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Suppose $f$ and $g$ are PDFs of real-valued random variables. Show that the convolution $f\ast g$ of $f$ and $g$ (defined below) is also a PDF. $$(f\ast g)(x)=\int_{-\infty}^\infty f(y)g(x-y)\,dy.$$

Firstly, I am not certain exactly what I need to show. In the text, we started with a random variable and defined from that the PDF, and then observed that it had the property that its integral over $\Bbb{R}$ was $1$. For this reason I am thinking that if I can show $\int_\Bbb{R} (f\ast g)(x)\,dx=1$, then I will be done.

Would the following be sufficient to prove that the convolution is a PDF? \begin{align} \int_{-\infty}^\infty\int_{-\infty}^\infty f(y)g(x-y)\,dy\,dx &= \int_{-\infty}^\infty f(y)\bigg(\int_{-\infty}^\infty g(x-y)\,dx\bigg)\,dy\\ &= \int_{-\infty}^\infty f(y)\cdot1\,dy =1, \end{align} Using that $f$ and $g$ are PDFs in the bottom line.

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Your proof is correct. You just need to justify the change in order of integration using Tonelli's theorem (the integrand is non-negative).

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  • $\begingroup$ Ahh thanks. And I guess I should also remark that the convolution is non-negative, as that's a necessary property of PDFs $\endgroup$
    – Szmagpie
    Jun 10 '16 at 6:45

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