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# Autoregressive effects of residuals #2

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opened this issue Sep 25, 2015 · 15 comments

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### paul-buerkner commented Sep 25, 2015

 Autoregressive (AR) effects as they are currently implemented refer to effects of the response on itself. Some other packages (such as nlme) use the AR term to refer to autoregression of the residuals. Thus, it would be nice to have AR effects of the residuals in addition to AR effects of the response.
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### paul-buerkner commented Oct 17, 2015

 AR effects of residuals are now implemented. They are replacing the previous AR effects to be in accordance with the nlme implementation of autocorrelation structures. AR effects of the response can still be modeled but are now called ARR effects to avoid using the term 'AR' for two different things. I will leave this issue open until I am confident that the implementation is stable.

### BERENZ commented Oct 18, 2015

 Could you provide an example how to specify such model? I would like to test the implementation. :) edit: I found cor_arr however, there is no information about that in the description of cor_brms
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### paul-buerkner commented Oct 18, 2015

 This is an example model using a data set that comes with brms: fit_brm <- brm(count ~ log_Age_c + log_Base4_c * Trt_c + (1|patient), data = epilepsy, autocor = cor_arma(~visit|patient, 1, 1), prior = c(set_prior("normal(0,0.5)", class = "ar"), set_prior("normal(0,0.5)", class = "ma"))) This model defines both, AR1 and MA1 effects (of residuals). If you want AR1 effects only, just replace cor_arma(~visit|patient, 1, 1) with cor_arma(~visit|patient, 1) or simpler cor_ar(~visit|patient). If you want to test AR effects of the response (previously called AR and now called ARR), you use cor_arr(~visit|patient) for ARR1. Note that the name 'ARR' was my (probably not ideal) idea, and you will not find it in the literature. The cor_brms is just the overall class of correlation structures and you should not worry about it. They only class that is currently relevant is cor_arma.

### BERENZ commented Oct 18, 2015

 Hi, results that I've got are different from those that supposed to be…(I've reinstalled package to github version 0.5.0.9000). It looks like the AR(1) still refers to the AR effect of the response not for the residuals. Below please find the code for the sae model that has two random effect - domain and time | domain (AR(1)). library(sae2) library(sae) library(brms) # Loading 'brms' package (version 0.5.0.9000). Useful instructions # can be found by typing help('brms'). A more detailed introduction # to the package is available through vignette('brms'). data(spacetime) # Load data set from sae package spacetime$sei <- spacetime$Var^.5 D <- nrow(spacetimeprox) # number of domains T <- length(unique(spacetime$Time)) # number of time instants # Fit model with AR(1) time effects for each domain (domain is already present) resultT.RY <- eblupRY(Y ~ X1 + X2, D, T, vardir = diag(spacetime$Var), data=spacetime, ids=spacetime$Area) # Fit the same model with brms fit_brm <- brm(Y | se(sei) ~ X1 + X2 + (1|Area), data = spacetime, autocor = cor_ar(~ Time | Area), prior = c(set_prior("beta(2,2)", class = "ar"), set_prior("normal(0,5)", class = "b"))) fit_brm_fitted <- fitted(fit_brm)  Results obtained from sae2 package (sig2_u refers to AR(1) effect, sig2_v refers to domain effect, rho is the autocorrelation). The same results I've obtained in metafor package. > resultT.RY$fit $model [1] "T: Rao-Yu, REML"$convergence [1] TRUE $iterations [1] 40$estcoef beta std.error tvalue pvalue (Intercept) 2.324605 0.4488063 5.179527 2.224491e-07 X1 -2.606220 0.8264804 -3.153396 1.613827e-03 X2 -1.830352 0.1989480 -9.200151 3.574481e-20 $estvarcomp estimate std.error sig2_u 0.0008172101 0.0004979779 sig2_v 0.0000000033 0.0009056348 rho 0.3331617406 0.7526800594$goodness loglike restrictedloglike 62.57172 55.87707  and here are results from brms package > fit_brm Family: gaussian (identity) Formula: Y | se(sei) ~ X1 + X2 + (1 | Area) Data: spacetime (Number of observations: 33) Samples: 2 chains, each with n.iter = 2000; n.warmup = 500; n.thin = 1; total post-warmup samples = 3000 WAIC: -115.39 Random Effects: ~Area (Number of levels: 11) Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat sd(Intercept) 0.08 0.02 0.04 0.14 541 1.01 Correlation Structure: arma(~Time|Area, 1, 0, 0) Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat ar[1] 0.19 0.12 0.03 0.47 1194 1 Fixed Effects: Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat Intercept 1.47 0.45 0.57 2.33 997 1 X1 -2.07 0.82 -3.59 -0.40 1050 1 X2 -0.53 0.21 -0.94 -0.14 1252 1 Samples were drawn using NUTS(diag_e). For each parameter, Eff.Sample is a crude measure of effective sample size, and Rhat is the potential scale reduction factor on split chains (at convergence, Rhat = 1).
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### paul-buerkner commented Oct 18, 2015

 Thanks for testing! brms is definitly fitting AR effects of residuals now when using cor_ar, so this cannot explain the differences. I have another idea though. The eblupRY function (what I great name by the way), seems to fit 2 variances (sig_u and sig_v) and 1 correlation (rho), whereas the brms model only has 1 variance / sd (for Area) and 1 correlation (ar[1]). So I think that these two models are not equivalent. From what you wrote, I think that the brms model should rather look like this fit_brm <- brm(Y | se(sei) ~ X1 + X2 + (1|Area) + (1|Time:Area), data = spacetime, autocor = cor_ar(~ Time | Area), prior = c(set_prior("normal(0,0.5)", class = "ar"), set_prior("normal(0,5)", class = "b")))  with an additional random effect (1|Time:Area). I also changed the prior of the AR effect, as beta priors restrict the parameter to [0,1], but autocorrelations may be between [-1,1] or even lie outside this interval if the process is not stationary. The results look very similar to that of eblupRY:  Family: gaussian (identity) Formula: Y | se(sei) ~ X1 + X2 + (1 | Area) + (1 | Time:Area) Data: spacetime (Number of observations: 33) Samples: 2 chains, each with n.iter = 2000; n.warmup = 500; n.thin = 1; total post-warmup samples = 3000 WAIC: -143.93 Random Effects: ~Area (Number of levels: 11) Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat sd(Intercept) 0.02 0.02 0 0.08 286 1.01 ~Time:Area (Number of levels: 33) Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat sd(Intercept) 0.03 0.01 0.01 0.04 551 1 Correlation Structure: arma(~Time|Area, 1, 0, 0) Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat ar[1] 0.21 0.4 -0.54 0.97 782 1 Fixed Effects: Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat Intercept 2.18 0.51 1.15 3.15 1059 1.00 X1 -2.50 0.94 -4.36 -0.64 1163 1.00 X2 -1.63 0.35 -2.17 -0.75 365 1.01 Samples were drawn using NUTS(diag_e). For each parameter, Eff.Sample is a crude measure of effective sample size, and Rhat is the potential scale reduction factor on split chains (at convergence, Rhat = 1).  When using the autocor argument in brm, only the autocorrelations itself is added to the model (in this case ar[1]), and one has to specify the random effects seperately. So when I refer to the AR[1] effect, I always mean the autocorrelation not the residual variance (which seems to be sig2_u in the sae2 model). For the present brms model, the residual variance is modeled through the random effect of Time:Area.

