How to find the mean and variance of minimum of two dependent random variables?

Question

Assume that $X$, $Y$, and $Z$ are identical independent Gaussian random variables. I'd like to compute the mean and variance of $S=\min\{P, Q\}$, where : $$Q=(X-Y)^2$$ and $$P=(X-Z)^2$$ Any help is appreciated.

Answer 1

HINT

$$ \begin{split} 1 - F_S(s) &= 1 - \mathbb{P}[\min\{P,Q\} \le s] \\ &= \mathbb{P}[\min\{P,Q\} > s] \\ &= \mathbb{P}[P > s, Q > s] \\ &= \mathbb{P}[(X-Y)^2 > s, (X-Z)^2 > s] \end{split} $$ which is a triple integral over the correct domain of $(X,Y,Z)$... can you finish it?

Answer 2

This is what I've done so far: $$F_S(s) = \mathbb{P}[\min\{P,Q\} \le s]$$ $$ = \mathbb{P}[P \le s~ \text{or}~ Q \le s] $$ $$ = \mathbb{P}[(X-Y)^2 \le s ~\text{or}~ (X-Z)^2 \le s]$$ $$ = \mathbb{P}[(X-Y)^2 \le s ]+\mathbb{P}[(X-Z)^2 \le s]$$ $$ = \mathbb{P}[-\sqrt{s} \le X-Y \le \sqrt{s}]+ \mathbb{P}[-\sqrt{s} \le X-Z \le \sqrt{s}]$$ $$ = \mathbb{P}[Y-\sqrt{s} \le X \le Y+\sqrt{s}]+ \mathbb{P}[Z-\sqrt{s} \le X \le Z+\sqrt{s}]$$ $$ =\int_{-\infty}^{\infty} \int_{y-\sqrt{s}}^{y+\sqrt{s}}f_{X,Y}(x,y)dxdy + \int_{-\infty}^{\infty} \int_{z-\sqrt{s}}^{z+\sqrt{s}}f_{X,Z}(x,z)dxdz $$

Is that correct?