I'm struggling to prove the following:
Let $X,Y,Z$ be iid random variables (with pdf $f$) that are unimodal and symmetric around 0. Then $X \mid (X = Z)$ is more peaked than $X \mid \left(\tfrac12 X + \tfrac12 Y = Z \right)$.
Definitions:
- A random variable with probability distribution function $f$ is unimodal and symmetric around 0 iff $f(t)=f(-t)$ and $f(t)$ is decreasing for $t\geq 0$.
- Random variable $A$ is more peaked than $B$ iff $$\mathrm{P}(|A|<\alpha) \geq \mathrm{P}(|B|<\alpha)$$ for all $\alpha >0$.
I tested the hypothesis numerically on randomly generated pdfs $f$, but I have difficulty proving it. I found that the assumption that $X$ and $Y$ have the same distribution is essential. I also realized that it generally is not true that $X\mid(X=U)$ is more peaked than $X\mid (X=V)$ whenever $U$ is more peaked than $V$ (where $U$ and $V$ are two unimodal distributions symmetric around 0 and independent of $X$).
Is there a known technique to solve this type of problems?
