How to estimate E$[\alpha-\alpha']$ and Var$[\alpha-\alpha']$ conditioned on $\alpha>\alpha'>0$ where $\alpha$ and $\alpha'$ are iid standard normal?

Question

Let $\alpha$ and $\alpha'$ be independent standard normal random variables, how to estimate the conditional mean $\mathbb{E}[\alpha-\alpha'|\alpha > \alpha' >0]$ and variance Var$[\alpha-\alpha'|\alpha > \alpha' >0]$? (a bound would be ok) It's clear that, due to $\alpha > \alpha'$, $\alpha-\alpha'$ is a normal random variable sampled from the positive part, but I am not sure how to deal with the condition $\alpha'>0$.

Answer 1

Let's rewrite the problem with $X$ and $Y$ instead of $\alpha$ and $\alpha'$. We need to calculate $E(X-Y|X>Y>0)$ and $E((X-Y)^2|X>Y>0)$ (because $\text{Var}(X-Y|X>Y>0) = E((X-Y)^2|X>Y>0) - (E(X-Y|X>Y>0))^2 $).

We have $$ \begin{align} E(X-Y|X>Y>0) &= \frac{E((X-Y)\mathbb{I}_{\{X>Y>0\}})}{P(X>Y>0)}\\ &=8E((X-Y)\mathbb{I}_{\{X>Y>0\}}) \\ &=8\int_{x>y>0}{\frac{x-y}{2\pi}\exp{(-\frac{x^2+y^2}{2}})}dxdy \\ \end{align} $$ Make a change of varibles $x=r\cos(t)$ and $y=r\sin(t)$ then

$$ \begin{align} E(X-Y|X>Y>0) &= 8\int_{r>0,0<t<\frac{\pi}{8} }{\frac{r(\cos(t)-\sin(t))}{2\pi}\exp{(-\frac{r^2}{2}})}rdrdt \\ &= \frac{4}{\pi}\int_{r>0}r^2\exp{(-\frac{r^2}{2}})dr \int_{0<t<\frac{\pi}{8}}(\cos(t)-\sin(t))dt \\ &= \frac{4}{\pi}\int_{r>0}r^2\exp{(-\frac{r^2}{2}})dr \int_{0<t<\frac{\pi}{8}}(\cos(t)-\sin(t))dt \\ &= \frac{4}{\pi}*\sqrt{\frac{\pi}{2}} *(-1 + \frac{\sqrt{2 - \sqrt{2}}}{2} + \frac{\sqrt{2 + \sqrt{2}}}{2}) \\ \end{align} $$

Now, we work with the second term $E((X-Y)^2|X>Y>0)$, we have $$E((X-Y)^2|X>Y>0) = \frac{E((X-Y)^2\mathbb{I}_{\{X>Y>0\}})}{P(X>Y>0)} = 8E((X-Y)^2\mathbb{I}_{\{X>Y>0\}})$$ But by symmetry, we have $E((X-Y)^2\mathbb{I}_{\{X>Y>0\}}) = \frac{1}{8}E((X-Y)^2)$, then $$E((X-Y)^2|X>Y>0) = E((X-Y)^2)$$ It's obvious that $$E((X-Y)^2) = E(X^2) +E(Y^2) = 1+1=2$$ Hence, $E((X-Y)^2|X>Y>0) = 2$.