3
$\begingroup$

The question states:

A random variable $X$ is called symmetric about 0 if for all $x \in \mathbb R$, $\mathbb P(X \geq x) = \mathbb P(X \leq -x)$.

Prove that if $X$ is symmetric about 0, then for all $t > 0$, its distribution function $F$ satisfies the following relations: (I'm only going to give one example so I can do the rest myself)

a) $\mathbb P(|X|\leq t) = 2F(t)-1$.

How do I prove this?

And also what does it mean that the random variable $X$ is symmetric about 0?

  • 1
    It's ok. What properties about probability do you know? For example, do you know that $\mathbb P(A) = 1 - \mathbb P(A^c)$ for any event $A$? Do you know that if $A$ and $B$ are disjoint, then $\mathbb P(A \cup B) = \mathbb P(A) + \mathbb P(B)$? I've shown you the relationship between $F(t)$ and $\mathbb P(X \leq t)$. Try to use properties similar to the above by rewriting the event $\{|X| \leq t\}$ in different ways where you can use these properties. :-)2012-10-10

1 Answers 1

1

Try this:

$ LHS=P(|X| \leq t)=P(-t \leq X \leq t)=F(t)-F(-t) $ by symmetry, $F(-t)=P(X \leq -t)=P(X \geq t)=1-F(t)$, hence $ LHS=F(t)-1+F(t)=2F(t)-1=RHS $

  • 0
    @cardinal: thanks, I sort of assumed $X$ is absolutely continuous2012-10-10