### BERENZ commented Oct 18, 2015

 Great! Thanks for clarification, I should noticed that my results from brm function don't have second variance component. However, the I am not sure whether specifying (1|Time:Area) and cor_ar(~ Time | Area) separately would give the same results. In that model the second component is assumed to follow AR(1) process for residuals given by: u_{t} = \rho u_{t-1} + e where e \sim N(0,\sigma^2). Thus, the variance component and rho are estimated together. The model Y | se(sei) ~ X1 + X2 + (1 | Area) + (1 | Time:Area) would result in: random effect for Area random effect in Time/Area autocorrelation of residuals independent from (1) and (2). Nice example of different variance structure provide proc mixed documentation. Toepliz matrix is nice for specifying AR effects, that's pitty that stan do not have that... In addition, thanks for commend on prior. I've used beta(2,2) because I know that in this case \rho would be between (0,1). For other models I would use normal prior.
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### paul-buerkner commented Oct 18, 2015

 The autocorrelation is not independent from the random effects components, even though they are defined in seperate parts of the model specification. This is because fixed effects, random effects, and the autocorrelation components all go into the same linear regression for Y and thus will all be estimated together. Since the random effect of Time:Area has the same number of levels as observations in the data, it corresponds to the residuals e, and leads to the equation u_{t} = \rho u_{t-1} + e with e \sim N(0,\sigma^2), when combined with an AR(1) autocorrelation (even if they are not explicitly defined together). Thanks for the link! I will have a look into this. :-)

### BERENZ commented Oct 18, 2015

 I've looked into the code that brm provide for the above model, in particular transformed parameters transformed parameters { vector[N] eta; # linear predictor row_vector[Karma] E[N]; # ARMA design matrix vector[N] e; # residuals vector[N_1] r_1; # REs vector[N_2] r_2; # REs # compute linear predictor eta <- X * b + b_Intercept; E <- E_pre; r_1 <- sd_1 * (pre_1); # scale REs r_2 <- sd_2 * (pre_2); # scale REs # if available add REs to linear predictor for (n in 1:N) { eta[n] <- eta[n] + Z_1[n] * r_1[J_1[n]] + Z_2[n] * r_2[J_2[n]]; # calculation of ARMA effects e[n] <- (Y[n]) - eta[n]; for (i in 1:Karma) { if (n + 1 - i > 0 && n < N && tgroup[n + 1] == tgroup[n + 1 - i]) { E[n + 1, i] <- e[n + 1 - i]; } } eta[n] <- (eta[n] + head(E[n], Kar) * ar); } }  and as I see here Z_1[n] * r_1[J_1[n]] refers to the first random effect, Z_2[n] * r_2[J_2[n]] refers to the second random effect and the AR is calculated for the residual given by e[n] <- (Y[n]) - eta[n]. That means the model calculate rho after simulating the random effects and sigma2_2 (for the second component) is not connected with the rho itself. While, AR(1) effect should have variance structure as in the proc mixed link. I have found some example at Stan google group https://groups.google.com/forum/#!searchin/stan-users/AR/stan-users/Cq85_k0QcRo/3mEmkO3p5pgJ data { int N; // length of data set vector[N] y; // data } parameters { real mu; real sigma; real rho; } transformed parameters { // declare mu, Sigma, chol_Sigma vector[N] Mu; matrix[N,N] Sigma; matrix[N,N] chol_Sigma; // define mu for(i in 1:N) Mu[i] <- mu; // define Sigma for(i in 1:N) for(j in 1:N) Sigma[i,j] <- pow(sigma,2)*pow(rho,abs(i-j)); // define chol_Sigma chol_Sigma <- cholesky_decompose(Sigma); } model { // prior distribution rho ~ uniform(0.0,1.0); // likelihood y ~ multi_normal_cholesky(Mu, chol_Sigma); }  EDIT: Maybe there is misunderstanding between us in the sense what I refer as AR process in residuals. In mixed models the difference between observed values and fixed effects are residuals for which we assume various covariance structure. Then, after estimating the model with fixed and random effect we have conditional residuals which are free of the covariance structure and follow N(0,\sigma) where \sigma in general is unknown, in small area is sampling variance. Then, the way how the AR process is implemented now is not the same as is assumed in mixed model theory. EDIT2: In November I will have finally enough time to work on that in Stan. Probably the best way is to code as on the Stan google group.
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### paul-buerkner commented Oct 18, 2015

 You are right. My implementation becomes problematic as soon as the user defines known SEs. If this is not the case, the computation order should be correct (at least the models can accurately recover parameters from simulations). I will have a deeper look at this in the next days and see how I can implement AR structures for known SEs. Thanks for taking the time to test this! :-)

### BERENZ commented Oct 18, 2015

 I've adapted the code from Stan google group for SAE model. Hope it will be useful for you. :) diag_matrix(rep_vector(sd_1,N)) could be used to specify user-defined matrices for covariance structure. library(sae) library(brms) data_in <- list(N=nrow(milk), Y = milk$yi, K = 4, X = model.matrix(~as.factor(MajorArea),milk), samperr = milk$SD^2) model <- ' data { int N; vector[N] Y; vector[N] samperr; int K; matrix[N, K] X; } parameters { vector[K] b; # fixed effects real sd_1; # RE standard deviation } transformed parameters { vector[N] eta; matrix[N,N] Sigma; matrix[N,N] chol_Sigma; eta <- X * b; Sigma <- diag_matrix(rep_vector(sd_1,N)) + diag_matrix(samperr); chol_Sigma <- cholesky_decompose(Sigma); } model { # prior specifications sd_1 ~ cauchy(0,5); # likelihood contribution Y ~ multi_normal_cholesky(eta, chol_Sigma); } ' fit <- stan(model_code = model, data= data_in)
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### paul-buerkner commented Oct 20, 2015

 I just pushed some commits to github allowing to fit AR1 processes in models with predefined SEs using AR1 covariance matrices. For the spacetime example, the following code should work. fit_brm <- brm(Y | se(sei) ~ X1 + X2 + (1|Area), data = spacetime, autocor = cor_ar(~ Time | Area), prior = c(set_prior("normal(0,1)", class = "ar"), set_prior("normal(0,5)", class = "b")))  For the time being, this implementation of AR1 effects will only be used in models with predefined SEs and I will run more tests in the next few days to make sure that the models work as intended.

### BERENZ commented Oct 20, 2015

 Thanks for info! I will test it during this week. :)
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### paul-buerkner commented Nov 2, 2015

 MA(1) and ARMA(1,1) effects using covariance model formulation are now also implemented. In addition, I added the logical argument cov to the cor_arma / cor_ar / cor_ma functions, indicating if covariance model formulation should be used. The code for the spacetime example with an AR(1) effect now looks like this: fit_brm <- brm(Y | se(sei) ~ X1 + X2 + (1|Area), data = spacetime, autocor = cor_ar(~ Time | Area, cov = TRUE), prior = c(set_prior("normal(0,1)", class = "ar"), set_prior("normal(0,5)", class = "b")))

### paul-buerkner closed this Nov 23, 2015

referenced this issue Oct 2, 2016
referenced this issue Nov 30, 2016

### asantosp commented Apr 4, 2017 • edited

 Hey guys, Do you know how may I specify an unstructured covariance matrix such as in lme? lme(C~ 1,random= ~B|A, correlation=corCompSymm(form=~1|A), weights=varIdent(form=~1|B), data=d)  Thank you so much!
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### paul-buerkner commented Apr 4, 2017

 To what part of the model specification are you refering to? The compound symmetry correlation structure is not yet implemented in brms, but you are invited to open a new issue so that I don't loose track of it. 2017-04-04 11:11 GMT+02:00 Adrian Santos Parrilla : … Hey guys, Do you know how may I specify an unstructured covariance matrix such as in lme? lme(C~ 1,random= ~B|A, correlation=corCompSymm(form=~1|A), weights=varIdent(form=~1|B), data=d) Thank you so much! — You are receiving this because you modified the open/close state. Reply to this email directly, view it on GitHub <#2 (comment)>, or mute the thread .
referenced this issue Feb 7, 2018


