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Say $f_X(x), f_Y(y)$ are two random independent PDF's with domains $D_X$ and $D_Y$. Let $Z = X + Y$. I know $f_Z(z) = f_X \star f_Y(z)$ where $\star$ means convolve. How do I determine how to integrate this given only the domains of the random variables?

I am trying to solve this for $X - U(0,1)$ and $f_Y(y) = a^{-y}$ but I realized I don't understand how to choose the bounds in general.

For my example I would think to do $\int_0^1a^{-(z-x)}dx$ but I don't understand how $z$ would have anything to do with this.

Anyone have any ideas?

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  • $\begingroup$ Laplace transforms give one way to evaluate convolutions. $\endgroup$ Oct 10 '18 at 1:34
  • $\begingroup$ Maybe you can explain your problem a bit more because what you wrote is not really clear to me. My problems start with: "How do I determine how to integrate this given only the domains of the random variables?" If you do not have the densities $f_X$ and $f_Y$ you cannot integrate. Do you like to know the area over which to integrate given the two domains? Next what is $X - U(0,1)$ supposed to mean? Tell us at least what $U(0,1)$ is. For your example: State clearly the random variables, their domains and what you would like to assume about their density. $\endgroup$
    – g g
    Oct 12 '18 at 14:45
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    $\begingroup$ For convolution, there is a universal formula $\displaystyle f_Z(z) = \int_{-\infty}^{\infty} f_X(x)f_Y(z-x)dx$. Your problem arises when the support of $X, Y$ is not the set of all real numbers. In such case the marginal pdfs $f_X, f_Y$ is a piecewise function, as it equals to $0$ outside the support and with other definition inside the support. So most importantly you need to find out the set where the integrand is non-zero. $\endgroup$
    – BGM
    Oct 13 '18 at 3:29
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    $\begingroup$ ...but I don't understand how $z$ would have anything to do with this. As BGM commented and as the answer by Bjørn Kjos-Hanssen very kindly displayed, $z$ comes into play always, at least by restricting the support (often more than that). $\endgroup$ Oct 13 '18 at 6:58
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$f_X(x)=1_{(0,1)}(x)$ and I'll assume $f_Y(y)=a^{-y}\cdot 1_{(0,\infty)}(y)$ and moreover $a=e$. Then for $z\ge 0$, $$f_Z(z)=\int_{-\infty}^{\infty} f_X(x)f_Y(z-x)\,dx=\int_0^1 a^{-(z-x)}\cdot 1_{(0,\infty)}(z-x)\,dx$$ $$=\int_0^{\min(z,1)} a^{-(z-x)}\,dx=e^{-z}\int_0^{\min(z,1)}e^{x}\,dx$$ $$=e^{-z}(e^{\min(z,1)}-1)=\begin{cases}1-e^{-z}&\text{if }z<1,\\ e^{-z}(e-1)& \text{if }z\ge 1.\end{cases}$$ So maybe the point that you were missing is how to work with the domains, like $$\int_{-\infty}^\infty 1_{(a,b)}(x)1_{(c,d)}(x)f(x)\,dx=\int_{-\infty}^\infty 1_{(a,b)\cap(c,d)}(x)f(x)\,dx=\int_{\max(a,c)}^{\min(b,d)}f(x)\,dx$$ if $\max(a,c)<\min(b,d)$.

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