Conditional density function of a sum of two independent random variables

Question

I want to find the conditional density function $f_{X+Y|X>Y}$. Where $X,Y \sim N(0,1)$ and independent.

This is what I've tried so far:

Let $Z=X+Y$, we can calculate the cumulative distribution $F_{Z|X>Y}(z) = \frac{P(Z\leq z, X>Y)}{P(X>Y)}$.

Now since $X,Y$ are i.i.d $P(X>Y)=\frac{1}{2}$,

so $F_{Z|X>Y}(z) = 2 P(Z \leq z, X>Y)$.

How do I go from here? Not even sure it's the right thing do either, is there a better way?

Answer 1

Since $X,Y$ are i.i.d $N(0,1)$, we have $X+Y\sim N(0,2)$.

Now, $$P(X+Y\le x)=\frac{1}{2}\left[P(X+Y\le x\mid X>Y)+P(X+Y\le x\mid X<Y)\right]$$

Let $\Phi$ and $\phi$ be the CDF and PDF of $N(0,1)$. And since $X,Y$ are i.i.d, $X+Y\mid X>Y$ has the same distribution as $X+Y\mid X<Y$. The above then reduces to

$$\Phi\left(\frac{x}{\sqrt 2}\right)=P(X+Y\le x\mid X>Y)$$

So the density of $X+Y\mid X>Y$ for all real $x$ is

$$f_{X+Y\mid X>Y}(x)=\frac{1}{\sqrt 2}\phi\left(\frac{x}{\sqrt 2}\right)=\frac{1}{2\sqrt \pi}e^{-x^2/4}$$