I have alredy proved: $f, g$ two density functions. Prove $h(x)=$$\int_{-\infty}^{\infty} g(x-y)f(y) dy$ define a new density function.
Then is asked: $f, g$ are probability density functions of an exponential distribution, prove h is $\gamma (\lambda ,2)$.
I tried: $g(x-y)f(y)=\lambda e^{-\lambda (x-y)}\lambda e^{-\lambda y}=\lambda^{2}e^{-\lambda x}1_{[0,\infty)}(x)$ and $\gamma (\lambda ,2)=\lambda^{2}x e^{-\lambda x}$.
The distribution is $H=\int_{-\infty}^{x} h(x)dx=\int_{-\infty}^{x} (\int_{-\infty}^{\infty} g(x-y)f(y)dy)dx=\int_{-\infty}^{x}(\int_{-\infty}^{\infty}\lambda^{2}e^{-\lambda x}1_{[0,\infty)}dy)dx $.
I've to do $\int_{-\infty}^{\infty}\lambda^{2}e^{-\lambda x}1_{[0,\infty)}dy=\gamma (\lambda ,2)$. I think I'm wrong because none integral gives me $\gamma (\lambda ,2)=\lambda^{2}x e^{-\lambda x}$
Best Answer
For $x\leq0$ notice that $g\left(x-y\right)f\left(y\right)=0$ for every $y$.
For $x>0$ notice that $g\left(x-y\right)f\left(y\right)=0$ if $y\notin(0,x)$.
So for $x\leq 0$ we find that $h(x)=0$ and for $x>0$ we find:$$h(x)=\int g\left(x-y\right)f\left(y\right)dy=\int_{0}^{x}\lambda e^{-\lambda\left(x-y\right)}\lambda e^{-\lambda y}dy=\lambda^{2}e^{-\lambda x}\int_{0}^{x}dy=\lambda^{2}xe^{-\lambda x}$$