text:
# [NTG-context] Need for \buildtextaccent, was Re: latest beta

Hans Hagen pragma at wxs.nl
Fri Jan 17 19:51:49 CET 2014

```On 1/17/2014 7:39 PM, Mojca Miklavec wrote:
> On Fri, Jan 17, 2014 at 7:11 PM, Hans Hagen wrote:
>>
>> do we need a module with predefined 'missing from unicode' characters?
>
> I don't think so. We can end up with a neverending and always
> incomplete list (with theoretically unlimited number of entries),
> usable only to those few people who will care to contribute. In my
> opinion such a list is pretty much useless.
>
> Unicode already specifies that one could use "q" followed by
> "combining acute". And TeX also has support for {\'q}. Everything else
> is private use, something that users might not even use in more than a
> single document. I don't see any advantage in creating an endless
> incomplete list. In all honesty it's a lot more
> useful/readable/straightforward to use {\'q} or proper Unicode in
> these cases. And if it's not comfortable enough for the user, he can
> always provide his own private definitions.

> Thomas and Rik most probably need just the existing mechanism to work
> properly. Well, it might be slightly different for Thomas where
> Ancient Greek has a limited set of letters, but then such a list
> should better be defined in an "ancient greek predefined characters
> module".

the mechanisms works ok (i just made it a bit more efficient, not that
it matters much)

> Just my 2 cents.

it depends on to what extend such characters play a different role too
and need to travel around (but as there never was any demand for that i
assume 2 cents will do)

Hans

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

format: plain


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\includegraphics within braces with metropolis beamer theme: image doesn't appear

I am using the Metropolis theme in beamer, and wondering why having \includegraphics within braces {...} doesn't work (image doesn't get added in). I get a blank slide (with title) with the code below:

\documentclass[10pt]{beamer}
\usetheme[progressbar=frametitle]{metropolis}
\begin{document}
\begin{frame}{test}
{\hfill\includegraphics[width=1cm]{example-image}\hfill}
\end{frame}
\end{document}

Commenting out the metropolis line, (thus, switching to using the default beamer theme), the image appears as expected. So I know this behaviour has something to do with the metropolis theme.

I know I can remove the braces, but you can imagine a scenario where I want some effect localized/grouped, e.g. \color{blue}. I want to know why does this happen?

• If you leave an empty line after \begin{frame}{test} the image shows up. I prefer the cleaner syntax \begin{frame} \frametitle{test} {\hfill\includegraphics[width=1cm]{example-image}\hfill} \end{frame}, with which the problem is absent. – marmot Feb 6 at 0:23
• @marmot does this mean \begin{frame} takes a second argument? do you know what this second argument means/refers to? – ksgj1 Feb 6 at 0:26
• Right, but the default beamer theme doesn't have this issue, so I'm wondering what's the difference introduced by metropolis here. – ksgj1 Feb 6 at 0:27
• I agree that one may suspect this but I do not know for sure whether this is the case, simply since I always use \frametitle because of problems I had with the other syntax. – marmot Feb 6 at 0:28

It's not a problem with beamer-metropolis, it's a problem with your code.

beamerdefines this syntax \begin{frame}{frame title}{frame subtitle} where title and subtitle are optionals. Therefore your figure is considered as a subtitle, but metropolis has no subtitle commands, then it's ignored.

Compare this

\documentclass[10pt]{beamer}
\begin{document}
\begin{frame}{test}
{\hfill\includegraphics[width=1cm]{example-image}\hfill}
\end{frame}
\end{document}

with this:

\documentclass[10pt]{beamer}
\begin{document}
\begin{frame}{test}

{\hfill\includegraphics[width=1cm]{example-image}\hfill}
\end{frame}
\end{document}

As you can see, even without metropolis your syntax does not produce the expected result.

If you don't want to leave an empty line between title and contents, use this alternative syntax (or at least don't start new line with '{')

\documentclass[10pt]{beamer}
\usetheme[progressbar=frametitle]{metropolis}
\begin{document}
\begin{frame}{test}
\centerline{\includegraphics[width=1cm]{example-image}}
%{\hfill\includegraphics[width=1cm]{example-image}\hfill}
\end{frame}
\end{document}

• i would mentioned that for titles and subtitles is better to use \frametitle{...} and \framesubtitle{...} (as recommended in beamer documentation). using them the op error will not happen. – Zarko Feb 8 at 13:22


text:
# Math Help - set and relation

1. ## set and relation

Define the the relation R on the integers Z by xRy if and only if either x/y or y/x.
For each n E Z, define An to be the set of integers related to n. An={xEZ:xRn}
Proving the quality below
Intersection An= {-1,0,1}

So here my prove
An is a subset of {-1,0,1}, so if xEAn then XE{-1,0,1} ==> Intersection An is a subset of {-1,0,1}.

Suppose that m is not belong of {-1,0,1} then m is not belong to Intersection An ===> {-1,0,1} is a subset of Intersection An

I a little confusing , i know im missing something out there,
Any help or suggestions would be greatly appreciated. Thanks

2. Originally Posted by logglypop
Prove
An={xEZ:xRn}
Intersection An= {-1,0,1}
nEZ

An is a subset of {-1,0,1}, so if xEAn then XE{-1,0,1} ==> Intersection An is a subset of {-1,0,1}.

Suppose that m is not belong of {-1,0,1} then m is not belong to Intersection An ===> {-1,0,1} is a subset of Intersection An

I a little confusing , i know im missing something out there,
Any help or suggestions would be greatly appreciated. Thanks
I'm not sure what the question is actually asking. Could you type it in LaTeX or at least more carefully?

3. I edited abit
i hope u understand

4. Originally Posted by logglypop
I edited abit
i hope u understand
What is the relation?

5. What is the relation?
sorry there no relation.
I just have to prove the Intersection of the set An = {-1,0,1}

6. Suppose you and me are talking and I ask, "By the way, did you know that Peterson is a bullplooper?" You would naturally ask, "Who is Peterson? And what is a bullplooper?"

It's the same way when you say "An={xEZ:xRn}" but don't say what R is.

7. Define the the relation R on the integers Z by xRy if and only if either x/y or y/x.

8. Originally Posted by logglypop
Define the the relation R on the integers Z by xRy if and only if either x/y or y/x.
I presume you mean $xRy \Leftrightarrow x/y \in \mathbb{Z} \text{ or } y/x \in \mathbb{Z}$.

I will write $a\sim b$ instead of $aRb$ as this is easier to read.

For your proof, firstly note that $0\sim n$ for all $n \in \mathbb{Z}$ as $0/n = 0 \in \mathbb{Z}$. Also, $\pm1 \sim n$ as $n/\pm 1 = \pm n \in \mathbb{Z}$. Therefore, $\{-1, 0, 1\}$ is in the intersection.

(It's late, so this bit is a bit rough around the edges.) To prove that this is precisely the intersection, choose $a \in \mathbb{Z} \setminus \{-1, 0, 1\}$. Then there exists a number, $b$, such that $a$ and $b$ are coprime (for example, if a is positive choose a+1, while if a is negative choose a-1). Clearly $a \not\sim b$, and so $a$ is not contained in the intersection. As a is arbitrary, we have the result.

