Conditional probability of random variables

Question

Say I have random variables ~$Exp$ and lets call them $X$ with rate $\lambda$ and $Y$ with rate $\mu$. How do I find $\mathbb{P}\{X>Y|Y>4\}$? I know that $\mathbb{P}\{X>Y\} = \frac{\mu}{\mu + \lambda}$ but I'm not sure where to go from there.

Answer 1

As there is no answer till now let me write it down.

$F(x) = \begin{cases} \lambda e^{- \lambda x} & \text{if $x \ge 0$} \\ 0 & \text{if $x < 0$} \end{cases}$.

$F(y) = \begin{cases} \mu e^{- \mu y} & \text{if $y \ge 0$} \\ 0 & \text{if $y < 0$} \end{cases}$.

Now $X$ and $Y$ are independent and so the joint probability will be

$F(x, y) = \begin{cases} \lambda \mu e^{-\lambda x} e^{-\mu x} & \text{if $x \ge 0$ and $y \ge 0$} \\ 0 & \text{Otherwies} \end{cases}$.

$P(X > Y | Y > 4) = \frac{P(X > Y , Y > 4)}{P(Y > 4)}$

You can easily find out the value of $P(Y > 4)$ by integrating $F(Y = y)$ in $(4,\infty)$. The value will be $e^{-4 \mu}$. Please verify.

To find the value of $P(X > Y | Y > 4)$ integrate the joint probability function for $X > Y$ and $Y > 4$ (Double Integration). You shall get $\frac{\mu}{\lambda + \mu}e^{-4(\lambda + \mu)}$. Please check.

Hope it is done. Answer will be $\frac{\mu e^{-4\lambda}}{\lambda + \mu}$