Expectation value of symmetric random variableCan someone please help me answer this question?
From the definition of symmetry it follows that: $P(X≥b+x)=P(X≤b-x)$. What I did: $E[X]$$=$$∑$$xi$$P(X=xi)=$$∑$$(b+(xi-b))$$P(X=xi)$$=$$∑$$b$$P(X=xi)$$+$$∑$$(xi-b)$$P(X=xi)$$=$$b⋅1$ $+$ $(a-b)$$P(X=a)$ $+$ $(b-a)$$P(X=2b-a)$ $=$ $b$ due to the symmetry of the random varible $X$ since it follows that $P(X=2b-a)$$=$$P(X=a)$. Is this correct?
Thanks in advance!
To show that $EX=b$ the only information you need is the fact that the distribution is symmetric around $b$. You don't have to use the values of $X$ at all.
This follows from the fact that $X-b$ and $b-X$ have the same distribution (as you can verify from the definition of symmetry). If two random variables have the same distribution the they have the same mean (assuming that the mean exists). Hence $E(X-b)=E(b-X)$ and this gives $EX=b$.
ACCEPTED]
We have $a<0<b<2b-a$. Thus $$\mathbb{P}(X=a)=\mathbb{P}(X<b)$$and $$\mathbb{P}(X=2b-a)=\mathbb{P}(X>b)$$
By symmetry of $X$ about $b$:
$$\mathbb{P}(X\geq b)=\mathbb{P}(X\leq b)=1-q$$
for some $q\in(0,1)$
Thus $$\begin{align*}\mathbb{P}(X<b)=\mathbb{P}(X>b)&=q\\\mathbb{P}(X=b)&=1-2q\end{align*}$$
It follows that
$$\mathbb{E}[X]=qa+q(2b-a)+(1-2q)b=b.$$