9. OK. Does "x / y" mean x divides y? Usually this is denoted by "x | y".

So here my prove
An is a subset of {-1,0,1},
If you prove this, you solve the whole problem. The whole difficulty of the problem is to show that for every integer $n$, $A_n\subseteq\{-1,0,1\}$.

Now, $x\in\bigcap_{n\in\mathbb{Z}}A_n$ iff $x\in A_n$ for every $n\in\mathbb{Z}$. In turn, this happens iff $xRn$ for all $n\in\mathbb{Z}$, i.e., for every $n$, either $n$ divides $x$ or vice versa.

You can check that $x = -1$, $0$ or $1$ satisfy this property. This means that $\{-1,0,1\}\subseteq\bigcap_{n\in\mathbb{Z}}A_n$. On the other hand, when $x$ is none of the three, it is easy to find an $n$ such that neither $x\mid n$ nor $n\mid x$. This means that $x\notin\{-1,0,1\}$ implies $x\notin A_n$ (where $n$ was found in the previous sentence) and so $x\notin\bigcap_{n\in\mathbb{Z}}A_n$, which is the same thing to say as $\bigcap_{n\in\mathbb{Z}}A_n\subseteq\{-1,0,1\}$.


text:
## Results (displaying matches 1-50 of 183607) Next

LMFDB label dimension base field Weil polynomial $p$-rank isogeny factors
4.2.ai_bg_adc_fg 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )^{4}$ $0$ 1.2.ac 4
4.2.ah_z_acg_ds 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )^{2}( 1 - 3 x + 5 x^{2} - 6 x^{3} + 4 x^{4} )$ $2$ 1.2.ac 2 $\times$ 2.2.ad_f
4.2.ah_ba_ack_ea 4 $\F_{2}$ $( 1 - x + 2 x^{2} )( 1 - 2 x + 2 x^{2} )^{3}$ $1$ 1.2.ac 3 $\times$ 1.2.ab
4.2.ag_s_abk_ce 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )^{2}( 1 - 2 x + 2 x^{2} - 4 x^{3} + 4 x^{4} )$ $0$ 1.2.ac 2 $\times$ 2.2.ac_c
4.2.ag_t_abq_cr 4 $\F_{2}$ $( 1 - 3 x + 5 x^{2} - 6 x^{3} + 4 x^{4} )^{2}$ $4$ 2.2.ad_f 2
4.2.ag_t_abp_co 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )( 1 - 4 x + 9 x^{2} - 15 x^{3} + 18 x^{4} - 16 x^{5} + 8 x^{6} )$ $3$ 1.2.ac $\times$ 3.2.ae_j_ap
4.2.ag_t_abo_cm 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )^{2}( 1 - 2 x + 3 x^{2} - 4 x^{3} + 4 x^{4} )$ $2$ 1.2.ac 2 $\times$ 2.2.ac_d
4.2.ag_u_abt_cw 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )( 1 - x + 2 x^{2} )( 1 - 3 x + 5 x^{2} - 6 x^{3} + 4 x^{4} )$ $3$ 1.2.ac $\times$ 1.2.ab $\times$ 2.2.ad_f
4.2.ag_u_abs_cu 4 $\F_{2}$ $( 1 + 2 x^{2} )( 1 - 2 x + 2 x^{2} )^{3}$ $0$ 1.2.ac 3 $\times$ 1.2.a
4.2.ag_v_abw_dc 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )^{2}( 1 - x + 2 x^{2} )^{2}$ $2$ 1.2.ac 2 $\times$ 1.2.ab 2
4.2.af_k_aj_g 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )( 1 - 3 x + 2 x^{2} + x^{3} + 4 x^{4} - 12 x^{5} + 8 x^{6} )$ $3$ 1.2.ac $\times$ 3.2.ad_c_b
4.2.af_l_ao_q 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )^{2}( 1 - x - x^{2} - 2 x^{3} + 4 x^{4} )$ $2$ 1.2.ac 2 $\times$ 2.2.ab_ab
4.2.af_m_au_bd 4 $\F_{2}$ $1 - 5 x + 12 x^{2} - 20 x^{3} + 29 x^{4} - 40 x^{5} + 48 x^{6} - 40 x^{7} + 16 x^{8}$ $4$ simple
4.2.af_m_as_y 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )^{2}( 1 - x - 2 x^{3} + 4 x^{4} )$ $1$ 1.2.ac 2 $\times$ 2.2.ab_a
4.2.af_n_aba_bq 4 $\F_{2}$ $( 1 - 3 x + 5 x^{2} - 6 x^{3} + 4 x^{4} )( 1 - 2 x + 2 x^{2} - 4 x^{3} + 4 x^{4} )$ $2$ 2.2.ad_f $\times$ 2.2.ac_c
4.2.af_n_az_bn 4 $\F_{2}$ $1 - 5 x + 13 x^{2} - 25 x^{3} + 39 x^{4} - 50 x^{5} + 52 x^{6} - 40 x^{7} + 16 x^{8}$ $4$ simple
4.2.af_n_ax_bi 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )( 1 - 3 x + 5 x^{2} - 7 x^{3} + 10 x^{4} - 12 x^{5} + 8 x^{6} )$ $3$ 1.2.ac $\times$ 3.2.ad_f_ah
4.2.af_n_aw_bg 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )^{2}( 1 - x + x^{2} - 2 x^{3} + 4 x^{4} )$ $2$ 1.2.ac 2 $\times$ 2.2.ab_b
4.2.af_o_abd_bv 4 $\F_{2}$ $( 1 - 3 x + 5 x^{2} - 6 x^{3} + 4 x^{4} )( 1 - 2 x + 3 x^{2} - 4 x^{3} + 4 x^{4} )$ $4$ 2.2.ad_f $\times$ 2.2.ac_d
4.2.af_o_abc_bs 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )( 1 - x + 2 x^{2} )( 1 - 2 x + 2 x^{2} - 4 x^{3} + 4 x^{4} )$ $1$ 1.2.ac $\times$ 1.2.ab $\times$ 2.2.ac_c
4.2.af_o_abb_bq 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )( 1 - 3 x + 6 x^{2} - 9 x^{3} + 12 x^{4} - 12 x^{5} + 8 x^{6} )$ $3$ 1.2.ac $\times$ 3.2.ad_g_aj
4.2.af_o_aba_bo 4 $\F_{2}$ $( 1 + x + 2 x^{2} )( 1 - 2 x + 2 x^{2} )^{3}$ $1$ 1.2.ac 3 $\times$ 1.2.b
4.2.af_p_abg_bz 4 $\F_{2}$ $( 1 - x + 2 x^{2} )( 1 - 4 x + 9 x^{2} - 15 x^{3} + 18 x^{4} - 16 x^{5} + 8 x^{6} )$ $4$ 1.2.ab $\times$ 3.2.ae_j_ap
4.2.af_p_abg_ca 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )( 1 + 2 x^{2} )( 1 - 3 x + 5 x^{2} - 6 x^{3} + 4 x^{4} )$ $2$ 1.2.ac $\times$ 1.2.a $\times$ 2.2.ad_f
4.2.af_p_abf_by 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )( 1 - x + 2 x^{2} )( 1 - 2 x + 3 x^{2} - 4 x^{3} + 4 x^{4} )$ $3$ 1.2.ac $\times$ 1.2.ab $\times$ 2.2.ac_d
4.2.af_p_abe_bw 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )^{2}( 1 - x + 3 x^{2} - 2 x^{3} + 4 x^{4} )$ $2$ 1.2.ac 2 $\times$ 2.2.ab_d
4.2.af_q_abj_cf 4 $\F_{2}$ $( 1 - x + 2 x^{2} )^{2}( 1 - 3 x + 5 x^{2} - 6 x^{3} + 4 x^{4} )$ $4$ 1.2.ab 2 $\times$ 2.2.ad_f
4.2.af_q_abi_ce 4 $\F_{2}$ $( 1 - x + 2 x^{2} )( 1 + 2 x^{2} )( 1 - 2 x + 2 x^{2} )^{2}$ $1$ 1.2.ac 2 $\times$ 1.2.ab $\times$ 1.2.a
4.2.af_r_abl_ck 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )( 1 - x + 2 x^{2} )^{3}$ $3$ 1.2.ac $\times$ 1.2.ab 3
4.2.ae_e_h_av 4 $\F_{2}$ $1 - 4 x + 4 x^{2} + 7 x^{3} - 21 x^{4} + 14 x^{5} + 16 x^{6} - 32 x^{7} + 16 x^{8}$ $4$ simple
4.2.ae_e_i_ay 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )^{2}( 1 - 2 x^{2} )^{2}$ $0$ 1.2.ac 2 $\times$ 2.2.a_ae
4.2.ae_f_c_al 4 $\F_{2}$ $1 - 4 x + 5 x^{2} + 2 x^{3} - 11 x^{4} + 4 x^{5} + 20 x^{6} - 32 x^{7} + 16 x^{8}$ $4$ simple
4.2.ae_f_e_aq 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )^{2}( 1 - 3 x^{2} + 4 x^{4} )$ $2$ 1.2.ac 2 $\times$ 2.2.a_ad
4.2.ae_g_ae_c 4 $\F_{2}$ $1 - 4 x + 6 x^{2} - 4 x^{3} + 2 x^{4} - 8 x^{5} + 24 x^{6} - 32 x^{7} + 16 x^{8}$ $0$ simple
4.2.ae_g_ab_ag 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )( 1 - 2 x + 3 x^{3} - 8 x^{5} + 8 x^{6} )$ $3$ 1.2.ac $\times$ 3.2.ac_a_d
4.2.ae_g_a_ai 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )^{2}( 1 - 2 x^{2} + 4 x^{4} )$ $0$ 1.2.ac 2 $\times$ 2.2.a_ac
4.2.ae_h_ak_p 4 $\F_{2}$ $( 1 - 3 x + 5 x^{2} - 6 x^{3} + 4 x^{4} )( 1 - x - x^{2} - 2 x^{3} + 4 x^{4} )$ $4$ 2.2.ad_f $\times$ 2.2.ab_ab
4.2.ae_h_ah_h 4 $\F_{2}$ $( 1 - x + 2 x^{2} )( 1 - 3 x + 2 x^{2} + x^{3} + 4 x^{4} - 12 x^{5} + 8 x^{6} )$ $4$ 1.2.ab $\times$ 3.2.ad_c_b
4.2.ae_h_ag_e 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )( 1 - 2 x + x^{2} + 2 x^{4} - 8 x^{5} + 8 x^{6} )$ $2$ 1.2.ac $\times$ 3.2.ac_b_a
4.2.ae_h_af_c 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )( 1 - 2 x + x^{2} + x^{3} + 2 x^{4} - 8 x^{5} + 8 x^{6} )$ $3$ 1.2.ac $\times$ 3.2.ac_b_b
4.2.ae_h_ae_a 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )^{2}( 1 - x^{2} + 4 x^{4} )$ $2$ 1.2.ac 2 $\times$ 2.2.a_ab
4.2.ae_i_aq_bc 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} - 4 x^{3} + 4 x^{4} )^{2}$ $0$ 2.2.ac_c 2
4.2.ae_i_an_u 4 $\F_{2}$ $( 1 - 3 x + 5 x^{2} - 6 x^{3} + 4 x^{4} )( 1 - x - 2 x^{3} + 4 x^{4} )$ $3$ 2.2.ad_f $\times$ 2.2.ab_a
4.2.ae_i_am_r 4 $\F_{2}$ $1 - 4 x + 8 x^{2} - 12 x^{3} + 17 x^{4} - 24 x^{5} + 32 x^{6} - 32 x^{7} + 16 x^{8}$ $4$ simple
4.2.ae_i_al_o 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )( 1 - x + 2 x^{2} )( 1 - x - x^{2} - 2 x^{3} + 4 x^{4} )$ $3$ 1.2.ac $\times$ 1.2.ab $\times$ 2.2.ab_ab
