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  • 0 posts edited
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  • 26 votes cast
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.
Jun
13
accepted Inverse of partitioned matrix, checking result
Jun
13
asked Inverse of partitioned matrix, checking result
Jun
12
accepted Marginal pdf $f_2(y)$ is proportional to $g_2(y)$.
Jun
12
comment Marginal pdf $f_2(y)$ is proportional to $g_2(y)$.
Very helpful. Thanks.
Jun
12
asked Marginal pdf $f_2(y)$ is proportional to $g_2(y)$.
Jun
8
revised If F is a cumulative distribution function, then $\lim_{x\to\infty}F(x)=1$
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