5.53x^3,1221.5<=x<=3480}}]

If you're defining a function, it is important to use the SetDelayed command: f[x_]:= Piecewise[(* the rest *)] –  zentient Mar 20 '13 at 7:05
If this function stands alone, and you don't want to do anything else with f[x_] or x elsewhere, then it appears to produce the same result. However, among other things this may cause scoping issues. The difference can produce unexpected results elsewhere in your code. SetDelayed,f[x_] := Piecewise[(* the rest *)], will keep all x's in the RHS local (they stay green). If you use Set, f[x_] = Piecewise[(* the rest *)], then RHS x's are global (they're blue) and may conflict with your use of x elsewhere. See Documentation > SetDelayed > Properties > ...scope –  zentient Mar 20 '13 at 9:03
Try executing x = 1. before your definition using =, and see what happens. :) That is why you need :=. –  Ｊ. Ｍ. Mar 23 '13 at 16:17

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# Atmel AVR XMEGA-A USART MPCM (Multiprocessor communication mode) configuration with ASF 3.33

I'm using two XMEGA256-A3BU's on XMEGA-A3BU Xplained boards. My goal is to make MPCM (Multiprocessor communication mode) possible between the two using their USART in synchronous mode.
They are physically connected so that the master's PC1 (SCL), PC2 (RXD) and PC3 (TXD) are connected to the slave's PC1 (SCL), PC3 (TXD) and PC2 (RXD) respectively.
I use ASF 3.33's "USART - serial service" to make the communication work.

Master's initialization code:

static usart_serial_options_t usart_options = {
.baudrate = 1200;
.charlength = USART_CHSIZE_8BIT_gc,
.paritytype = USART_PMODE_ODD_gc,
.stopbits = false
};

sysclk_enable_module(SYSCLK_PORT_C, PR_USART0_bm);
usart_serial_init(&USARTC0, &usart_options);

usart_set_mode(&USARTC0, USART_CMODE_SYNCHRONOUS_gc);

// Pin to set as output for clock signal.
ioport_configure_pin(IOPORT_CREATE_PIN(PORTC, 1), IOPORT_DIR_OUTPUT);
// Port to output clock signal on.
PORTCFG.CLKEVOUT = PORTCFG_CLKOUT_PC1_gc;

usart_set_rx_interrupt_level(&USARTC0, USART_RXCINTLVL_OFF_gc);
usart_set_tx_interrupt_level(&USARTC0, USART_TXCINTLVL_OFF_gc);

Slave's initialization code:

static usart_serial_options_t usart_options = {
.baudrate = 1200;
.charlength = USART_CHSIZE_8BIT_gc,
.paritytype = USART_PMODE_ODD_gc,
.stopbits = false
};

sysclk_enable_module(SYSCLK_PORT_C, PR_USART0_bm);
usart_serial_init(&USARTC0, &usart_options);

usart_set_mode(&USARTC0, USART_CMODE_SYNCHRONOUS_gc);

// Pin to set as input for clock signal.
ioport_configure_pin(IOPORT_CREATE_PIN(PORTC, 1), IOPORT_DIR_INPUT);
// Do not output the clock signal.
PORTCFG.CLKEVOUT = PORTCFG_CLKOUT_OFF_gc;

usart_set_rx_interrupt_level(&USARTC0, USART_RXCINTLVL_OFF_gc);
usart_set_tx_interrupt_level(&USARTC0, USART_TXCINTLVL_OFF_gc);

I use usart_serial_putchar(&USARTC0, buffer[i]) and usart_serial_getchar(&USARTC0, &buffer[i]) to send and receive data for testing, the final design will use DMA.

In order to make MPCM work (and make it possible to add more slaves to the bus), I know that I need to switch to the 1 start bit + 9 data bits + parity + 1 stop bit format, and I also know that somehow I should use these constants I found in iox256a3bu.h file:

#define USART_MPCM_bm  0x02  /* Multi-processor Communication Mode bit mask. */
#define USART_MPCM_bp  1  /* Multi-processor Communication Mode bit position. */

Could someone help me out with some ASF sample code? What should I add to set the slave's own address, and where should I define on the master who I'm intending to send the message to?

I've spent a few hours on the problem, and fortunately it works now with one master and two slaves. It doesn't use DMA because I think it is impractical in this case, so I decided to use interrupts instead.
First I changed the USART's format to 9 bits, so the initialization method starts like this:

static usart_serial_options_t usart_options = {
.baudrate = 1200;
.charlength = USART_CHSIZE_9BIT_gc,
.paritytype = USART_PMODE_ODD_gc,
.stopbits = false
};

And I changed the interrupt level at its end:

usart_set_rx_interrupt_level(USART_SERIAL, USART_RXCINTLVL_OFF_gc);
usart_set_tx_interrupt_level(USART_SERIAL, USART_TXCINTLVL_MED_gc);

Then I defined two methods to replace the usart_serial_putchar(&USARTC0, buffer[i]) and usart_serial_getchar(&USARTC0, &buffer[i]) methods:

static void usart_send_mpcm_data(usart_if usart, uint8_t data, bool is_address)
{
/* Wait until we are ready to send the next 9 bits */
while (usart_data_register_is_empty(usart) == false) {}

/* First we need to set the 9. bit of the frame, which is the indicator for the address frame */
{
(usart)->CTRLB |= USART_TXB8_bm;
} else
{
(usart)->CTRLB &= ~USART_TXB8_bm;
}

/* Second, let's load the 8-bit data into the DATA register of the USART */
(usart)->DATA = data;
}

{
bool result = false;

/* Are we in MPCM mode? */
bool MPCM_mode = usart->CTRLB & USART_MPCM_bm;
/* Is this an address frame? */
bool address_frame = usart->STATUS & USART_RXB8_bm;

/* Read char, clearing RXCIF */

{
/* This is a data frame, and we are interested in it */
result = true;
}

{
}

/* Set the MPCM bit in the USART's control register */
if (MPCM_mode)
{
usart->CTRLB |= USART_MPCM_bm;
} else
{
usart->CTRLB &= ~USART_MPCM_bm;
}

/* Return true if we read some useful data */
return result;
}

After having these, I could setup the interrupt method on the slaves:

// USART-C Reception Complete Interrupt
ISR(USARTC0_RXC_vect)
{

On the master I can use the usart_send_mpcm_data(&USARTC0, usart_source[i], false); method to send data, or usart_send_mpcm_data(&USARTC0, slave_address, true); to change the target slave address. If the address is 0, then that's going to be a broadcast message, so all slaves will receive the data.


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# Thread: equation of line through point orthogonal to line

1. ## equation of line through point orthogonal to line

I am having a blank moment.

I have a point and a parametric equation for a line. I need to find the equation of the line through that point that is perpendicular to the line I already have.

