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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?

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Hint: $F(t) = \mathbb P(X \leq t)$. –  cardinal Oct 10 '12 at 0:15
    
@cardinal So you're saying $\Bbb P$(|X|=<t) = 2$\Bbb P$(|X|=<t)-1?? –  TheHopefulActuary Oct 10 '12 at 0:34
    
No. It's how I wrote it. :-) –  cardinal Oct 10 '12 at 0:37
    
I'm sorry I'm just not following this.. –  TheHopefulActuary Oct 10 '12 at 0:53
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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. :-) –  cardinal Oct 10 '12 at 0:59
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1 Answer 1

up vote 1 down vote accepted

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 $$ LHS=F(t)-1+F(t)=2F(t)-1=RHS $$

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Careful! The last "equality" on your first line is not (necessarily) an equality. –  cardinal Oct 10 '12 at 1:25
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Ohhhhhhhhhhhhh now I see what I'm looking for. Very big thanks to @cardinal and Alex, thats greatly appreciated!! I was looking for a numerical anser this whole time. –  TheHopefulActuary Oct 10 '12 at 1:35
    
@cardinal: thanks, I sort of assumed $X$ is absolutely continuous –  Alex Oct 10 '12 at 1:37
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