I am facing such a problem
Let $X$ and $Y$ be independent random variables and suppose $X$ has an absolutely continuous distribution. Show $X+Y$ also has an absolutely continuous distribution.
My understanding is that, $X$ has an absolutely continuous distribution, it means that $X$ has density $f_{X}(x)$, then let $Z=X+Y$, $P(Z \leq z)$=$P(X \leq z-Y)$=$\int_{-\infty}^{z-Y} f_{X}(x)dx$. But I don't know how to proceed, since I don't know if $Y$ has density so I can't use iterated integral. Could someone help me with it?
Thanks in advance!
Best Answer
$P(X+Y \in A)=\int P(X \in A-y) dF_Y(y)$ for any Borel set $A$ because of independence. If $A$ has Lebesgue measure $0$ then $P(X \in A-y)=0$ for each $y$ because $A-y$ also has Lebesgue measure $0$. Hence $P(X+Y \in A)=0$ whenever $A$ has Lebesgue measure $0$ proving that $X+Y$ has an absolutely continuous distribution.
In fact $g(x)=\int f_X(x-y)dF_Y(y)$ is the density of $X+Y$.