# Is there a general way to integrate for the convolution sum of two random variables?

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?

• Laplace transforms give one way to evaluate convolutions. Oct 10 '18 at 1:34
• 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.
– g g
Oct 12 '18 at 14:45
• 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.
– BGM
Oct 13 '18 at 3:29
• ...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). Oct 13 '18 at 6:58

$$f_X(x)=1_{(0,1)}(x)$$ 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 $$\max(a,c)<\min(b,d)$$.