Comparing two exponential random variable and expected values

Question

I have two independent exponential r.v. X and Y with rates $\lambda_1$ and $\lambda_2$. It is known that $\lambda_1 < \lambda_2$. I'm curios about the followings

• $E[X| X<Y]$
• $E[Y| X<Y]$ (I guess = $E[Y]$)
• $E[(X-Y)| X<Y]$

What I found:

$f_{X,Y}(x,y) = \lambda_1\lambda_2 e^{-\lambda_1x - \lambda_2y}$ , since X and Y are independent.

$P\{X<Y\} = \int_0^\infty dx \int_x^\infty f_{X,Y}(x,y)$

I'm not sure whether

$E[X| X<Y] = \int_0^\infty xdx \int_x^\infty f_{X,Y}(x,y)$

If you can help me on this, I will be appreciated.

Answer 1

I think I found the solution, but it is complicated than I thought.

$X$ and $Y$ are independent exponential r.v. with $\lambda$ and $\mu$.

Start with the easy one:

$$ E[Y|Y>X] = \int_0^\infty E[Y|Y>x]f_X(x)dx$$ $$ E[Y|Y>x] = \int_0^\infty y f_{Y|Y>x}(y)dy = \int_0^\infty \frac{yf_Y(y)I_{Y>x}}{P(Y>x)}dy$$ $$=\int_x^\infty \frac{y\mu e^{-\mu y}}{e^{-\mu x}}dy = E[Y] + x = \frac{1}{\mu} + x$$ --> Because of the memoryless property.

Back to the first equation.

$$E[Y|Y>X] =\int_0^\infty E[Y|Y>x]f_X(x)dx = \int_0^\infty \bigl( \frac{1}{\mu} + x\bigl) \lambda e^{-\lambda x} dx = \frac{1}{\mu} + \frac{1}{\lambda}$$

Note that it is same thing as $E[Y| Y>X] = \int_0^\infty dx \int_x^\infty y f_{X,Y}(x,y) dy$

Others will come...