Distribution of two dependent events from three independent random variables Say that we have three i.i.d random variables $X,Y,Z$. Each has pdf $f(\cdot)$ and cdf $F(\cdot)$, and furthermore, the difference of any two (e.g. $Y-X$) has pdf $f_d(\cdot)$ and cdf $F_d(\cdot)$. 
The problem is to calculate the probability of these two events occurring:
$Pr(Y-X<c\cap Z-X<d)$
for known $c,d$.
So, we want:
$Pr(Y-X<c)\cdot Pr(Z-X<d|Y-X<c)$ 
At the moment, I'm specifically working on $X\sim N(0,1)$, so $Y-X \sim N(0,2)$, but knowing the general answer would be useful.


*

*Does this calculation depend on knowing whether $c>d$?

*What's the answer? :)


Any help would be greatly appreciated!
 A: Because $Y$ and $Z$ are independent, they are independent conditional on $X$, too: by definition, that means the conditional chance of the intersection of events is the product of their conditional chances.  Therefore, since adding $X$ to both sides of all inequalities does not change them,
$$\eqalign{
\Pr(Y-X\lt c,\ Z-X\lt d) &= \Pr(Y \lt c+X,\ Z \lt d+X) \\
&=\int_\mathbb{R}\Pr(Y\lt c+X,\ Z\lt d+X\mid X=x)\mathrm{d}F(x)\\
&=\int_\mathbb{R}\Pr(Y\lt c+X\mid X=x)\Pr(Z \lt d+X\mid X=x)\mathrm{d}F(x) \\
&=\int_\mathbb{R} F(c+x)F(d+x)\mathrm{d}F(x).
}$$
This does not simplify further in general.


*

*Evidently the order of $c$ and $d$ does not matter. (Since $Y$ and $Z$ are exchangeable, that conclusion doesn't require any calculation.)

*Even for Normal distributions this integral is likely to require numeric integration except in special cases.  For instance, with $c=d=0$ the integral clearly is $\int_\mathbb{R}\mathrm{d}\left(\frac{1}{3}F(x)^3\right)=1/3$.
A: An alternative calculation, applicable to the special case considered by the OP where $X,Y,Z$ are independent standard normal random variables (or, more generally, i.i.d. $N(\mu,\sigma^2)$ random variables for that matter) is to note
that $Y-X$ and $Z-X$ are $N(0,2\sigma^2)$ random variables with correlation coefficient 
$$\rho = \frac{\operatorname{cov}(Y-X,Z-X)}{2\sigma^2} = \frac{\operatorname{var}(X)}{2\sigma^2} = \frac 12$$
and so for this special case,
$\Pr(Y-X < c, Z-X < d)$ can be expressed in terms of the 
bivariate
normal distribution function $L(h,k,\rho)$ as 
$$\Pr(Y-X < c, Z-X < d) =L\left(-\frac{c}{\sigma},-\frac{d}{\sigma},\frac 12 \right).$$
The function $L(h,k,\rho)$ is known to have value
$L(0,0,\rho) = \frac 14+\frac{\arcsin \rho}{2\pi}$
which equals $\frac 14 + \frac{\pi/6}{2\pi} = \frac 13 ~ \text{when}~ \rho = \frac 12$
as in whuber's answer.