2. ## Re: equation of line through point orthogonal to line

the point is (3,1,-2) and the line is x=-1, y = -2+t, z = -1+t

I know the dot product will be zero meaning the are orthogonal and I know it has something to do with find an intersection point but drawing a blank on how Ifind the intersection point that gives me the proper coordinates to have orthogonal lines

3. ## Re: equation of line through point orthogonal to line

"Perpendicular" in three dimensions is different from perpendicular in two dimensions. In two dimensions, perpendicular lines intersect. In three dimensions, they don't need to. Hence, there are an infinite number of lines that are perpendicular to the given line and pass through the given point. There is either exactly one line or an infinite number of lines perpendicular to a given line that passes through BOTH that line AND a given point. For this particular given line and point, there is exactly one. I assume you want the equation for that line.

Since the line passes through both the given point and the given line, a vector in the direction of this perpendicular line would be the vector $\begin{pmatrix}3-(-1) \\ 1-(-2+t) \\ -2-(-1+t)\end{pmatrix} = \begin{pmatrix}4 \\ 3-t \\ -1-t\end{pmatrix}$.

Let's find a vector in the direction of the given line by finding two points along the line (we can choose any two points, so let's say $t=0, t=1$). So, a vector in the direction of the given line is $\begin{pmatrix}-1-(-1) \\ -2+1 - (-2+0) \\ -1+1 - (-1+0)\end{pmatrix} = \begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}$. Since we want these vectors to be perpendicular, we want the dot product to be zero. Hence:

$(4\cdot 0) + (1\cdot (3-t)) + (1\cdot (-1-t)) = 2-2t = 0$ implies $t=1$. Hence, a vector in the direction of the perpendicular line is $\begin{pmatrix}4 \\ 3-1 \\ -1-1\end{pmatrix} = \begin{pmatrix}4 \\ 2 \\ -2\end{pmatrix}$.

So, the perpendicular line would have parametric equation $x=3+4t, y=1+2t, z=-2-2t$.

4. ## Re: equation of line through point orthogonal to line

so the point (does this mean you are treating the point as a bounded vector) minus the line gives a direction vector along the perpendicular line then I take the given direction vector of the line which is the same you got setting t values <0,1,1>

then set up the dot product and solve for t and input t into the direction vector I got from taking the point minus the line, then write the new parametric equations based off my original point and the direction vector

I'll do that way

but for sake of completeness

(1) the book suggests find the intersection

(2) can I use projection to find it as well??

5. ## Re: equation of line through point orthogonal to line

I got two points along the given line by plugging in $t=0$ and $t=1$. When $t=0$, I get the point $(-1,-2+0,-1+0) = (-1,-2,-1)$. When $t=1$, I get the point $(-1,-2+1,-1+1) = (-1,-1,0)$. So, the vector from the first point to the second point is $\begin{pmatrix}-1-(-1) \\ -1-(-2) \\ 0-(-1)\end{pmatrix} = \begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}$. That is a vector in the direction of the given line (Essentially, this a vector containing some multiple of the coefficients of $t$ in the parametric equation of the given line). You can rewrite the line as $\begin{pmatrix}-1 \\ -2 \\ -1\end{pmatrix} + t\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}$. This immediately gives a vector in the direction of the given line to be $\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}$.

Essentially, what I did was I found all possible lines that intersect both the given line and pass through the point $(3,1,-2)$. The point of intersection (in reference to your question 1) would be some point on the line, so it intersects your given line at $(-1,-2+t,-1+t)$ for some value of $t$. We are looking to find what value of $t$ will make the line perpendicular. That's pretty much all I did.

For (2), I am not sure what you mean. What projection are you trying to use to find it?

6. ## Re: equation of line through point orthogonal to line

orthogonal projection:

$\vec{p} = \frac{\vec{a} * \vec{v}}{||a||^2}\vec{a}$

'a' being the direction vector on the given line, 'v' being a vector from a point on the line to the point then from the tip of v to the tip of p there is an orthogonal line

7. ## Re: equation of line through point orthogonal to line

Originally Posted by Jonroberts74
orthogonal projection:

$\vec{p} = \frac{\vec{a} * \vec{v}}{||a||^2}\vec{a}$

'a' being the direction vector on the given line, 'v' being a vector from a point on the line to the point then from the tip of v to the tip of p there is an orthogonal line
Yes, that will work, as well.


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# Bug: Non-ascii URLs get mangled after updating interesting/ignored tags

Basel Shishani
197 7 2 14

There is buggy behavior when updating the tags in the interesting/ignored tags on the main page:

If the window URL contains URL encoded non ascii chars (as in internationalized url's where the 'question' segment has been translated) then the refreshed URL after the Ajax call would get mangled (the non-ascii segment goes through incorrect encode/decode it seems). Also, the newly added/removed tags would not show immediately on the page, but would show on subsequent page updates.

This seems to be an issue not with Askbot but with the History.js library - when calling History.pushState.

I modified the code in the 'livesearch.js' file to call history.pushState directly (notice the lower case 'h') bypassing the History.js library, and the issue went away as far as I can tell. (Of course by doing so we lose the supposed benifit of History.js)

I provide more details in my recent StackOverflow posting, so check it there please.

delete close retag edit

[hide preview]


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# Prove xy=0 <=> (x=0 or y=0)

1. Jan 20, 2012

### Rayquesto

Our teacher is teaching calculus 1 with a logic approach. So, he wants us to practice our math logic skills, however, I lack in providing formal methods of validity in operation and to some degree, application. So, here is my solution outline (please tell me what you think)

if operation axiom translational invariance of order:

for all x,y,z€ℝ
x<y => x + z < y + z then the statement:

xy=0 <=> (x=0 or y=0) is true

xy=0 => for all x,y,0€ℝ:[x<y => (x + 0)< (y + 0) or y<x => (y + 0)<(x+0)] by the translation invariance of order

y>0[x€ℝ => 0x = 0] => x>0[y€ℝ => 0y = 0] => for all x,y,0€ℝ:[x<y => (x + 0)< (y + 0) or y<x => (y + 0)<(x+0)] by the multiplication by 0 theorem

Anything else to the proof?

Last edited: Jan 21, 2012
2. Jan 20, 2012

### Stephen Tashi

We'd have to know what definitions and assumptions your text materials make to know how to prove this using them. It's not clear what you mean by "the translational invariance of order". How does your book define that?

3. Jan 20, 2012

### Rayquesto

operation axiom translational invariance of order:

for all x,y,z€ℝ
x<y => x + z < y + z

4. Jan 20, 2012

### Dickfore

To prove the necessary condition, use contraposition. Suppose $x \neq 0 \wedge y \neq 0$. Then, you can divide by either $x$ or $y$ in $x \, y = 0$. What do you get?

5. Jan 20, 2012

### Rayquesto

If neither x nor y can be zero, then the statement that x or y in xy=0 is false.

6. Jan 20, 2012

### Rayquesto

So, I'm sorry. I'm really trying to hard to move on with this proof. I'm just having a tough time with it.

When it comes to proofs, I hardly know when to start and when I do, I don't know when It's complete or when it needs to be complete. I had 80 percent confidence in my answer, but apparently, I'm wrong. So, as far as this contraposition method, it's sounds like you are suggesting I use symbols to work out the logic, which I have done. Perhaps I am wrong in that interpretation. What exactly do you mean?

