Dilip Sarwate

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I am a retired professor of electrical and computer engineering with a lifetime of experience teaching probability and statistics to reluctant engineering undergraduates.

	4,931 reputation	bio	website		visits	member for	1 year, 7 months
1921 badges		location	Urbana, IL		seen	12 hours ago

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13h	comment	Identifying argmax result in information theory Is the result you need something like: If $p_i, q_i \geq 0$ and $\sum_{i=1}^n p_i = \sum_{i=1}^n q_i = 1$ are two different pmfs, then $$\sum_{i=1}^n p_i \log p_i > \sum_{i=1}^n p_i \log q_i\quad ?$$ If so, read about Kullback-Liebler divergence and the Gibbs inequality.
22h	comment	How can I visualise a distribution that is univariate normal but bivariate non-normal? See this fantastic answer for a huge collection of bivariate nonnormal distributions that are marginally normal.
May 19	comment	How to compute expectation for poisson distributed variable in context of binomial? Actually, I assumed that you had just copied what Mathematica had produced, and was wondering why Mathematica had not used the simpler form.
May 18	comment	How to compute expectation for poisson distributed variable in context of binomial? Interesting! I wonder why the fourth argument is not written as $\frac{p}{1-p}$.
May 18	answered	How to compute expectation for poisson distributed variable in context of binomial?
May 15	comment	Combining covariances Put another way, if you simply merged the two samples into a single sample $(n=50)$, the covariance that you will compute for the merged sample will typically not work out to be $0.12$.
May 15	comment	Combining covariances You need to take the means of the two samples into account. Consider the simpler case of one variable. If different samples give different (conditional) means and variances, then the combined variance is the (weighted) average of the conditional variances plus the variance of the conditional mean. A similar effect will arise with covariance: the two covariances are likely computed using different means, $(\mu_{X,2},\mu_{Y,2})$ instead of $(\mu_{X,1},\mu_{Y,1})$, and so just a weighted sum of the covariances will not do either.
May 12	comment	What is the Mean and Standard Deviation of the division of two random variables? The mean of $Z = \frac{X}{Y}$ is not $\frac{\mu_X}{\mu_Y}$. Whatever gave you that idea? Even for one random variable, $E[g(X)] \neq g(E[X])$ in general. This is a fundamental notion that you would do well to learn well. Engrave it on your heart in letters of gold....
May 11	answered	How to show that $E[(\hat\theta -\theta)^2]<Var(\bar X)=\dfrac{1}{n}$?
May 10	revised	How to show that $E[(\hat\theta -\theta)^2]<Var(\bar X)=\dfrac{1}{n}$? corrected typo in definition of $\hat{\theta}$
May 10	comment	How to show that $E[(\hat\theta -\theta)^2]<Var(\bar X)=\dfrac{1}{n}$? First of all, get your notation correct. You are asked to show that $E[(\hat{\theta}-\theta)^2] < \operatorname{var}(\bar{X})$ and not $E[\hat\theta -\theta]^2$ as you have it. Your expression would be taken to mean $\left(E[\hat\theta -\theta]\right)^2$ the square of the mean, whereas what is wanted is the mean of the square. Then argue that the difference between the distributions of $\bar{X}$ and $\hat{\theta}$ is that all the probability mass to the left of $\theta_0$ (to the right of $\theta_1$) has been replaced by a point mass at $\theta_0$ (at $\theta_1$) and this reduces the variance.
May 9	comment	Function with variable having gaussian distribution I changed the placement of the word "only" in the sentence about nonlinear transformations possibly resulting in normal random variables. I think it makes the sentence read better. Please put back the way it was if you don't think so too.
May 9	revised	Function with variable having gaussian distribution changed placement of one word towards the end
May 9	comment	Function with variable having gaussian distribution The answer to the question "Is the result actually a Gaussian distribution?" is that $f(X)$ is exactly (as opposed to approximately or asymptotically) a Gaussian random variable whenever $f(X)=aX+b$ for constants $a$ and $b$ (which includes as a special case a degenerate Gaussian random variable with variance $0$ (commonly called a constant in non-statistical non-probabilistic circles) when $a=0$ as in the comment by @gung)
May 9	comment	Function with variable having gaussian distribution @whuber Thanks for pointing out the erroneous statement. I remember the counterexample but it was not at the forefront of my mind. I am deleting my previous comment and replacing it with a corrected one.
May 9	comment	Function with variable having gaussian distribution You are welcome. However, please note that @whuber has just pointed out in a comment on the main question that "if and only if" is incorrect; it should be "if" only.
May 9	comment	Function with variable having gaussian distribution Perhaps you should add a fourth case right after the approximate normality of $X^2$ pointing out that the distribution is exactly normal if and only if $f(X) = aX+b$.
May 9	comment	Underlying physical basis of an exponential distribution Perhaps physics.SE would be a better place for this question which doesn't seem to have much to do with statistics. You can ask the moderators to migrate it to the other site by clicking on the `flag` link to make the request.
May 7	comment	Asymptotic probability concerning the largest absolute value in an iid Gaussian sample @Tarantula Thanks. Glad to have been of help. With regard to your last query, I just followed up on Moderator whuber's comment which was there already and had been emended slightly by me. I could have used $\{v_\max > z\}$ from the beginning, and that would have been my inclination if I had been answering ab initio, but I wanted to link up matters to what had been said before.
May 7	comment	Show that $G(x)$ is a distribution function and find mean @whuber Thanks for pointing out the mistake. I was mixing up $\mu^{-1}[1-F(t)]$ which is a valid pdf and $G(x)$ which is a valid CDF. I have deleted the incorrect statements. But I do think that for $x < 0$, setting $$G(x)=\frac{1}{\mu}\int_0^x [1-F(t)]dt = -\int_x^0 [1-0]\,\mathrm dt = x$$ as is seemingly implied by the problem is incorrect.