Expected value of $\max(x,y) $ given that $x<y$

Question

What is the expected value of $\max(X,Y)$ given that $X<Y$? $X$ and $Y$ are independent and exponential random variables.

Answer 1

$$\begin{align} \mathsf E(\max(X,Y)\mid X<Y) & = \mathsf E(Y\mid X<Y) \\[1ex] & = \int_0^\infty\int_0^\infty y\, f_{X,Y}(x,y\mid X<Y)\operatorname d x\operatorname d y \\[1ex] & = \int_0^\infty\int_0^\infty y\cdot \frac{f_{X,Y}(x,y)\mathbf 1_{x<y}}{\mathsf P(X<Y)}\operatorname d x\operatorname d y \\[1ex] & = \frac{\int_0^\infty\int_0^y y\,f_{X,Y}(x,y)\operatorname d x\operatorname d y}{\mathsf P(X<Y)} \\ & \ddots \end{align}$$

The rest is left to you.

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PS:

$\mathbf 1_{x<y}$ is an indicator function; having the value of one when the subscript is true, and zero elsewhere.

$$\mathbf 1_{x,y}=\begin{cases}1 & : x<y \\ 0 & : \text{elsewhere}\end{cases}$$