We have $X\sim \operatorname{Exp}(\lambda_1), Y\sim \operatorname{Exp}(\lambda_2)$. I need to find out $P\{Y>t|X>Y\}$ and $E[Y|X>Y]$. I am wondering the correctness of my approach.
For the first part, since I have found that $P\{X>Y\} = \dfrac{\lambda_2}{\lambda_{1}+\lambda_{2}}$. Can I write $$P\{Y>t|X>Y\} \\ = \dfrac {P\{X>t\}}{P\{X>Y\}}\\ $$
And for the expectation, can I just compute $$P\{Y|X>Y\}$$ then use the definition of expectation, that is $$\int_{0}^{\infty}yP\{Y|X>Y\}$$ Is there any faster way?
$P(Y>t|X>Y) = \dfrac {P(X>t)}{P(X>Y)}$ is not correct.
Rather, $P(Y>t|X>Y) = \dfrac {P(X \gt Y>t)}{P(X>Y)}$
Here is how I would approach this. Knowing $X$ and $Y$ are independent,
$ \displaystyle f_{Y | X \gt Y} (y|x\gt y) = \frac{\int_y^{\infty} f(x) f(y) ~ dx}{\int_0^{\infty}\int_y^{\infty} f(x) f(y) ~ dx ~ dy}$
$ \displaystyle = (\lambda_1 + \lambda_2) e^{-(\lambda_1 + \lambda_2)y} ~$, which is an exponential distribution with mean $(\lambda_1 + \lambda_2)$ so $E(Y | X \gt Y) = \dfrac{1}{\lambda_1 + \lambda_2}$.
Now, $ \displaystyle P(Y \gt t | X \gt Y) = \int_t^{\infty} f_{Y | X \gt Y} (y|x\gt y) ~ dy$
$ \displaystyle = e^{- (\lambda_1 + \lambda_2) t}$