468 reputation
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bio website msemac.redwoods.edu/~darnold/…
location Eureka, California
age
visits member for 1 year, 6 months
seen Jun 26 at 0:44

Jul
2
awarded  Curious
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
awarded  Informed
Jun
19
asked $E[U^2]=0$ implies $Pr(U=0)=1$
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
asked $X<Y$ implies $E[X]<E[Y]$?
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
accepted $X_1$, $X_2$ independent implies $e^{tX_1}$, $e^{tX_2}$ independent
Jun
18
asked $X_1$, $X_2$ independent implies $e^{tX_1}$, $e^{tX_2}$ independent
Jun
18
accepted Existence of kth moment
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
18
asked Existence of kth moment
Jun
15
accepted x in the range of a random variable $X$ implies the pdf $f_X(x)>0$
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
15
asked x in the range of a random variable $X$ implies the pdf $f_X(x)>0$
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.