7. Jan 20, 2012

### Stephen Tashi

It's fine to use symbols to abbreviate thought but you if you treat them as some sort of magical substitute for thought, you only get in trouble.

For example, in your first attempt at the proof you write down the step:
It isn't clear what you thought you were deducing this statement from. If you were trying to deduce it from the statement
then you were trying to deduce from the statement that you are attempting to prove, which wouldn't make sense.

To prove a statment with an "if and ony if" condition in it, it is often simplest to break this into two implications. So "for each pair of real numbers x,y , xy = 0 if and only if x = 0 or y = 0" can be proved by proving these two statements separately:

1) "for each pair of real numbers x,y, if (x = 0 or y = 0 ) then xy = 0 "
2) "for each pair of real numbers x,y, if (xy = 0) then x = 0 or y = 0.

To prove an "if.... then...." statement, we are allowed to assume the "if..." part is true.

To prove statement 1, you need to refer to some assumption or theorem that tells you that 0 times another number is 0. You can assume x = 0 or y = 0. You can break that into two cases if you need to: Case 1) x = 0 Case 2) y = 0 . (The case of both x= 0 and y = 0 is treated by either one of those cases. The "or" in mathematics is not exclusive.)

To prove statement 2. Dickfore suggests you use the contrapositive. The contrapostive of a statement of the form "If A then B" is the statement "If not B then not A". These two forms of a statement are logically equivalent, so if you can prove the contrapositive of a statement, it proves the original statement.

The contrapostive of the if-then part of the statement "for each pair of real numbers, if (xy = 0) then x = 0 or y = 0" is "for each pair of real numbers if( it is not true that (x = 0 or y = 0) then it is not true that xy = 0)"

To prove this we may assume "it is not true that (x = 0 or y = 0)". This is equivalent to assuming "x is not zero and y is not zero" by one of DeMorgans laws of logic. This law states that a statement of the form "not(A or B)" is equivalent to the statement "(not A) and (not B)".

Dickfore did not mention this, but his suggested method of proof employs proof by contradiction. In this method of proof we assume the negation of the then-part of the if-then statement and show that this implies a statement that contradicts the if-part or some other assumption or theorem.

Our if-part is equivalent to the statement that "x is not zero and y is not zero".
The negation of the then-part is "not (it is not true that xy = 0)" , i.e. "xy is 0".

In summary, we may take these facts as given:
x is not zero
y is not zero
xy = 0

You need to find the assumption or theorem in your text that states the fact that a non-zero number has a multiplicative inverse. Using that as justification, we may multiply the equation xy = 0 on both sides by the multiplicative inverse of x. This gives the new equation (1)y = 0 so y = 0. This contradicts the fact that y is not 0. Hence we have accomplished proof by contradiction.

--------------

Many people write proofs without stating that they are using DeMorgan's laws or contraposition or proof by contradiction or other principles of logic. Many people acquire the skill of using these rules simply by reading proofs in books. They come to think of such laws as "common sense". I don't know whether you teacher expects explain your use of logic in detail or whether you are allowed to proceed with "common sense". It might help if you glance at a book on elementary symbolic logic. I don't think you need go deeply into it in order to get the hang of things like contrapostiives and DeMorgan's laws.

8. Jan 21, 2012

### Rayquesto

"then you were trying to deduce from the statement that you are attempting to prove, which wouldn't make sense.

To prove a statment with an "if and ony if" condition in it, it is often simplest to break this into two implications. So "for each pair of real numbers x,y , xy = 0 if and only if x = 0 or y = 0" can be proved by proving these two statements separately:

1) "for each pair of real numbers x,y, if (x = 0 or y = 0 ) then xy = 0 "
2) "for each pair of real numbers x,y, if (xy = 0) then x = 0 or y = 0.

To prove an "if.... then...." statement, we are allowed to assume the "if..." part is true.

To prove statement 1, you need to refer to some assumption or theorem that tells you that 0 times another number is 0. You can assume x = 0 or y = 0. You can break that into two cases if you need to: Case 1) x = 0 Case 2) y = 0 . (The case of both x= 0 and y = 0 is treated by either one of those cases. The "or" in mathematics is not exclusive.)"

Thanks for the response. So, based on what I just tried interpreting, I'm going to do a few things. First off, I want to get rid of this statement:'(x=0 or y=0) <=> xy=0 (What is the postulate defining the operation I just made?),' since I realize that that's pretty much what you are proving. It's like looking at something at the opposite yet equal side of a spectrum or 180 degrees. See what I'm saying. So, I'm getting rid of it.

now, I'm going to switch the two sets I used for the last operation:

xy=0 => for all x,y,0€ℝ:[x<y => (x + 0)< (y + 0) or y<x => (y + 0)<(x+0)] by the translation invariance of order (does this statement make sense to you? I'm reading it and it does to me.)

Now, I will show that the theorem that proves 0 times any other number is 0 (we call it the multiplication by 0 theorem which he proved in class and I know how to prove, but is unnecessary)

The Multiplication by 0 theorem states that x€ℝ => 0x = 0

So, now,

y>0[x€ℝ => 0x = 0] => x>0[y€ℝ => 0y = 0] => for all x,y,0€ℝ:[x<y => (x + 0)< (y + 0) or y<x => (y + 0)<(x+0)]

I have a feeling my interpretation of your suggestive logic was very off, but I'm still trying to get this. So, please bare with my intermediate level of this logic stuff. I hardly knew about these symbols until a few days ago, basically. I suppose I was familiar with the symbols to some degree, but hardly. I'm surprised I'm actually getting it enough to write statements, regardless of genuine logic. Thanks!

9. Jan 21, 2012

### Stephen Tashi

I interpret you notation to mean:

(For all real number x,y) if xy = 0 then [ for all real numbers x,y, if x < y then (x+0)< (y+0) ] or if (y < x) then (y+0) < (x+0)].

That may be a true statement, but it makes no sense to me as a step in a proof, because you don't need to say "if (xy = 0)" to deduce the part that says "for all all real numbers x,y, if (x < y) then ..... " etc. As you indicated, that part of the statement follows from what you call "the translation invariance of order".

What do you mean by "the theorem"? Are you saying that you are proving the "multiplication by 0 theorem? If this is already known to be a theorem, you don't have to try to prove it.

Try writing in words what you wrote in symbols. To me, what you wrote doesn't make much sense.

For example "y > 0{ if (x in R) then (0x = 0)". You don't need the fact that (y > 0) to conclude that "if (x in R) then (0X= 0)" if we are assuming the theorem that says 0 times any number is 0.

If you are trying to prove that theorem, the theorem isn' proved by the condition "if (y > 0)".

Unless you are studying a very advanced form of symbolic logic, you cannot write this proof completely in symbols. My advice is to think about the proof in words and then translate some of your verbal statements to symbols. Don't try to translate all of them to symbols.

