pdf of exponential Random Variable.

Question

I'm studying for my exam and confronted with a problem.

Suppose we have two RV's $X_1$ and $X_2$ both are exponential. Now we define $Y=X_1+X_2$ and trying to find pdf of $Y$.

I have this equation but I don't know how do we derive the right side from the left side. $$=\int P( X_2 \leq x - X_1 ).\lambda e^{-\lambda X_1} dx \\=\int (1-e^{-\lambda(x-X_1)} ).\lambda e^{-\lambda X_1} dx $$ The complete proof is this: Sum of two exponential random variables

I appreciate for answers in advance and I am sorry if my question is not good enough according to the rules.

Answer 1

That step uses the fact that $X_2$ follows an exponential distribution with parameter $\lambda$. The cdf of an exponential distribution is well known and equals $F(y) = 1-\exp(-\lambda y)$. That means $P(X \leq y) = 1-\exp(-\lambda y)$. If you plug in $y=x-X_1$ you obtain the desired result.