Bounds of sum of independant random variables

Question

If X is a random variable with the following densitiy $f_X(x) = \lambda e^{-\lambda x}$ for $ x \ge 0$ else $0$ (Exponential distribution)

Y is a random variable with density $f_Y(y) = 1$ if $0 \le y \le 1$ else $0$ (Uniform distribution with bounds [0, 1])

Assume X and Y are independent.

Now goal is to compute $f_{X + Y}(z)$.

The solution now has 3 cases.

Case 1: $z < 0$: $f_{X + Y}(z) = ...$

Case 2: $z < 0 \le 1$: $f_{X + Y}(z) = ...$

Case 3: $z \ge 1$: $f_{X + Y}(z) = ...$

What I don't understand is how these 3 bounds ($z < 0$, $z < 0 \le 1$, $z \ge 1$) are chosen.

Answer 1

Perhaps this plot will help you:

Answer 2

To answer your question, let's actually derive the convolution formula. Don't worry, we don't have to complete the computation. We only need to do the beginning of it.

$f_Z(z)= \int_{-\infty}^{+\infty}f_X(x)f_Y(z-x)dx$

We have boundaries $$0 \leq x$$ $$0 \leq y \leq 1 $$

Therefore, we get to change the lower boundary of the integral to $0$ since we are integrating regarding $x$.

$f_Z(z)= \int_{0}^{+\infty}f_X(x)f_Y(z-x)dx$

Now, we have to consider the boundary given by $Y$, which is $z-x$ in our case.

Since $0 \leq y \leq 1 $, we have $0 \leq z-x \leq 1 $, $x \leq z \leq x+1 $, this is the only region that the function won't yield 0. Integration over 0 doesn't change the value, so we ignore what's outside of this region.

I am not good at generating graph so let me verbally explain this graph:

y axis : z

x axis : x

blue line : $z = x+1$

red line : $z = x$

As you can see in the graph, when $x\leq 1$, we need to integrate from $z$ to $0$. Whereas, when $x\geq 1$, we need to integrate from $z$ to $z-1$.