You better take my advice about breaking the proof of a statement with a "<=>" condition into two statements, one with "<=" and one with "=>".

The proof I indicated makes no use of "the translation invariance of order".

It looks you are approaching this work as person would "prove" a trigonometric identity. In that type of problem, you write down a symbolic expression and then manipulate individual parts of it. This is not a good way to write most proofs and the so-called "proofs" people create for trigonometric identities are not real proofs.

10. Jan 21, 2012

### Rayquesto

Yes! You are probably correct on that idea. I am so much more used to proving trigonometric identities, since I practiced my pre-calculus a lot and this is a calculus 1 class, that I have approached this form of mathematics the same way I approach trigonometric proofs. Most of the time, It's about substitution from identities and also algebraic operation. What type of approach should one go for trigonometric proofs over other proofs? It sounds like I'm truly not grasping the firm idea of proving statements to be correct.

11. Jan 21, 2012

### Dickfore

Starting from the axioms of addition and multiplication, can you prove the following:
$$\left( \forall x \in \mathbb{R} \right) 0 \cdot x = 0$$

12. Jan 21, 2012

### Rayquesto

0x=0
<=> 0x - 0x same as 0
<=> (0+0)x - 0x substitution
<=> 0x + (0x - 0x) associative law
<=> 0x transitive property of equality

13. Jan 22, 2012

### Stephen Tashi

If you are trying to prove (as Dickfore suggested) that "for each real number x, (x)(0) = 0", you can't begin the proof by assuming "0x = 0".

As you said, you are copying the method used on trigonometric identities, where the tradition is to begin by writing down the identity to be proven and then manipulate it. But trigonometric identities are proven by a rote process that is not an actual proof. Forget about using that approach.

Also, don't bother to prove "for each real number x, (x)(0) = 0" if you your book has already established that as a theorem. Just use the theorem as justification if you need that fact.

Think about the problem in ordinary language. If you were having a political argument and you said "If Grod Fressington is elected, he'll hire his brother-in-law as the city attorney", you couldn't convincing argue you case by beginning with "Let's assume if Grod Fresston is elected then he will hire his brother-in-law as city attorney."

The pattern for proving the statement "If A then B" directly goes like this:

Assume A
Show B is true
----------------------
Conclude "if A then B".

If you want to prove "if x is a real number then (x)(0) = 0" then you are only allowed:

Assume x is a real number.

You must show x(0) = 0 using that assumption plus all the theorems and assumptions already established

We can't tell you exactly how to do this proof in a manner that is consistent with your textbook unless we know exactly how your textbook states the assumptons and theorems that are relevant. (There are many different ways of formulating the the ideas needed to develop the properties of real number.)

As I said, before, don't worry about proving this if your textbook as already presented this as a theorem.

14. Jan 22, 2012

### Dickfore

He didn't assume $0 \cdot x = 0$. He just wrote it in the first line. All the other steps are, in fact, a correct way of proving it. So, you are done with the
$$x = 0 \vee y = 0 \Rightarrow x \cdot y = 0$$

Now, you are left with proving the direction:
$$x \cdot y = 0 \Rightarrow x = 0 \vee y = 0$$
I suggest you use the method "reductio ad absurdum".

Your implication is equivalent to the following statement:
$$\neg (x \cdot y = 0) \vee \left( x = 0 \vee y = 0 \right)$$
and its negation, by De Morgan's rules, is:
$$\neg \left[ \neg (x \cdot y = 0) \vee \left( x = 0 \vee y = 0 \right) \right]$$
$$x \cdot y = 0 \wedge \neg (x = 0) \wedge \neg (y = 0)$$
$$x \cdot y = 0 \wedge x \neq 0 \wedge y \neq 0$$
Can you prove this statement incorrect?

Last edited: Jan 22, 2012
15. Jan 22, 2012

### Stephen Tashi

I disagree. For example, by that interpretation, the first step is then:

"<=> 0x - 0x same as 0"

Which in words (considering the previous line), says:

"( for each real number x, 0x = 0) if and only if ( 0x -0x = 0)"

There isn't any reason given for this step. We'd have to know what Rayquesto's textbook has defined and proven up to this point in order to advise him what reason to give.

It isn't clear that Rayquesto needs to prove (0)(x) = 0. Prior to assigning this proof, his textbook may have defined a zero to be a number such at for any real number x, (x)(0) = 0 and assumed or proved that zero is unique ( thus establishing the fact that the symbol "0" can be used unambiguously). If that work has already been done, then it would not be appropriate to re-do it in writing the proof for the theorem in the original post.

I agree with the content of the rest of your post, but I think Rayquesto would do better to think in verbal terms unless his class is employing that level of symbolism already. (We are all being sloppy about the quantifiers that could be included in symbolic expressions to make them correct.)

16. Jan 22, 2012

### Dickfore

And, if you went through the rest of the argument, you would see that he never used that "assumption". I agree that he made a sloppy notation, but his logic deduction is solid.

I am interpreting what has been written. It was asked to prove an equivalence. This means you need to prove both directions of the implication.

I don't think verbal terms are any easier to understand than shorthand notation. For example, no one uses "necessary and sufficient condition" everyday talk without being a student of logic. Also, De Morgan's laws as well as the identity $p \Rightarrow q \equiv \neg p \vee q$ are rules of (formal, mathematical) logic, and not language and I don't see any "common-sense way" to deduce their general validity. They are crucial in obtaining a negation of the expression he is trying to verify. Now, the expression is one step away from demonstrating it is a contradiction (I already have given a hint how to do it in a previous post).

17. Jan 22, 2012

### Rayquesto

I just received a very large amount of text from you tutors and I thank you deeply for this. I will read it and try to understand it, but I feel compelled to state something. My calculus 1 book does not teach us the language of logic or go over it at all. My teacher, however, gave us a 50 page packet that includes information pertaining to this language of mathematics. He did not cover De Morgan's law or any of that. He covers general concepts and general applications so we can take calculus 1 into a different approach. Technically, I don't need to know this. I just have a strong curious nature to me that leads me to feel compelled to learn certain material even if it isn't necessary. So, I'm going to read what you guys wrote and I'll get back to you soon. Thanks for the help!


text:
# Is there a meaningful distinction between “inclusion” and “monomorphism”?

The title pretty much says it all. As far as I can tell, the terms "inclusion" and "monomorphism" are equivalent. (Ditto for $\hookrightarrow$ and $\rightarrowtail$.) Is this the case?

Edit: Judging by the answer and comments I've received so far, I realize I'm working with definitions that may not be universal. My source is Mitchell's Theory of Categories (1965). On p. 6, he writes (emphasis in the original):

If $\alpha:A' \to A$ is a monomorphism, we shall call $A'$ a subobject of $A$, and we shall refer to $\alpha$ as the inclusion of $A'$ in $A$.

