# Dilip Sarwate

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bio website location Urbana, IL age member for 2 years, 2 months seen 5 hours ago profile views 1,155

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

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 6h comment What does it mean for variances to be equal? Your answer and response to my comment are mostly gobbledygook. A $N(2/3, 1/18)$ random variable has its mass spread evenly about the mean $2/3$, and mostly in the interval $(-0.04, 1.37)$. How is this comparable to a random variable $X$ with density $2x$ for $0 \leq x \leq 1$, mean $E[X]= \int_0^1 2u^2 du = 2/3$ and variance $E[X^2] - (E[X])^2 = \int_0^1 2u^3 du -\left(\frac{2}{3}\right)^2 = \frac{1}{2} - \frac{4}{9} = \frac{1}{18}$?? Now the mass is only between $0$ and $1$ and is clearly not "spread out equally" from the mean in any sense of the words. 11h comment What does it mean for variances to be equal? You mean that the mean is the same for both variables? And what does "spread out equally from the mean" really mean? Nov30 comment How to solve the expectation of continuous random variable Hint: $u=ax^2, v = -e^{-ax}, dv = ae^{-ax}$, $\int u\, dv = \cdots$ Nov29 comment Is a Stationary VAR Process with Zero Mean Gaussian Innovations a Gaussian Stationary Process? Duplicate of question already asked on dsp.SE Nov27 comment The joint pdf of two random variables defined as functions of two iid chi-square Note that $V = X+Y$, being the sum of i.i.d. $\chi^2$ random variables with $4$ degrees of freedom, is itself a $\chi^2$ random variable with $8$ degrees of freedom. Since $X$ and $Y$ are nonnegative random variables, we note that for any $\beta > 0$, conditioned on $V = X+Y = \beta$, $X-Y$ can take on values in $[-\beta,+\beta]$. Thus, $U = \frac{X-Y}{X+Y}$ takes on values in $[-1,+1]$, and this holds true regardless of the choice of positive number $\beta$. This hints at the possibility that $U$ and $V$ might be independent random variables (as is indeed shown in wolfie's answer). Nov27 comment The joint pdf of two random variables defined as functions of two iid chi-square It is interesting to see that $U$ and $V$ turn out to be independent random variables. $V$ is, of course, a $\chi^2$ random variable, while $U$ is a centered and scaled Beta random variable? Perhaps this latter too can be deduced directly? Nov23 comment Must the SD of a 'bell shaped' distribution be less than the SD of a skewed distribution with the same range? Can you think of a reason why d) is not true? Nov22 comment Expected value of an indicator function Errr.... not quite. Your answer is correct if $X$ is a continuous random variable but not if $X$ is a discrete random variable. If the CDF is defined as $F(x) = P\{X \leq h\}$, then $$E[I(x)] = \lim_{\delta \to 0} F(x+h-\delta) - F(x-h)$$ where $\delta > 0$. Some people write the limiting value as $F((x+h)^-)$ meaning the limit as we approach the point $(x+h)$ from the left. These people also say that $F(x) = F(x^+)$ and emphasize that the CDF might be discontinuous in which case the limit from the right is larger than the limit from the left. Nov21 comment Total probability theorem with normal probability density functions There is no such thing as the random variable $Y|X$; what you are saying is that the conditional density of $Y$ given the value of $X$ is a normal density. Nov20 answered Variance of product of 2 independent random vector Nov11 comment I want a book about Normal distribution and Multivariate normal distribution What is your question? Demands for full explanation are far too broad to be answered here: people have even written whole books on multivariate normal distributions. Nov10 comment How to simulate rare events with extremely low probability Search for "importance sampling" Nov10 comment Finding conditional pdf of $Z$ given $X=x$ where $Z=X+Y$ This question is a duplicate of this one that the OP asked on math.SE about one hour earlier. Nov10 revised Unsure how to calculate mean square error of a variable with a joint distribution added 644 characters in body Nov9 comment Unsure how to calculate mean square error of a variable with a joint distribution $\frac{1}{24}$ is the correct answer. Nov9 revised Unsure how to calculate mean square error of a variable with a joint distribution added 208 characters in body Nov9 answered Unsure how to calculate mean square error of a variable with a joint distribution Nov7 answered Joint cdf of extreme values Nov6 answered Flaw in a conditional probability argument Nov1 comment How can we find the decision boundary for two overlapping continuous uniform distribution? @kunfu If the threshold is $\gamma\in [c,b]$ then an $X$ is mis-classified as a $Y$ with probability $\frac{b-\gamma}{b-a}$ while a $Y$ is mis-classified as an $X$ with probability $\frac{\gamma-c}{d-c}$. Thus, the average probability of mis-classification, being a weighted sum of these two probabilities, is a linear function of $\gamma$ for $\gamma\in [c,b]$. Where do you think its minimum might be?