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bio website msemac.redwoods.edu/~darnold/…
location Eureka, California
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visits member for 1 year, 11 months
seen Dec 4 at 5:17

Jun
20
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
Jun
19
comment Existence of kth moment
$|-2|>1$, but $|-2+3|$ is not less than $4(-2)$.
Jun
19
comment Existence of kth moment
And that's only to the right of $x=1$.
Jun
19
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
Jun
19
comment $X<Y$ implies $E[X]<E[Y]$?
Nice replies everyone, very, very helpful.
Jun
18
comment $X_1$, $X_2$ independent implies $e^{tX_1}$, $e^{tX_2}$ independent
Absolutely wonderful. All of these answers are useful. Haven't got to the Jacobian yet, but it lies here to help me when I get there. Thanks for the help.
Jun
18
comment Existence of kth moment
Hmmm... If the integrand goes to zero as x gets large, the integral is not necessarily finite (e.g., $\int_1^{\infty}1/\sqrt x\,dx$). So, short answer, no, but your detailed solution is amazing, absolutely amazing. Great answer!
Jun
15
comment x in the range of a random variable $X$ implies the pdf $f_X(x)>0$
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.
Jun
13
comment Inverse of partitioned matrix, checking result
Very nice move.
Jun
13
comment Inverse of partitioned matrix, checking result
Excellent suggestion, and it worked.
Jun
12
comment Marginal pdf $f_2(y)$ is proportional to $g_2(y)$.
Very helpful. Thanks.
Jun
8
comment If F is a cumulative distribution function, then $\lim_{x\to\infty}F(x)=1$
Agree, making a correction.
Jun
6
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 $y=v$?
Jun
6
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.
Jun
6
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 $y=x^2$.
Jun
3
comment What is a bounded discrete random variable
And this means that the normal random variable Z is unbounded?
Apr
13
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?
Jul
13
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.
Jul
13
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 $E\subset X$.
Jul
9
comment Constructing a complete measure space
In your latest proof, why is $\mu(U\cup G)=0$?