As far as this quote goes, I'd say that "inclusion" and "monomorphism" are essentially synonymous, differing only in the fine grammatical structures of the English phrases that can be formed with each; e.g. "$f$ is the inclusion of $X$ in $Y$" vs. "$f$ is a monomorphism from X to Y". The fact that Mitchell uses the definite article for "the inclusion", even though it was defined by the existence of "a monomorphism" (indefinite article), is a bit disconcerting, but Mitchell addresses this concern a few lines later:

...it is important to remember that in general there is more than one monomorphism from $A'$ to $A$, and that whenever we speak of $A'$ as a subobject of $A$ we shall be referring to a specific monomorphism $\alpha$.

On the basis of this additional clarification, I'd "fill out" Mitchell's definition of "inclusion" slightly: if $A'$ is a subobject of $A$ (as defined above), then the inclusion of $A'$ in $A$ is

the monomorphism $\alpha$ that our designation of $A'$ as a subobject of $A$ is based on. (✽)

In fact, I imagine that, given this context, one could use a phrase like "the monomorphism of $A'$ to $A$" as shorthand for the formula (✽) I just proposed. If so, "inclusion" and "monomorphism" would be quite synonymous. (At the very least, there is a bijection between the class of monomorphisms and the class of inclusions.)

Due to these considerations, I was puzzled by random comments I'd found online that implied some meaningful distinction between these two terms (e.g. statement of preference as to which type of arrows to assign in diagrams to inclusions and which to monomorphisms), and this prompted my original question.

I am now particularly interested in learning of examples of monomorphisms that are not inclusions, as alluded to by Zhen Lin in one of the comments below. This will give me a clue about the various non-synonymous meanings given to the term "inclusion".

I regret that this clarification of my original post is many times longer than the original, but after having erred perhaps on the side of brevity, now I'm making up for it by going verbose.

-
Well, "monomorphism" is a term from category theory. In certain categories the idea of "inclusion" also makes sense (because of the inner nature of the objects there). In some cases these two notions coincide, but in other cases they do not coincide. –  GEdgar Sep 13 '11 at 14:35
Not every monomorphism is an inclusion (e.g. in the category of topological spaces), but I certainly hope every inclusion is a monomorphism... –  Zhen Lin Sep 13 '11 at 14:40
@kjo: Regarding your elaborated post, you may some of the discussion here about subobjects and monomorphisms useful. (In particular, Mitchell's definition of subobject is not standard.) –  Zhen Lin Sep 14 '11 at 3:01
It would be very much easier to answer your query if you gave a few examples of categories you're comfortable with. From your previous questions I'm not entirely sure if you have a good stock of examples to draw from. So: could you give a short list of categories you like to think about? E.g. topological spaces, modules, groups, rings, fields, etc? –  t.b. Sep 14 '11 at 3:02
If using “inclusion” in Mitchell's sense, these terms are different from the formal point of view. Monomorphism is a property of morphisms. Inclusion is a component of a subobject, i.e. the term “inclusion” extracts a morphism from any subobject. These terms have very different nature. –  beroal Sep 17 '11 at 11:19

Monomorphism is a categorical notion. Sere this wikipedia article. Its characterising property is left-cancellation.

The word inclusion doesn't have a strictly defined meaning. It often refers to the natural injection which is associated with a subobject, like a subgroup of a group or a subspace of a topological space or a vector space. This is in accordance with this wikipedia article.

As for the notation: I think both $\hookrightarrow$ and $\rightarrowtail$ are used to denote both inclusions and monomorphisms. Personally, I'd prefer $\hookrightarrow$ for inclusions and $\rightarrowtail$ for monomorphisms.

-

As Rasmus points out, monomorphism is a purely categorial notion.

On the other hand, inclusion is "multi-paradigmatic" - its designed to bridge the classical and categorial approaches. In particular, if $A$ and $B$ are sets and $A \subseteq B$, then according to category theory, it would not be technically correct to say that $A$ is a subobject of $B$, because we don't know how $A$ "sits inside" $B$ until we specify an injection $\phi : A \rightarrow B$.

The idea behind the inclusion map is that there's an obvious choice of $\phi$. Namely, we can choose the unique $\phi : A \rightarrow B$ such that for all $a \in A$ it holds that $\phi(a)=a$, and we call this the inclusion of $A$ in $B$. In this way, we see that subsets in the classical sense of the word correspond naturally to subsets in the categorial sense of the word.

The same goes for subgroups, subrings etc.

-


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## Elementary Statistics (12th Edition)

8.104 km < $\mu$ < 11.512 km Because the sample size is greater than 30, the confidence interval yields a reasonable estimate of m, even though the data appear to be from a population that is not normally distributed.
8.104 km < $\mu$ < 11.512 km Because the sample size is greater than 30, the confidence interval yields a reasonable estimate of m, even though the data appear to be from a population that is not normally distributed.


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 1 - 2 / 21 1.Decomposition of skew-morphisms of cyclic groupsIstván Kovács, Roman Nedela, 2011, original scientific articleAbstract: A skew-morphism of a group ▫$H$▫ is a permutation ▫$\sigma$▫ of its elements fixing the identity such that for every ▫$x, y \in H$▫ there exists an integer ▫$k$▫ such that ▫$\sigma (xy) = \sigma (x)\sigma k(y)$▫. It follows that group automorphisms are particular skew-morphisms. Skew-morphisms appear naturally in investigations of maps on surfaces with high degree of symmetry, namely, they are closely related to regular Cayley maps and to regular embeddings of the complete bipartite graphs. The aim of this paper is to investigate skew-morphisms of cyclic groups in the context of the associated Schur rings. We prove the following decomposition theorem about skew-morphisms of cyclic groups ▫$\mathbb Z_n$▫: if ▫$n = n_{1}n_{2}$▫ such that ▫$(n_{1}n_{2}) = 1$▫, and ▫$(n_{1}, \varphi (n_{2})) = (\varphi (n_{1}), n_{2}) = 1$▫ (▫$\varphi$▫ denotes Euler's function) then all skew-morphisms ▫$\sigma$▫ of ▫$\mathbb Z_n$▫ are obtained as ▫$\sigma = \sigma_1 \times \sigma_2$▫, where ▫$\sigma_i$▫ are skew-morphisms of ▫$\mathbb Z_{n_i}, \; i = 1, 2$▫. As a consequence we obtain the following result: All skew-morphisms of ▫$\mathbb Z_n$▫ are automorphisms of ▫$\mathbb Z_n$▫ if and only if ▫$n = 4$▫ or ▫$(n, \varphi(n)) = 1$▫.Found in: ključnih besedahSummary of found: ...groups in the context of the associated Schur rings. We prove the following decomposition theorem...Keywords: cyclic group, permutation group, skew-morphism, Schur ringPublished: 15.10.2013; Views: 1935; Downloads: 55 Full text (0,00 KB) 2.CI-property for decomposable Schur rings over an Abelian groupIstván Kovács, Grigory Ryabov, 2019, original scientific articleFound in: ključnih besedahKeywords: isomorphism, CI-group, Schur ringPublished: 17.12.2018; Views: 216; Downloads: 45 Full text (0,00 KB)
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# How can you implement a two-way ANOVA with nesting in R or SPSS?