4.2.ae_i_ak_m 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )( 1 - 2 x + 2 x^{2} - 2 x^{3} + 4 x^{4} - 8 x^{5} + 8 x^{6} )$ $0$ 1.2.ac $\times$ 3.2.ac_c_ac
4.2.ae_i_aj_k 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} )( 1 - 2 x + 2 x^{2} - x^{3} + 4 x^{4} - 8 x^{5} + 8 x^{6} )$ $3$ 1.2.ac $\times$ 3.2.ac_c_ab
4.2.ae_i_ai_i 4 $\F_{2}$ $( 1 + 2 x + 2 x^{2} )( 1 - 2 x + 2 x^{2} )^{3}$ $0$ 1.2.ac 3 $\times$ 1.2.c
4.2.ae_j_as_be 4 $\F_{2}$ $( 1 - 2 x + 2 x^{2} - 4 x^{3} + 4 x^{4} )( 1 - 2 x + 3 x^{2} - 4 x^{3} + 4 x^{4} )$ $2$ 2.2.ac_c $\times$ 2.2.ac_d
4.2.ae_j_ar_bb 4 $\F_{2}$ $1 - 4 x + 9 x^{2} - 17 x^{3} + 27 x^{4} - 34 x^{5} + 36 x^{6} - 32 x^{7} + 16 x^{8}$ $4$ simple

Next


text:
# Could Someone Check This Answer?

1. Jun 16, 2005

### kitz

Hello!!

I did this problem and have gotten it wrong-- but my math is right. Maybe I'm missing something? Here's the question:

We want to support a thin hoop by a horizontal nail and have the hoop make one complete small-angle oscillation each 2.0 s. What must the hoop's radius be?

So this is a physical pendulum, and in order to find the radius, I need to know the period. The period is found by:

$$T=2\pi\sqrt{\displaystyle{\frac{I}{mgd}}}$$
Where I is the moment of inertia, g is the acceleration due to gravity, and d is the distance from the pivot point to the center of gravity.

For a hoop, the moment of inertia is: $$I=MR^2$$
And the center of gravity is in the center of the hoop, so d is the radius.

So with that information, I set up the following:

$$T=2\pi\sqrt{\displaystyle{\frac{MR^2}{MgR}}}$$
$$T=2\pi\sqrt{\displaystyle{\frac{R}{g}}}$$

And in plugging in the values, I get:

$$2=2\pi\sqrt{\displaystyle{\frac{R}{9.8}}}$$
Solving for R, I get .9929475

Which, when I plug into my equation, I get very close to 2 for my period. However, this is incorrect...

Any ideas as to why?

Thanks!!!!

2. Jun 16, 2005

### kitz

That makes sense. How would I go about finding the moment of inertia about the nail?

...This is a longshot, but would the moment of inertia be 2MR^2??
(using the parallel-axis theorem...)

3. Jun 16, 2005

### kitz

...There was a reply here a minute ago... ...That suddenly disappeared...

4. Jun 16, 2005

### StatusX

Sorry, that was me. I wasn't sure about my answer because I haven't done this stuff in a while, but you can try it if you want. And yes, that is the correct moment of inertia around the nail.

5. Jun 17, 2005

### CarlB

Yes, you're right, you have to adjust the MI by adding the appropriate amount because you're spinning the hoop around a spot on the rim instead of its center.

If I recall, the formula for adjusting the MI is to take the formula you look up, and add to it $$md^2$$, where "m" is the mass of the object, and "d" is the distance away from the center of mass. So you get $$MR^2+MR^2 = 2MR^2$$, like you suspected.

Carl

Last edited: Jun 17, 2005


text:
# How do you find all the zeros of  x^4-x^3-20x^2?

May 7, 2016

You must set the expression to 0 and factor.

#### Explanation:

${x}^{2} \left({x}^{2} - x - 20\right) = 0$

${x}^{2} \left(x - 5\right) \left(x + 4\right) = 0$

$x = 0 , 5 , - 4$

Hopefully this helps!


text:
## Elementary Algebra

$\dfrac{5\sqrt{2x}}{2x}$
Using the properties of radicals, the given expression, $\dfrac{5}{\sqrt{2x}} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{5}{\sqrt{2x}}\cdot\dfrac{\sqrt{2x}}{\sqrt{2x}} \\\\= \dfrac{5\sqrt{2x}}{2x} .\end{array} Note that all variables are assumed to represent positive real numbers.


text:
# Writing equations with unknowns II

Lesson

## Two-step problems

We have looked at how to write equations with unknowns, using a problem that only has one part. What if we were to look at a problem involving more than one part? We can still write an equation with unknowns, but we need to remember the order of operations.

In Video 1, we'll look at problems with two different operations in them. We have multiplication and addition, followed by addition and subtraction.

## Division

When our problems include division, we are able to think about what we are trying to solve, and what information we already know. By doing this, we can fill in the gaps, and create our equation. In Video 2, we look at a division and addition problem, as well as why brackets can be so important when it comes to the order of operations.

## The whole equation!

What if we have no equation at all, and need to start from scratch? That's okay, we can work through the information we know, and think about what we are trying to create. Let's take a look how in our final video.

Remember!

It's really important to think about the order of operations when you're working with equations.

#### Worked Examples

##### Question 1

Rochelle has boxes of pears. Each box contains $19$19 pears, and there are $8$8 pears left over that didn't fit into any of the boxes. In total, she has $160$160 pears.

Let $m$m be the number of boxes of pears that Rochelle has. Fill in the blanks to make an equation in terms of $m$m.

1. $19\times\editable{}+8=160$19×+8=160

##### Question 2

Kenneth is a rock climber, and is scaling a cliff face. Part way up he slips, falling $4$4 m down before his harness catches him. He then continues, climbing the remaining $13$13 m to the top of the cliff. The cliff is $23$23 m tall.

Let $a$a be the height at which Kenneth slipped. Fill in the blanks to make an equation.

1. $a-\editable{}+\editable{}=23$a+=23

##### Question 3

Xanthe is having dinner with four of her friends, and the $5$5 of them decide to split the cost equally. On the way home, Xanthe also buys a drink for $3$3 pounds. Altogether, she spends $21$21 pounds that evening.

Let $r$r be the total cost of their dinner. Write an equation for the situation using division and addition.


text:
# colour_checker_detection.colour_checkers_coordinates_segmentation#

colour_checker_detection.colour_checkers_coordinates_segmentation(image: ArrayLike, additional_data: bool = False, **kwargs: Any) Union[DataColourCheckersCoordinatesSegmentation, Tuple[ndarray, ...]][source]#

Detect the colour checkers coordinates in given image $$image$$ using segmentation.

This is the core detection definition. The process is a follows:

• Input image $$image$$ is converted to a grayscale image $$image_g$$.

• Image $$image_g$$ is denoised.

• Image $$image_g$$ is thresholded/segmented to image $$image_s$$.

• Image $$image_s$$ is eroded and dilated to cleanup remaining noise.

• Contours are detected on image $$image_s$$.

• Contours are filtered to only keep squares/swatches above and below defined surface area.

• Squares/swatches are clustered to isolate region-of-interest that are potentially colour checkers: Contours are scaled by a third so that colour checkers swatches are expected to be joined, creating a large rectangular cluster. Rectangles are fitted to the clusters.

• Clusters with an aspect ratio different to the expected one are rejected, a side-effect is that the complementary pane of the X-Rite ColorChecker Passport is omitted.

• Clusters with a number of swatches close to the expected one are kept.

Parameters:
• image (ArrayLike) – Image to detect the colour checkers in.

• aspect_ratio – Colour checker aspect ratio, e.g. 1.5.

• aspect_ratio_minimum – Minimum colour checker aspect ratio for detection: projective geometry might reduce the colour checker aspect ratio.

