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 Jun20 comment Existence of kth moment I finally figured out your answer, another great answer, but you are right, a huge overestimate. Here is some Mathematica code comparing the actual integral and your bound for $k=6$. In[3]:= k = 6 Out[3]= 6 In[7]:= Integrate[ Abs[x]^k*Exp[-(x - 3)^2], {x, -[Infinity], [Infinity]}] // N Out[7]= 2551.67 In[8]:= Integrate[Abs[x + 3]^k, {x, -1, 1}] + 2^(2*k + 1)*(k + 1)^k // N Out[8]= 9.63783*10^8 Jun19 comment Existence of kth moment $|-2|>1$, but $|-2+3|$ is not less than $4(-2)$. Jun19 comment Existence of kth moment And that's only to the right of $x=1$. Jun19 comment Existence of kth moment Not sure why the second integral is finite. Here is what I got in Mathematica with $k=20$. Integrate[Abs[x + 3]^k*(k + 1)^(k + 1)/x^(2 (k + 2)), {x, 1, [Infinity]}] // N 1.69501*10^38 Jun19 comment $X0$ I am trying to read Casella's Statistical Inference, and up to the page I am on, he had been using $\mathcal X$ as the range of a random variable $X$. Then he says: "When the transformation is from $X$ to $Y=g(X)$, it is most convenient to use $\mathcal X=\{x: f_X(x)>0\}$. The pdf of the random variable $X$ is positive only on the set $\mathcal X$ and is 0 elsewhere. Such a set is called the support set of a distribution or, more informally, the support of a distribution." So it seems like I misunderstood that an adjustment was being made. Jun13 comment Inverse of partitioned matrix, checking result Very nice move. Jun13 comment Inverse of partitioned matrix, checking result Excellent suggestion, and it worked. Jun12 comment Marginal pdf $f_2(y)$ is proportional to $g_2(y)$. Very helpful. Thanks. Jun8 comment If F is a cumulative distribution function, then $\lim_{x\to\infty}F(x)=1$ Agree, making a correction. Jun6 comment Transform rectangular region to region bounded by $y=1$ and $y=x^2$ OK, I do see that if you fix $v$ then $u$ varies between $u=-\sqrt v$ and $u=\sqrt v$, but why does that make you pick $x=u\sqrt v$ and $y=v$? Jun6 comment Transform rectangular region to region bounded by $y=1$ and $y=x^2$ This answer appears to be correct. I'm sorry, I'm just not seeing how you did this so quickly and easily. Jun6 comment Transform rectangular region to region bounded by $y=1$ and $y=x^2$ Two questions: 1) How did you figure this out? 2) I don't think it is correct as I believe it maps the region $(-1,1)\times(0,1)$ into the region $(-1,1)\times(0,1)$ minus the region bounded by the curves $y=1$ and $y=x^2$. Jun3 comment What is a bounded discrete random variable And this means that the normal random variable Z is unbounded? Apr13 comment Case C: Euler's equation in Simmon's textbook I've adjusted my work above to reflect what I learned from your comment. Am I understanding this properly? Jul13 comment Folland, Chapter 1 Problem 17 Eric, nope, you have it incorrect. It might be because we're using different letters and it's getting you confused. Here is the definition, direct from Folland: If $\mu^*$ is an outer measure on $X$, a set $A\subset X$ is called $\mu^*$-measurable if $\mu^*(E)=\mu^*(E\cap A)+\mu^*(E\cap A^C)$ for all $E\subset X$. Now if I change my letters it would read: A set $E$ is called $\mu^*$-measurable if $\mu^*(A)=\mu^*(A\cap E)+\mu^*(A\cap E^C)$ for all $A\subset X$. Hope this helps. Jul13 comment Folland, Chapter 1 Problem 17 $B$ is the union of a countable number of disjoint $\mu^*$-measurable sets, so $B$ is measurable, meaning $\mu^*(E)=\mu^*(E\cap B)+\mu^*(E\cap B^C)$ for any $E\subset X$. Jul9 comment Constructing a complete measure space In your latest proof, why is $\mu(U\cup G)=0$?