I currently am conducting a study where I have three variables: one that is binary, and two numerical ratio variables. Each of the subjects in my study has values for each of the three variables.

Variables:

• Condition (binary): Values 0 and 1
• Pre (ratio)
• Post (ratio)

I want to test if there is a significant difference between the the pre and post variables of the 0 control group and the 1 experimental group. Both groups have 103 subjects. The data meet all typical ANOVA assumptions such as normality and the like. I was thinking of nesting the variables as follows and then running a two-way ANOVA.

Variable 1 is exposed / not and variable 2 is pre / post. People are nested in Variable 1 (meaning that each person gives both pre and post information for either exposed or not exposed conditions)

Would this be the correct way to approach this problem? Also how would I implement this statistical analysis, preferably in R or SPSS?

Pre-post designs are quite common and the standard method is to use the pre-measure as a control variable and the post-measure as the response with the binary variable as the covariate.

This is quite simple:

m0 <- lm(Post ~ Pre, data=dat)
m1 <- lm(Post ~ Pre+Condition, data=dat)
anova(m1,m0) # test for difference in mean(Post) after control for Pre
m2 <- lm(Post ~ Pre*Condition, data=dat)
anova(m2, m1) # test for interaction: different slopes in the two Conditions

If this had been with repeated measures within subject (i.e, pre-post measures in each Condition, you would need to specify the error structure in the aov function: You could compare this model in R where the errors for each Condition are nested in subject-ID:

 m1 <- aov(post ~ pre + Condition+Error(subjID/Condition), data=dat)

Ratio variables are notorious for being poorly behaved statistically (i.e., often having significant skew and blowing up as the denominator goes near zero.) However, if the denominator is far from zero, you should not make the mistake of looking at the raw distributions and deciding that the data fail the assumptions needed for validity of linear methods. You need to look at the residuals before conducting any test. (And even then, there is some doubt whether minor departures from normality invalidate inferences.) The help page for aov warns you to use orthogonal contrasts. There was an article in R-News several years ago on multi-variate (as opposed to multi-variable) methods and this would provide further options for model comparisons and measures of sphericity.

Another option is offered by the Anova function in John Fox's car package.

Another approach is to analyze the pairwise differences, but that approach has flaws.

• By "ratio variable", I suspect she / he means a variable measured on a ratio scale. – gung Jul 9 '13 at 1:33
• Do you really need to specify an error structure here? I think there is only 1 post measurement for each subject. Given that you are using the pre measure as a covariate (& I agree w/ you about that) & not as a 2nd response variable, I'm not sure if the Error(subjID/Condition) part is necessary. – gung Jul 9 '13 at 1:41
• That's what I thought initially and had even composed an answer using lm(). You may still be correct. – DWin Jul 9 '13 at 1:47
• @gung I apologize if I was unclear. I did mean a variable measured on a ratio scale, not a ratio variable. Also there is only one "pre" and one "post" measurement for each subject. – ikemblem Jul 9 '13 at 2:01
• OK. I'll post code appropiate to that data setup. – DWin Jul 9 '13 at 6:40

The standard way of handling repeated measures under the linear regression framework (which encompasses ANOVAs) is to reduce the repeated measures into a series of difference scores, which represent the contrasts of interest. So, in your example, you might do what is essentially a t-test comparing the difference from pre to post in the control and the experimental conditions. You can find the details of how to implement this in R in my answer to this question.

Edit:

As described by DWin, an alternative way to analyze these data is to use the baseline measurement as a covariate in your model (i.e., basically just enter the pre-test measurement as a predictor in your linear model). As described by Van Breukelen (2006), this method will have more power than the difference score approach if you have (or expect) no pre-test differences on your dependent variable. If you do have pre-test differences on your dependent variable, the covariate approach will introduce bias into your results, and the difference score approach will therefore be preferable.


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# American Institute of Mathematical Sciences

November  2007, 1(4): 489-507. doi: 10.3934/amc.2007.1.489

## Public key cryptography based on semigroup actions

 1 Department of Mathematics, University of Zürich, Winterthurerstr 190, CH-8057 Zürich, Switzerland 2 Department of Mathematics and Statistics, Texas Tech University, Lubbock, TX 79409-1042, United States 3 Institut für Mathematik, Universität Zürich, Zürich, CH-8057

Received  June 2007 Revised  October 2007 Published  October 2007

A generalization of the original Diffie-Hellman key exchange in $(\mathbb Z$∕$p\mathbb Z)$* found a new depth when Miller [27] and Koblitz [16] suggested that such a protocol could be used with the group over an elliptic curve. In this paper, we propose a further vast generalization where abelian semigroups act on finite sets. We define a Diffie-Hellman key exchange in this setting and we illustrate how to build interesting semigroup actions using finite (simple) semirings. The practicality of the proposed extensions rely on the orbit sizes of the semigroup actions and at this point it is an open question how to compute the sizes of these orbits in general and also if there exists a square root attack in general.
In Section 5 a concrete practical semigroup action built from simple semirings is presented. It will require further research to analyse this system.
Citation: Gérard Maze, Chris Monico, Joachim Rosenthal. Public key cryptography based on semigroup actions. Advances in Mathematics of Communications, 2007, 1 (4) : 489-507. doi: 10.3934/amc.2007.1.489
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2019 Impact Factor: 0.734


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# Differences between homeomorphic and topologically conjugate dynamical systems.

I have begun studying homeomorphisms between dynamical systems. I have a few related questions about homeomorphic and topologically conjugate dynamical systems.

Questions

• How can I determine if a homeomorphism is orientation presevering?
• If two dynamical systems are Topologically Conjugate (that is $f \circ h=h \circ g$), does this imply that $h$ is an orientation preserving homeomorphism? (Answered see comment below this post)
• What properties are preserved between two topologically conjugate dynamical systems, compared to two dynamical systems which are only homeomorphic to each other?

Definitions

$f:X \rightarrow X$, $g:Y \rightarrow Y$ and $h:Y \rightarrow X$ are continuous functions on smooth orientable manifolds, $X$ and $Y$.

• Topologically Conjugate: $f \circ h=h \circ g$ and $h$ is homeomorphsim.
• Homeomorphism
• What it means for a manifold to be orientable.