• aspect_ratio_maximum – Maximum colour checker aspect ratio for detection: projective geometry might increase the colour checker aspect ratio.

• swatches – Colour checker swatches total count.

• swatches_horizontal – Colour checker swatches horizontal columns count.

• swatches_vertical – Colour checker swatches vertical row count.

• swatches_count_minimum – Minimum swatches count to be considered for the detection.

• swatches_count_maximum – Maximum swatches count to be considered for the detection.

• swatches_chromatic_slice – A slice instance defining chromatic swatches used to detect if the colour checker is upside down.

• swatches_achromatic_slice – A slice instance defining achromatic swatches used to detect if the colour checker is upside down.

• swatch_minimum_area_factor – Swatch minimum area factor $$f$$ with the minimum area $$m_a$$ expressed as follows: $$m_a = image_w * image_h / s_c / f$$ where $$image_w$$, $$image_h$$ and $$s_c$$ are respectively the image width, height and the swatches count.

• swatch_contour_scale – As the image is filtered, the swatches area will tend to shrink, the generated contours can thus be scaled.

• cluster_contour_scale – As the swatches are clustered, it might be necessary to adjust the cluster scale so that the masks are centred better on the swatches.

• working_width – Size the input image is resized to for detection.

• fast_non_local_means_denoising_kwargs – Keyword arguments for cv2.fastNlMeansDenoising() definition.

• adaptive_threshold_kwargs – Keyword arguments for cv2.adaptiveThreshold() definition.

• interpolation_method – Interpolation method used when resizing the images, cv2.INTER_CUBIC and cv2.INTER_LINEAR methods are recommended.

• kwargs (Any) –

Returns:

Tuple of colour checkers coordinates or DataColourCheckersCoordinatesSegmentation class instance with additional data.

Return type:

colour_checker_detection.detection.segmentation.DataColourCheckersCoordinatesSegmentation or tuple

Notes

• Multiple colour checkers can be detected if presented in image.

Examples

>>> import os


text:
Group with subgroups proof

1. Mar 30, 2009

kathrynag

1. The problem statement, all variables and given/known data

Let G be a group with subgroups H and K. Prove HintersectK is a subgroup.

2. Relevant equations

3. The attempt at a solution
G is a group with subgroups H and K. Then H and K are closed, have identity elements in G and have inverses.
HintersectK is a subgroup because H and K must satisfy the same properties of G.

2. Mar 30, 2009

NJOsment

let $$a,b \epsilon H$$ and $$a,b \epsilon K$$.Then $$a*b \epsilon H$$ and $$a*b \epsilon K$$.Let $$a \epsilon H$$ and $$a \epsilon K$$.Then $$a^{-1} \epsilon H$$ and $$a^{-1} \epsilon K$$.

3. Mar 30, 2009

kathrynag

that makes sense


text:
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# If P and Q are non-zero integers, is the equation Py + Qx = 7, exactly

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If P and Q are non-zero integers, is the equation Py + Qx = 7, exactly  [#permalink]

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04 Jul 2018, 23:34
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If P and Q are non-zero integers, is the equation Py + Qx = 7, exactly the same equation as 9x - 6y = 21 ?

(1) P + Q = 1

(2) P*Q = -6
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Joined: 07 Dec 2017
Posts: 592
Re: If P and Q are non-zero integers, is the equation Py + Qx = 7, exactly  [#permalink]

### Show Tags

04 Jul 2018, 23:44
amanvermagmat wrote:
If P and Q are non-zero integers, is the equation Py + Qx = 7, exactly the same equation as 9x - 6y = 21 ?

(1) P + Q = 1

(2) P*Q = -6

Two equations are equal if and only if all of their coefficients are equal.
We'll look for statements that give us this information, a Logical approach.

Rewriting our equations so they both equal 7 gives:
Py + Qx = 7
3x - 2y = 7

So our equations are equal when P = -2 and Q = 3.

(1) One equation, two variables --> infinite solutions.
Insufficient.

(2) One equation, two variables --> infinite solutions.
Insufficient

Combined:
Two equations, two variables usually gives one unique solution but in this case both of our equations are symmetrical.
So, while we can find the right numbers (they happen to be 3 and -2), we don't know which is P and which is Q.
Insufficient

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If P and Q are non-zero integers, is the equation Py + Qx = 7, exactly  [#permalink]

### Show Tags

05 Jul 2018, 05:02
amanvermagmat wrote:
If P and Q are non-zero integers, is the equation Py + Qx = 7, exactly the same equation as 9x - 6y = 21 ?

(1) P + Q = 1

(2) P*Q = -6

Given, Py + Qx = 7----------(a)
& 9x - 6y = 21 Or, 3x + -(2)y =7 ---(b)

Equating x & y co-efficient in (a) and (b) we have P=-2 & Q=3

So, when P=-2 and Q=3, both the equations become same.

Re-phrased Question stem:- Is P=-2 ,Q=3 ?

Statement-2
P + Q = 1
When P=-2 & Q=3, our answer to question stem is Yes, but there are 'n' number of combinations for which P+Q=1.
Hence, insufficient.

Statement-2
P*Q=-6
When P=-2 & Q=3, our answer to question stem is Yes, but there are 'n' number of combinations for which P*Q=-6.
Hence, insufficient.

(1)+(2), we have

$$(P-Q)^2=(P+Q)^2-4*PQ$$

we know the value of P+Q & PQ from st1 & st2 respectively, therefore, we can calculate P-Q & subsequently, value of P & Q can be determined.

So, ($$P-Q)^2=1^2-4*(-6)=1+24=25$$

So, P-Q=5 Or, P-Q=-5

P-Q=5 & P+Q=1. yields P=3, Q=-2...In this case answer to question is No.

P-Q=-5 & P+Q=1. yields P=-2, Q=3...In this case answer to question is Yes.

So, insufficient.

Ans (E)
_________________

Regards,

PKN

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text:
# How to interpret factor analysis transformation?

In the following discussion on Stack Overflow, How to create factors from factanal? (see answer from @Joris), what is the interpretation of the $F=ML$ transformation, where $M$ is the original data matrix, and $L$ is the loadings matrix? Should it not be $L^{-1}$?

-


text:
4 Pages

## RisksEdit

Risk is future uncertainty, split into two classifications:

• Pure risk

Exposure to adverse outcomes where insurance should have been normally used.

• Speculative risk

Exposure to BOTH adverse AND favourable outcomes.

In life, both are seen, and is managed by: Holistic Risk Management

Enterprise Risk Management is used by banks, insurers, companies to manage risks rigorously.

## Expected ValuesEdit

The expected value of a risk is called the Actuarial Value of Risk or the Pure Premium

For DISCRETE Random Variables: $E[X] = \sum_{all~x} P[X=x]\times x$

But, Expected Value just returns the average of return. It doesn't show any variation in results. So, Variability is used instead.

## Variance Edit

For DISCRETE Random Variables: $Var[X] = \sum_{all~x} P[X=x]\times (x-E[X])^2$

For example,

$\begin{array} {|c|c|c|c|} Outcome & P(Win) & Opt~A & Opt~B\\ \hline Good & \frac{1}{10} & 50,000 & 26,000\\ Middle & \frac{22}{25} & 12,500 & 15,000\\ Bad & \frac{1}{50} & 0 & 10,000\\ E[X] & & 16,000 & 16,000\\ Var[X] & & 131,500,000 & 11,600,000\\ \end{array}$

NOTE: $E[A]=\frac{1}{10}50,000+\frac{22}{25}12,500+\frac{1}{50}0=16,000$

