We know that if both $X$ and $Y$ are continuously distributed with PDFs $f^X$ and $f^Y$, then $X+Y$ is continuously distributed as well with density function $f^{X+Y}(x) = \int f^X(z)f^Y(x-z)\text{ d}\lambda(z)$ where $\lambda$ denotes the Lebesgue measure. Now, since $X,Y \sim \text{Exp}(\lambda)$, $f^{X+Y}(x) = \int_{0}^{\infty} \lambda e^{-\lambda z}\lambda e^{-\lambda (y-z)}\text{ d}z = \lambda^2\int_0^{\infty} e^{-\lambda z-\lambda y+\lambda z}\text{ d}z = \lambda^2\int_0^{\infty} e^{-\lambda y}\text{ d}z$ and this integral shouldn't exist. What did I do wrong here?
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Be aware that both densities take value $0$ on $(-\infty,0]$. – drhab Apr 17 '21 at 09:18
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I know, that's why I only integrated from $0$ to $\infty$. Edit: I found my mistake. $f^Y(x-z)$ is $0$ whenever $x \leq z$. Thank you. – Nicolas Apr 17 '21 at 09:30