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  • 0 posts edited
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  • 24 votes cast
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 $\lim_{x\to\infty}F(x)=1$
added 66 characters in body
Jun
8
comment If F is a cumulative distribution function, then $\lim_{x\to\infty}F(x)=1$
Agree, making a correction.
Jun
8
asked If F is a cumulative distribution function, then $\lim_{x\to\infty}F(x)=1$
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
6
asked Transform rectangular region to region bounded by $y=1$ and $y=x^2$
Jun
3
asked The p.d.f as a derivative of the c.d.f.
Jun
3
accepted What is a bounded discrete random variable
Jun
3
comment What is a bounded discrete random variable
And this means that the normal random variable Z is unbounded?