And: $Var[A]=\frac{1}{10}(50,000-16,000)^2+\frac{22}{25}(12,500-16,000)^2$ $+\frac{1}{50}(0-16,000)^2=131,500,000$

Now, since the $Var[B]$ is less than that of $Var[A]$, even though both expected values are = to 16,000, the lower the variability, the better the choice.

So B is better. [Lower variability around expected value]

## Utility Edit

In Economics, to represent preferences over alternatives, we use: $U(\bullet)$ which is the Utility Function.

Note, the following conditions hold:

• Each individual has OWN $U(\bullet)$

Eg: Very rich and very poor. Very poor has higher $U(\bullet)$ to get $1 when compared to rich person (lower $U(\bullet)$) • ASSUME companies have NO $U(\bullet)$ • If wealth increases, $U(\bullet)$ increases • Individual prefers X more than Y: $X \succ Y$ $U(X) > U(Y)$ ## Expected Utility Edit But before, we dealt with known utilities. How about if they are UNKNOWN? Then: we use Expected Utility $U(W) = E[v(W)]$ $U(W) = \sum_{i=1}^n P(W=w_i)\times v(w_i)$ 1. $n$ is n states of the world 2. $w_i$ is wealth in state i 3. $v(\bullet)$ is: 1. The expected $U(\bullet)$ 2. Continuous non-decreasing ## Expected Utility Axioms Edit All preferences are ASSUMED to be: • Complete All risks can be compared, ranked • Reflexive $X \succsim X$ meaning risk is at least as good as itself • Transitive If $X \succsim Y$ and $Y \succsim Z$ then $X \succ Y \succ Z$ • Independent Considering other risks to be equal, then if $X \succ Y$, then any other risks attached to either option (but same), will still result to $X \succ Y$ ## Risk Aversion Edit This is when the expected risk is preferred to the real risk. Remember $W$ is Wealth $E[W] \succ W$ or $v(E[W])>U(W)$ where $U(W)=E[v(W)]$ so $v(E[W])>E[v(W)]$ So, the graph is steeper at the beginning, then becomes smoother at higher wealth values • So higher utility for$1 then goes down at \$1,000.

Risk aversion shows that:

1. Utility function is CONCAVE
2. ${\partial\over\partial W}v(W)>0$
3. ${\partial^2\over\partial W^2}v(W)<0$

Side Note: Prove algebraically (not graphically) that $v(E[W])>E[v(W)]$ This is called JENSEN'S INEQUALITY

We know that $v(x) \approx v(y) + (x-y)v'(y) + \frac{1}{2} (x-y)^2 v''(y)$

Let $y = E[W]$ and also $x=W$

So, $v(W) \approx v(E[X]) + (W-E[W])v'(E[W]) + \frac{1}{2} (W-E[W])^2 v''(E[W])$

The condition is that $\partial ^2 \over \partial W^2 U(W) < 0$ so that $v''(W) <0$.

This entails that $\frac{1}{2} (W-E[W])^2 v''(E[W]) < 0$

Thus, $v(W) \approx v(E[X]) + (W-E[W])v'(E[W]) + \frac{1}{2} (W-E[W])^2 v''(E[W])$ $<= v(W) \approx v(E[X]) + (W-E[W])v'(E[W])$

## Risk Loving Edit

This is when the risk is preferred to the expected risk: eg. Lottery $E[W] \prec W$ or $v(E[W]) where $U(W)=E[v(W)]$ so $v(E[W])

## Risk NeutralityEdit

This is when the risk is indifferently preferred to the expected risk. $E[W] = W$ or $v(E[W])=U(W)$ where $U(W)=E[v(W)]$ so $v(E[W])=E[v(W)]$

This means that ${\partial^2\over\partial W^2}v(W)=0$.

Thus, $v(W) = aW + b$

## Insurance Applications Edit

Some notation:

1. Loss is $I$
2. Sum insured is $I$
3. Insurance premium is $\pi$
4. Insurance premium paid is $\pi I$
5. Individual has wealth $W$
6. $p$ is loss probability

$\begin{array} {|c|c|c|c|c|c|} Wealth & Outcome & P & If~Lose & Wealth(No Ins) & Wealth(Yes Ins)\\ \hline W & Loss & p & -I & W-I & W-\pi I \\ W & No~Loss & 1-p & 0 & W & W-\pi I \\ \end{array}$

So, with FULL Insurance, Loss: $W = W-I+I-\pi I~at~p$

No Loss: $W = W-\pi I~at~1-p$

Thus, Expected Utility is just addition: $v(W- \pi I) = p \times v(W- \pi I) + (1-p) \times v(W- \pi I)$

So, with NO Insurance, Loss: $W = W-I ~at~p$

No Loss: $W = W~at~1-p$

Thus, Expected Utility is just addition: $p \times v(W-I)+(1-p)\times v(W)$

Thus, someone would prefer Insurance if:

$v(W- \pi I) > p \times v(W-I)+(1-p)\times v(W)$

Now, what is the optimum amount of insurance premiums at $\pi \times X$? Where $X$ is the amount of insurance.

Loss: $W = W-I+X - \pi X ~at~p$

No Loss: $W = W - \pi X~at~1-p$

Thus, Expected Utility is just addition: $p \times v( W-I+X - \pi X) + (1-p) \times v(W - \pi X)$

To get the optimum amount, find the turning point: So find ${\partial v\over\partial X}=v'(\bullet)$

So $p(1-\pi) \times v'( W-I+X - \pi X) - \pi^2(1-p) \times v'(W - \pi X)=0$

Thus, $\frac{p \times v'( W-I+X - \pi X)}{(1-p) \times v'(W- \pi X)} = \frac{\pi}{(1-\pi)}$

1. LHS is individual's trade-off between Marginal Utility if Loss occurs and Marginal utility if Loss DOESNT Occurs.
2. RHS is insurer's trade-off between loss and no loss.
3. If $\pi = p$, then Risk Averse individual will choose $X=I$

Side Note: How do we know whether the turning point is in fact a Maximum Turning Point?

We need to show that $\partial v \over \partial X < 0$

So, $p(1-\pi)^2 \times v''(W-I+X- \pi X) + \pi^2 (1-p) \times v''(W- \pi X) < 0$

## Common Utility Functions Edit

1. $v(w) = w-\tau w^2$
2. $v'(w) = 1- 2\tau w$
3. $v''(w) = -2\tau$

Exponential Utility F(x)

1. $v(w) = -e^{\alpha w}$
2. $v(w) = \alpha e^{-\alpha w} > 0$
3. $v(w) = - \alpha^2e^{-\alpha w} < 0$

## Time Preference Edit

Now, how do we consider the time value of money?

1. $C_0$ is consumption
2. This means $W-C_0$ is used for investment
3. Now invest: $C_1=[W-C_0](1+r)$
4. Since $W = C_0 + (W-C_0)$

Then, $W=C_0 + \frac{C_1}{(1+r)}$

Thus, $W=U(C_0)+U(C_1)$

To find maximum utility, find ${\partial U(C_n)\over\partial C_0}$

So, $\frac{U'(C_1)}{U'(C_0)}=\frac{1}{(1+r)}$

Eg: $U(w)=-e^{-0.005w}$ If Individual has $W=100,000$ and $r=8%$, find optimal consumption.

$\frac{1}{1.08}=\frac{U'(C_1)}{U'(C_0)}$ $U'(w)=0.005e^{-0.005w}$

Knowing $C_1=[W-C_0](1+r)$ Then, $\frac{1}{1.08}=e^{-540+0.0104C_0}$

Thus, $C_0(max)=51,915.68$

So how are the $r$ or interested determined?

1. Determined by marginal Utility of additional consumption in future
2. Market IR is det by D for Investment. Reflects preferences of individuals for future vs current consumption.


text:
Inverse Lorentz transformations

Let's say we are in a frame of reference S and we see a clock moving at velocity $c/2$. Let's say that a clock in our frame of reference reads $t=0$ when the moving clock reads $t'=0$. I want to then find out what time our clock will read when the moving clock reads $t=100$.

Here's what I did: We have $$t=\gamma\left(t'+\frac{vx'}{c^2} \right)$$ and we know that $t'=100$, $v=c/2$ and I suppose that $x'=vt'=ct'/2$ and so to give $$t=\gamma\left(t'+\frac{c^2t'}{4c^2} \right)=\frac{5}{4}\gamma t'$$ so here's the problem: shouldn't it be simply $$t=\gamma t'$$ what did I overlook? Thank you.

$x'$ and $t'$ is the value measured in $S'$-frame.(i.e. frame moving at velocity $c/2$ in $S$-frame) So, $x' = 0$ in $S'$-frame.
More precisely, we should assume that $x'=0$ when $t'$ is 0


text:
Friedrich Hirzebruch passed away at the age of $$84$$ on  May$$27$$, $$2012$$.

Hirzebruch was born on  October$$17$$, $$1927$$ in Germany, and a leading figure in his generation.  他被认为是德国战后最重要的数学家。

For his manifold achievements Friedrich Hirzebruch received a number of awards and prizes. Among others, the Grand Merit Cross with Star of the Federal Republic of Germany, the Wolf Prize for Mathematics, the Seki Takakazu Prize, the Lomonossov Gold Medal, the Albert Einstein Medal, and the Georg Cantor Medal of the Deutsche Mathematikervereinigung. He held honorary doctorates from over 14 universities. He was member of a number of Academies of Science and of the Order pour le mérite.

