Convolution of 2 uniform random variables $U(0,a)$ and $U(0,b)$

Question

Let $X$ have a uniform distribution on $(0, a)$ and let $Y$ be independent of $X$ with a uniform distribution $(0, b)$. Let's say that $b$ is greater than $a$.

Q. How can I determined the distribution of $S=X+Y$?

I'm trying to solve this, but I'm having trouble on how to handle the integral limits. I'm not really understanding how to build this problem properly.

Answer 1

First of all, we know that the pdf of $X$ and $Y$ are respectively: $$ f_x(x)=\frac{1}{a}I_{(0,a)}(x) \quad f_y(y)=\frac{1}{b}I_{(0,b)}(y) $$ We want to know the pdf of $S = X + Y$. By using convolution, we get: $$ f_s(s)=\int f_x(s-y)f_y(y)dy $$

Now, we have 3 cases:

• If $0<s<a$ then:

$$ f_s(s)=\int_0^s \frac{1}{ab}dy = \frac{s}{ab} $$

• If $a<s<b$ then:

$$ f_s(s)=\int_{s-a}^s \frac{1}{ab}dy = \frac{1}{b} $$

• If $b<s<a+b$ then:

$$ f_s(s)=\int_{s-a}^b \frac{1}{ab}dy = \frac{1}{ab}(b+a-s) $$

In order to help visualing these cases:

Therefore, the pdf is: $$ f_s(s) = \frac{s}{ab} I_{(0,a)}(s) + \frac{1}{b} I_{(a,b)}(s) + \frac{1}{ab}(b+a-s) I_{(b,a+b)}(s) $$