Finding $P[X+Y > 1, X > 1]$

Question

I'm trying to solve the following problem: i have two independent exponentially distributed r.v. $X$ and $Y$ both with $\lambda = 1$. I want to know the probability $P[X+Y > 1, X > 1]$. Since they are independent, i wrote the joint p.d.f. as $f_{X,Y}(x,y) = f_X(x) \cdot f_Y(y) = e^{-x}\cdot e^{-y}$. Then i've tried to solve $$ \int_{1}^{\infty} \int_{1 - x}^{\infty} e^{-x}\cdot e^{-y}\, dy\, dx $$ so, $$ \int_{1}^{\infty} e^{-x}\cdot e^{x - 1} dx = e^{-1} \cdot \int_{1}^{\infty} dx = \infty $$ but i get as result $\infty$. What am i doing wrong? I think the integral is correctly soved so the initial fomula of the joint may be wrong...

Answer 1

Since $X$ and $Y$ are positive random variables the event $(X>1, X+Y>1)$ is same as $X>1$. So the value is $\int_1^{\infty} e^{-x} dx=\frac 1 e$.

The mistake in your calculation is the integral w.r.t $y$ should, start from $0$ and not $1-x$ because $1-x <0$.