Notes

• Are you assuming that $X$ and $Y$ are orientable manifolds? Otherwise, talking about orientation preserving doesn't make sense. In any case, it feels like you're maybe asking these questions a bit too early in your study, because you very quickly encounter invariants of topological conjugacy soon after the term is introduced - similarly with homeomorphisms (which is a purely topological notion, and is divorced from any dynamics). – Dan Rust Jul 25 '18 at 12:51
• For a counterexample to your second bullet point, consider the circle $X = [0,1]/{\sim}$ with clockwise rotation by some number $t$, so $f \colon X \to X \colon x \mapsto x+t$ and a circle $Y$ with anti-clockwise rotation $x \mapsto x-t$. Clearly, the two systems are topologically conjugate by just mapping $x \mapsto -x$, but this is not an orientation preserving map. – Dan Rust Jul 25 '18 at 12:51


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# Differentiation of the value function in Burdett Mortensen (1998)

I'm currently making my way through Burdett and Mortensen's classic paper of on the job search. What should be an easy task of finding an expression for the reservation wage is made slightly more complicated by the presence of the max operator. We are faced with the following Bellman equation for the value of a job paying a wage $w$. The bellman equations are standard. The value of a job paying $w$ consists of the wage $w$ plus the expected gain from searching and finding a better job discounted by the probability a job offer comes along $\lambda_1$ plus the loss of due to becoming unemployed when the job is destroyed at rate $\delta$. The value of unemployment $V_0$ consists of unemployment benefit $b$ plus the expected gain from becoming employed discounted by the probability an offer comes along $\lambda_0$. Note the probability an offer is made is different depending on whether someone is already employed or unemployed. The distribution of offers is given by $F$ $$rV_1(w)=w+\lambda_1\bigg[\int \max\{V_1(w),V_1(\tilde{x})\}-V_1(w)\bigg]\;dF(\tilde{x})+\delta [V_0-V_1(w)]$$ $$rV_0=b+\lambda_0 \bigg[\int \max\{V_0,V_1(\tilde{x})\}\;dF(\tilde{x})-V_0\bigg]$$ Since $V_1(w)$ is increasing in $w$ and $V_0$ is independent of it we know a reservation wage exists such that if $w>R\implies V_1(w)>V_0$, $w<R\implies V_1(w)<V_0$ and $V_1(R)=V_0$. Standard arguments (integration by parts) shows that $$R-b=(\lambda_0-\lambda_1)\int_R^\infty V_1'(\tilde{x})[1-F(\tilde{x})]\;d\tilde{x}$$ from here I would like to take the derivative of the first equation and solve for $V_1'(w)$. However If I use Leibniz integration rule I need the integrand to be differentiable. The max of two continuous functions is usually not differentiable where they are equal so I have a problem. If I assume that I integrate over all $\tilde{x}\geq w$ then $V_1(\tilde{x})\geq V_1(w)$ (wage offers that will be induce a worker to switch jobs) and the result follows by Leibniz rule. But there are wages in the distribution that won't be accepted and this derivative wont hold. The derivative is $$V'(\tilde{x})=\frac{1}{r+\delta+\lambda_1(1-F(\tilde{x}))}$$ I imagine I'm missing something but I'm not sure what. If anyone could give me any advice I would really appreciate it.

When you take the integral of a $\max_{\{\cdot\}}$ operator, I think you have to split the integral into two separate integrals with different supports on them.
Here is my attempt, where I assume an absolute upper limit on the support of $F$, $F(\overline{w})=1$, for simplicity.
Rewrite the first equation as $$rV_1(w)= w+\lambda_1\int_w^{\overline{w}}V_1(\tilde{x})dF(\tilde{x}) +\underbrace{\lambda_1\int_0^{w}V_1(w)dF(\tilde{x})}_{I} -\lambda_1\int_0^{\overline{w}}V_1(w)dF(\tilde{x}) +\delta[V_0-V_1(w)] \ ,$$ whereby $$-\lambda_1\int_0^{\overline{w}}V_1(w)dF(\tilde{x})= -\lambda_1\int_w^{\overline{w}}V_1(w)dF(\tilde{x}) -\underbrace{\lambda_1\int_0^{w}V_1(w)dF(\tilde{x})}_{II} \ .$$
The terms $I$ and $II$ cancel out, so that arranging gives $$(\delta +r)V_1(w)= w+\lambda_1\int_w^{\overline{w}}[V_1(\tilde{x})-V_1(w)]dF(\tilde{x}) +\delta V_0 \ .$$ If we apply Leibniz' rule know, we get $$(\delta +r)V_1'(w)= 1-\lambda_1\int_w^{\overline{w}}V_1'(w)dF(\tilde{x})=1-\lambda_1V_1'(w)[1-F(w)] \ ,$$ whereby the last equality follows from $F(\overline{w})=1$. Solving for $V_1'(w)$ gives the desired solution.


text:
470 views

If $A_i = \{-i, \dots , -2, -1, 0, 1, 2, \dots , i \}$ then $\cup_{i=1}^\infty A_i$ is

1. Z
2. Q
3. R
4. C

recategorized | 470 views

Option: A
Option (A) - as this is the union of all Integers from -$\infty$ to +$\infty$


text:
# Solving 2D convection-conduction equation via using Fourier integral transform: the disappearance of the convection term?(with code)

I am currently solving a 2D convection-conduction equation. The convection is only working on the x direction. The governing equation and its associated conditions are given as

where T is the temperature, x and z are the spatial distances, t is the time, and v is the velocity.

Firstly, I applied the Laplace transform method to eliminate t. These equations then become

where T with a bar represents the function in the Laplace domain and s denotes the Laplace parameter.

Later, I used the function called FourierTransform in MMA12 to transform the aforementioned equations to the Laplace-Fourier domain. The results are obtained as

in which T with the double bars represents the function in the Laplace-Fourier domain, a is the Fourier parameter, i is the imaginary unit, and 𝛿 is the dirac delta function.

After applying the DSolve function, I got this result

It can be observed that the convection term is disappeared (i.e., v = 0). Am I missing something here? Or is the convection term not important in this PDE? Thank you for the help.

The associated code I used to obtain the function in Fourier domain is

FourierTransform[T''[x], x, a]

gives

-a^2 FourierTransform[T[x], x, a]

The convection term with applying Fourier transform

FourierTransform[-v*T'[x], x, a]

gives

I a v FourierTransform[T[x], x, a]

and

FourierTransform[1/s, x, a]

gives

(Sqrt[2 \[Pi]] DiracDelta[a])/s

Applying the DSolve

DSolve[{s*T[z] == -a^2*T[z] + I a v*T[z] + T''[z],
T[0] == (Sqrt[2 \[Pi]] DiracDelta[a])/s, T[1] == 0}, {T[z]},
z] // Simplify

one gives

T[z] -> (E^(-Sqrt[s] z) (E^(2 Sqrt[s]) - E^(2 Sqrt[s] z)) Sqrt[
2 \[Pi]] DiracDelta[a])/((-1 + E^(2 Sqrt[s])) s)

• Can you show your code to obtain these? LaTeX is good, but if we can see the code you used it is really helpful in helping us help you :) which is the convection term originally also? If it is time dependent that may be your answer, as you eliminated t—but if it is a result of DSolve, then one could surmise that it disappeared due to it being dx/dt? I’m not as learned on this as I should be, admittedly. Aug 21 '19 at 3:43
• @CATrevillian hi, I added the code I used to obtain the result of the ODE in the Laplace domain. The convection term is the function multiplied by v. It seems that the use of the Laplace transform does not lead to such a result. The disappearance of v is confusing me. Aug 21 '19 at 3:56
• s is signal yeah? Is that not like the velocity? But also what about your T[0] not having the other terms in it? Aug 21 '19 at 4:06
• @CATrevillian Actually, s in the model is a Laplace parameter which can be found after applying the Laplace transform. Also, the convection term is -v dT/dx in the equation. T[0]=0 means the temperature in the investigated domain is zero initially. Aug 21 '19 at 4:26
• Thank you for the clarifications, in your final step, what you input into DSolve, the T(x=0) term does not include all of the items you list prior, could this be the cause? Aug 21 '19 at 4:30