“With Friedrich Hirzebruch, mathematics has lost one of his truly great personalites. He was a wonderful human being and an eminent researcher whose contributions have shaped the entire field”, said Peter Teichner, managing director of the Max Planck Institut for Mathematics in Bonn. “Our institute, which he founded, will always remain his institute, too.”


text:
ecm {ebGenotyping} R Documentation

## Genotyping and SNP Detection using Next Generation Sequencing Data

### Description

This function implements the method described in 'An Empirical Bayes Method for Genotyping and SNP detection Using Multi-sample Next-generation Sequencing Data'.

### Usage

ecm(dat,cvg,eps=1e-6,max.steps=500,eps.bisec=1e-6,ini.m=-7,ini.d=-7)

### Arguments

 dat a n*m matrix: the ith row, jth column of the matrix represents the non-reference counts of ith sample at jth position. cvg a n*m matrix: the ith row, jth column of the matrix represents the depth of ith sample at jth position. eps a single value: a threshold to control the convergence criterion. The default is 1e-06. max.steps a single value: the maximum steps to run iterative algorithm to estimate parameters. The default is 500.(Adjustment is needed according to the number of parameters to estimate and the initial value of them.) eps.bisec a single value: a threshold to control the convergence criterion of bisection criterion. The default is 1e-06. ini.m the initial value of each element of mu. We suggest users to use default -7. ini.d the initial value of each element of delta. We suggest users to use default -7.

This function implements the method described in 'An Empirical Bayes Method for Genotyping and SNP detection Using Multi-sample Next-generation Sequencing Data'. According to the paper, users can do genotyping with the estimated genotypes("geno.est"), and do SNP detection with the posterior probabilities of RR("post.probs$zRR"), based on the non-reference counts("dat")and depth("cvg"). ### Value  par.est a list including the estimate of position effect(mu), sample effect(delta), and the probability of RR and RV. post.probs 3 matrix: the estimate of the posterior probabilities of 3 genotypes for n samples at m positions. steps the total steps to run iterative algorithm. geno.est a n*m matrix: the estimated genotypes(0 for RR, 1 for RV and 2 for VV) of n samples at m positions. ### Author(s) Na You <youn@mail.sysu.edu.cn> and Gongyi Huang<53hgy@163.com> ### References Na You and Gongyi Huang.(2016) An Empirical Bayes Method for Genotyping and SNP detection Using Multi-sample Next-generation Sequencing Data. ### Examples #---------generate simulation data----------- #start:generate simulation data# set.seed(2016) m <- 100 m0 <- m*0.95 m1 <- m-m0 n <- 30 Q <- 0.8 z <- cbind(matrix(0,n,m0),matrix(rbinom(n*m1,1,Q),n,m1)) b <- which(z==1) R <- 0.8 # proportion of homozygous SNP w <- rbinom(length(which(z==1)),1,R) # z are genotypes z[b[which(w==0)]] <- 1 z[b[which(w==1)]] <- 2 mu <- rep(-3,m)# stands for no effect delta <- rep(-3,n)# stands for no effect er.p <- -abs(outer(delta,mu,"+")) p <- rlogit(er.p) p[which(z==1)] <- 1/2 p[which(z==2)] <- 1-p[which(z==2)] cvg <- matrix(rbinom(m*n,50,0.5),n,m) dat <- matrix(sapply(1:(m*n),function(i) rbinom(1,cvg[i],p[i])),n,m) #end:generate simulation data-# #-----genotyping and SNP detection---------------- res <- ecm(dat=dat,cvg=cvg) mean(z!=res$geno.est)#genotyping error
#----------call SNP---------
#start:call snp#
# define a function to calculate power, typeI error and FDR.
cutsnp <- function(fdr,alpha,true){
wh <- (true!=0)
tp <- sum((wh)&(fdr<alpha));
tn <- sum((!wh)&(fdr>=alpha));
fp <- sum((!wh)&(fdr<alpha));
fn <- sum((wh)&(fdr>=alpha));
pw  <- tp/(tp+fn);
t1 <- fp/(fp+tn);
fdr <- fp/(fp+tp);
return(c(TP=tp,TN=tn,FP=fp,FN=fn,power=pw,typeI=t1,FDR=fdr))
}
cutsnp(fdr=res$post.probs$zRR,alpha=0.001,true=z)
cutsnp(fdr=res$post.probs$zRR,alpha=0.01,true=z)
cutsnp(fdr=res$post.probs$zRR,alpha=0.05,true=z)
#end:call snp#

[Package ebGenotyping version 2.0.1 Index]


text:
# Amazing Bonus. 9. 9.74 All of the following methods can be used to prepare 2-butanone (CH3COCH2CH3)...

###### Question:

Amazing Bonus. 9. 9.74 All of the following methods can be used to prepare 2-butanone (CH3COCH2CH3) EXCEPT: HgSo H2SO..H20 1) RBH 2) H20, NaOH 1) R28H 2) H2O2. NaOH

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text:
Signal with Complex Gaussian noise

If I have:

$$y = x_r+jx_i + n_r +j n_i$$

with $$n_r$$ and $$n_i$$ Gaussian with mean 0 and variance $$\sigma^2$$, what is the pdf of the envelope |y| and phase(y)? Is it still Rayleigh-distributed and uniform, respectively?

• um, something's wrong with your equation: it says $y=jy+\text{other stuff}$, and I don't think that's what you meant. May 5 '20 at 7:07
• Am I right to assume that you define $x$ is a deterministic thing, and you only consider $n$ to be random? May 5 '20 at 7:46

2 Answers

No, it will be Rician Distribution.

Rayleigh distribution is a special case of Rician Distribution when the normal random variables involved are of zero mean and equal variances.

I have assumed you meant :$$y=x_r + j \cdot x_i + n_r + j \cdot n_i$$ where $$x_r, x_i$$ are deterministic variables.

Basically, you now have $$y_r = (n_r + x_r),\ \sim \mathcal N(x_r, \sigma^2)$$ and $$y_i = (n_i + x_i), \ \sim \mathcal N(x_i, \sigma^2)$$

Now, you can write $$y = y_r +j \cdot y_i = r \cdot e^{j\phi}$$, where $$r,\phi$$ are random variables and, $$y_r = r \cdot \cos(\phi)$$ and $$y_i = r \cdot \sin(\phi)$$.

Using method of transformations, Jacobian matrix can be given as follows: $$\begin{pmatrix} y_r \\ y_i \end{pmatrix} = \begin{pmatrix} r \cdot \cos(\phi) \\ r \cdot \sin(\phi)\end{pmatrix}$$ $$\mathbf J = \begin{pmatrix} \cos(\phi)&&-r \cdot \sin(\phi) \\ \sin(\phi)&& r \cdot \cos(\phi)\end{pmatrix}$$ $$|\mathbf J| = r$$ Hence, you see that Jacoboan does not change by changing the mean of the normal random variables from 0 to non-zero.

Joint PDF of derived RVs $$r, \phi$$ will be given by the following : $$f_{R,\Phi}(r, \phi) = f_{Y_r, Y_i}(r\cos(\phi), r\sin(\phi)) \cdot |\mathbf J |$$ $$f_{R,\Phi}(r, \phi) = \frac{r}{2\pi \sigma^2} \cdot \exp\left[-\frac{(r\cos(\phi) - x_r)^2+(r\sin(\phi)-x_i)^2)}{2\sigma^2}\right]$$ $$\forall r\in [0,\infty) \\ \phi \in[-\pi, \pi]$$

Solve for Marginal PDF of $$R$$ and $$\Phi$$ to see that it is not Rayleigh anymore. But a more generalized Rician Distribution.

• Also, $\Phi$ is not uniform but is peaked in the vicinity of the mean location, and $R$ and $\Phi$ are not independent. May 5 '20 at 21:14
• Hay, DSP Rookie, this is a friendly request. Please use \cdot instead of . And try to remember the backslash \ preceding function names. May 19 '20 at 2:08
• and you don't need to use so many dots for multiplication. we know when two expressions sit side-by-side, they're meant to be multiplied. i would suggest leaving the dot out except for dot product and maybe some expression of inner product (but that would be crappy notation). May 19 '20 at 2:12
• @robertbristow-johnson Sure, will keep in mind. Thanks. May 19 '20 at 10:13

The envelope |y| follows

The phase of y follows the PDF given in On the envelope and phase distributions for correlated gaussian quadratures.