How to find intervals while finding $PDF_{X+Y}$ by convolution

Question

$X, Y$ are continuous uniform random variables and I want to understand how to find $PDF_{X+Y}$ by convolution given various intervals for $X$ and $Y$.

I started with the simplest (I thought) case and read this answer but when both $X,Y\in[0,1]$ even though I do understand why $0<z<2$, I do not understand how we know in advance that we need to break this into 2 cases and how we know that we should consider $0<z<1$, $1<z<2$ (why not, for example, $0<z<1.5$, $1.5<z<2$?)

Now I have another similar problem, where $X\in[0,2]$ and $Y\in[3,4]$. I have a hunch that we need to break it all up into several cases as well, but I have no idea, again, how many cases there should be and what the limits of the intervals are.

How can I know that?

Answer 1

Let me try to explain. Write convolution and look at it: $$f_{X+Y}(z)=\int_{-\infty}^{\infty}f_X(t)\,f_Y(z-t)\,dt$$ We know that $f_X(t)=1$ for $t\in(0,1)$ only (and $0$ otherwise). It restricts integration to interval $(0,1)$: $$f_{X+Y}(z)=\int_0^1 f_X(t)\,f_Y(z-t)\,dt$$ Then look at $f_Y(z-t)$. It equals to $1$ only for the case when $z-t\in(0,1)$, or $z-1<t<z$. Otherwise we have $0$ in the integral.

Conclude that we should integrate over $t\in(0,1)\cap (z-1,z)$. But $z$ is the outer variable, and for different values of $z$ the intersection of the above intervals will be different. First look at the case when they cannot intersect. It is possible either for $z<0$ or for $z-1>1$ (that is $z>2$). Both cases the integration area is empty and $f_{X+Y}(z)=0$.

Next, let $z\in(0,2)$. You can imagine the intersection of intervals $(0,1)\cap (z-1,z)$ by moving the second unit interval from left to right. For small $z>0$ left point $z-1$ is smaller than zero, and the intersection of intervals will be $$(0,1)\cap (z-1,z)=(0,z).$$ It is valid until its right endpoint $z$ is less than $1$. As $z$ became $1$ or greater, $z-1$ became greater than $0$, and $$(0,1)\cap (z-1,z)=(z-1,1).$$

Therefore we have:

for $0<z<1$ the integration area is $(0,z)$

for $1<z<2$ the integration area is $(z-1,1)$.

Answer 2

First note that $Z=X+Y \in [0,2]$. So far so good.

Now let's look at the pdf of $Z$ given by the convolution:

$$\int_{-\infty}^{\infty} f_{X}(x)\cdot f_{Y}(z-x) \ dx$$

where

$$f_{X}(x) = \begin{cases} 1 \quad 0\leq x \leq 1 \\ 0 \quad \text{otherwise}\end{cases}$$

Inserting this definition for $f_{X}(x)$ into the convolution, we get

$$\int_{0}^{1} f_{Y}(z-x) \ dx$$

Similarly we define,

$$f_{Y}(z-x) = \begin{cases} 1 \quad 0\leq z-x \leq 1 \\ 0 \quad \text{otherwise}\end{cases}$$

This imposes some constraints. If $0\leq z-x$ then $z\geq x$. Similarly if $z-x \leq 1$ then $x\geq z-1$. So the integral can be re-written from these constraints as

$$\int_{0}^{1} f_{Y}(z-x) \ dx = \int_{0}^{z} 1 \ dx = \int_{z-1}^{1} 1 \ dx$$

But from the limits of integration $0$ to $1$, the first integral is valid only when $0\leq z \leq 1$ and the second only is only valid when $1\leq z \leq 2$