Finding $\mathbb{E}(X|X<Y)$ where $X,Y$ are iid $\exp(\lambda)$ random variables

Question

Question: $X,Y$ are two independent random variables following exponential distribution with mean $\lambda$. $Z$ is another random variable which takes value $1$ if $X<Y$ and $0$ otherwise. Find $\mathbb{E}(X|Z=1)$

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So basically I have to find $\mathbb{E}(X|X<Y)$ which is equal to $$\int x.f_{X|X<Y}(x)dx$$ I am having trouble finding this conditional pdf . I tried $$f_{X|X<Y}(x)=\frac{f_x(x)}{P(X<Y)}= 2f_x(x)$$ and if this is correct then the required expectation is $2\lambda$ but I am not sure if this is the correct way to approach the problem.

Any help would be very much appreciated!

Answer 1

$E(X|X<Y)=\frac 1 {P(X<Y)} E(XI_{X<Y})=\frac 1 {P(X<Y)}\int_0^{\infty} \int_0^{y} x\lambda e^{-\lambda x} \lambda e^{-\lambda y} dx dy$. Can you take it from here?.