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Let $\mathcal X$ be the range of a continuous random variable $X$. Next, let $x\in\mathcal X$. If $f_X$ is the pdf of the random variable $X$, how can I prove that $f_X(x)>0$?

David
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1 Answers1

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Let $\Omega=[0;1]$ with probability defined by the density $6x(1-x)$, i.e., for $A\subset \Omega$ we define $$P(A):=\int_A6x(1-x)dx$$

Consider random variable $X:\Omega\mapsto \Bbb R$ where $X(x)=x$. Then $X$ is continuous, its pdf is $f_X(x)=6x(1-x)$, and $f_X(0)=f_X(1)=0$ despite both 0 and 1 being in the range of $X$.

Therefore the statement is wrong.

sds
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  • 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. – David Jun 15 '14 at 19:53