Let me extend my comment to an answer. I think the solution is correct. It's the boundary conditions at infinity that filter out influence of v. I'm not that familiar with the theory of integral transforms, but only functions that satisfy certain criterion can be FourierTransformed, which isn't satisfied by solution of PDE involving convective terms in many cases. Somewhat related:

https://math.stackexchange.com/q/2084704/58219

To support my point from another side, let's solve the problem with a slightly different method. First write down the equation:

With[{T = T[t, x, z]}, eq = D[T, t] == D[T, x, x] - v D[T, x] + D[T, z, z];
ic = T == 0 /. t -> 0;
bc = {T == 1 /. z -> 0, T == 0 /. z -> 1}]

Then impose Laplace transform just as you've done:

{teq, tbc} =
LaplaceTransform[{eq, bc}, t, s] /. Rule @@ ic /.
HoldPattern@LaplaceTransform[a_, __] :> a

(* {s*T[t, x, z] == Derivative[0, 0, 2][T][t, x, z] - v*Derivative[0, 1, 0][T][t, x, z] +
Derivative[0, 2, 0][T][t, x, z], {T[t, x, 0] == 1/s, T[t, x, 1] == 0}} *)

I've made replacement HoldPattern@LaplaceTransform[a_, __] :> a because LaplaceTransform will cause trouble in subsequent programming. Just keep in mind T[t, x, z] in teq and tbc actually means Laplace transform of $$T$$.

Next, we use finite Fourier sine transform to eliminate derivative of z

tteq = finiteFourierSinTransform[teq, {z, 0, 1}, n] /. Rule @@@ tbc /.
HoldPattern@finiteFourierSinTransform[a_, __] :> a

(*
s*T[t, x, z] == (-n)*Pi*(-(1/s) + n*Pi*T[t, x, z]) - v*Derivative[0, 1, 0][T][t, x, z] +
Derivative[0, 2, 0][T][t, x, z]
*)

Similarly, I've made replacement HoldPattern@finiteFourierSinTransform[a_, __] :> a to avoid trouble in subsequent programming, keep in mind T[t, x, z] is actually finite Fourier sine transform of Laplace transform of $$T$$ in tteq.

Finally, directly solve tteq with DSolve:

DSolve[tteq, T[t, x, z], x][[1]]
(*
{T[t, x, z] -> -((n π)/((-n^2 π^2 - s) s)) +
E^(1/2 (v - Sqrt[4 n^2 π^2 + 4 s + v^2]) x) C[1] +
E^(1/2 (v + Sqrt[4 n^2 π^2 + 4 s + v^2]) x) C[2]}
*)

It's clear that C[…] can only be 0 if the boundary conditions at infinity are satisfied i.e. v doesn't play a role in the solution.

• This is the result of incompatible boundary conditions and initial data. It is necessary to solve the problem using FEM and see what happens when the scale of the region increases along x. Aug 21 '19 at 13:02
• @Alex According to my (limited) experience, integral transform should be able to find the solution in the sense of limitation when b.c.s are inconsistent, here is an example: mathematica.stackexchange.com/q/127081/1871 Aug 21 '19 at 14:11
• After applying the Fourier sine transform to dT/dx, it will become the function in the Fourier cosine domain. Can we solve the ODE consisting of the function of T in the Fourier since and cosines domains simultaneously? Aug 27 '19 at 7:46
• @LingLong I'm not aware of such technique. Also, notice imposing Fourier sine transform is amount to implicitly setting b.c. at infinity i.e. the system will be over-determined if you use it in $t$ direction. Aug 27 '19 at 8:30


text:
# Integral of Characteristic Function

## Theorem

Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

Let $E \in \Sigma$ be a measurable set, and let $\chi_E: X \to \R$ be its characteristic function.

Then $I_\mu \left({\chi_E}\right) = \mu \left({E}\right)$, where $I_\mu \left({\chi_E}\right)$ is the $\mu$-integral of $\chi_E$.

### Corollary

$\displaystyle \int \chi_E \, \mathrm d\mu = \mu \left({E}\right)$

where the integral sign denotes the $\mu$-integral of $\chi_E$.

## Proof

Let $a_1 = 1$ and $E_1 = E$.

As in the definition of standard representation, denote $a_0 = 0$ and $E_0 = X \setminus E_1$.

Then for $x \in X$, we have:

$\chi_E \left({x}\right) = 0 \cdot \chi_{E_0} \left({x}\right) + 1 \cdot \chi_{E_1} \left({x}\right)$

since $E_1 = E$.

Hence $\chi_E = a_0 \chi_{E_0} + a_1 \chi_{E_1}$ is a standard representation for $\chi_E$.

Thus, by definition of $\mu$-integral:

$\displaystyle \int \chi_E \, \mathrm d \mu = a_0 \mu \left({E_0}\right) + a_1 \mu \left({E_1}\right) = \mu \left({E}\right)$

as desired.

$\blacksquare$


text:
# Transforming Poisson

Let $N, Y_n,n\in\mathbb{N}$ be independent random variables, with $N \sim P(\lambda), \lambda \lt \infty$ and $\mathbb{P}(Y_n = j)=p_j$, for $j=1,\dots,k$ and all $n$. Set $$N_j = \sum_{n=1}^{N}\mathbb{1}(Y_n=j).$$ Show that $N_1,\dots,N_k$ are independent random variables with $N_j \sim P(\lambda p_j)$ for all $j$.

For distribution I used that $N_j | N \sim Bin(N, p_j)$ and so $$\mathbb{P}(N_j=k)=\sum_{N=k}^{\infty}{{N}\choose{k}}p_j^k(1-p_j)^k\frac{\lambda^N}{N!}e^{-\lambda}=\dots=\frac{(p_j\lambda)^k}{k!}e^{-\lambda p_j}$$ Need help with independence.

-
Let $T$, $T_1$, $\dots$, $T_k$ be indeterminates. Then the probability generating function of $N$ is $$f(T):={\Bbb E}[T^N]=e^{\lambda T}.$$ Now, if we condition on $N=n$, the joint distribution of the $N_j$'s will be multinomial and given by the coefficients of $(\sum_j p_j T_j)^n$. Therefore, setting $T=\sum_j p_j T_j$ in $f(T)$ will give the joint probability generating function of the $N_j$'s: $${\Bbb E}[\prod_j T_j^{N_j}]= f(\sum_j p_j T_j) = e^{\lambda (\sum_j p_j T_j)}.$$ This equals the product $$\prod_j e^{\lambda p_j T_j},$$ and so the $N_j$'s are independent with each $N_j\sim P(\lambda p_j)$.