 pouzdrotf

2022-07-04

So, let's say that I was given two independent random variables $\xi$ and $\eta$ and was told that $\xi$ has continuous distribution (basically, $P\left(\xi =c\right)=0\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall }\phantom{\rule{thickmathspace}{0ex}}c\in \mathbb{R}$)
How can I prove that in such case their sum also will have continuous distribution? Problem that I have here is that I know that if two variables are independent then we can say ${P}_{\xi +\eta }={P}_{\xi }\ast {P}_{\eta }$, where $\ast$ stands for measures convolution operation, but I am nor really sure how to compute such a convolution. Can you provide some explanation or intuition on how to calculate such integrals? In my case I need to know integral like this:
${\int }_{{\mathbb{R}}^{2}}{\mathbb{1}}_{\left\{c\right\}}\left(x+y\right)\phantom{\rule{thickmathspace}{0ex}}d{P}_{\xi }\left(x\right)\phantom{\rule{thickmathspace}{0ex}}d{P}_{\eta }\left(y\right)$
And I have no rigorous way of showing that it is actually zero. Keegan Barry

Expert

By Fubini/Tonelli Theorem we have
${\int }_{{\mathbb{R}}^{2}}{\mathbb{1}}_{\left\{c\right\}}\left(x+y\right)\phantom{\rule{thickmathspace}{0ex}}d{P}_{\xi }\left(x\right)\phantom{\rule{thickmathspace}{0ex}}d{P}_{\eta }\left(y\right)={\int }_{\mathbb{R}}{P}_{\xi }\left(c-y\right)d{P}_{\eta }y=0$
since ${P}_{\xi }\left(c-y\right)=0$ for every $y$.

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