I am having trouble applying the convolution formula for probability density functions when the pdfs of the random variables in question are defined over different intervals.
Here is an example:
Say $X \in Un([a,b])$ and $Y \in Exp(\theta)$ are independent random variables.
We want to find the probability density function of $Z = X + Y$
The convolution formula would give us $f_Z(z) = \int_{-\infty}^{\infty} f_X(x) f_Y(z-x) dx$ which would be the same as $\int_{-\infty}^{\infty} f_X(z-y) f_Y(y) dy$ (as far as I understand it).
Since $f_X(x) = 0$ for $x < a$ & $x > b$, and $f_Y(y) = 0$ for $y < 0$, we only need to integrate over a certain interval, but how does one determine that interval?
And how would the choice of whether to integrate $\int_{-\infty}^{\infty} f_X(x) f_Y(z-x) dx$ or $\int_{-\infty}^{\infty} f_X(z-y) f_Y(y) dy$ affect the interval of integration?
Am I missing something trivial here? Any help is greatly appreciated.
Best Answer
I would break the integral down into cases when the product is $0$ and then take minimums and maximums as needed, as demonstrated below.
The product $f_X(x) f_Y(z - x)$ is $0$ when
Because it is $0$ when $x > b$ or $x > z$ we know it is $0$ when $x > \min(b, z)$. So the integral is $$\int_{a}^{\min(b, z)} f_X(x) f_Y(z - x) dx.$$
Doing the $dy$ integral you would have $f_X(z - y) f_Y(y)$ is $0$ when
and you could use these three to work out the bounds on the $dy$ integral.