Problem with counting conditional expectation with PDF

Question

Let $(X,Y)$ be a random variable with PDF :

$$f(x,y)=\begin{cases} e^{-x-y} &if\;x,y\ge0 \\ 0&in \;other\;case \end{cases}$$

I want to calculate $E(X+Y|X<Y).$

$E(X+Y|X<Y)=E(X|X<Y)+E(Y|X<Y).$

Firstly let's count $P(X<Y)$.

$P(X<Y)=\int_0^{\infty}(\int_0^{y}e^{-x-y}dx)dy=\frac12.$

$E(X|X<Y)=\frac{1}{P(X<Y)}\cdot \int_{\{X<Y\}}X\;dP$

And now i have problem with meaning of $\int_{\{X<Y\}}X\;dP$. Im not sure what the exactly it is. I would say that it's some double integral of probability density function but i don't know what the upper and lower integration limit should be. Can you please help me with this ?

Answer 1

Shortcut:

The PDF betrays that $X$ and $Y$ are iid continuous random variables.

Then $\mathbb E[X+Y\mid X<Y]=\mathbb E[X+Y\mid Y<X]$ and $P(Y<X)=P(X<Y)=\frac12$

Now make use of:$$\mathbb E[X+Y]=\mathbb E[X+Y\mid X<Y]P(X<Y)+\mathbb E[X+Y\mid Y<X]P(Y<X)$$

Answer 2

It's useful to use that integral over some set is equal to integral over the entire space of function multiplied by indicator of this set. So $\int_{\{X < Y\}} X dP = \int X \cdot \mathbb{I}_{X < Y} dP$. And $X \cdot \mathbb{I}_{X < Y}$ is a function of $X$ and $Y$, so we can find it's expectation by integrating it's product with joint density over $\mathbb{R}^2$:

$\int X \cdot \mathbb{I}_{X < Y} dP = \int_\mathbb{R_+^2}\ dx \ dy\ e^{-x -y} x \cdot \mathbb{I}_{x < y} = \int_0^\infty\ dy \int_0^y\ dx\ e^{-x - y}$