Convolution of $3$ Independent Random Variable for Uniform Distribution

Question

I have a question on convolution of 3 random variables. Below is the given problem:

Let $X_i$ be independent random variables and have a uniform distribution on $(-0.5,0.5)$ for $i = 1, 2, 3$. Find the pdf and cdf of $X_1$ + $X_2$ and $X_1$ + $X_2$ + $X_3$

My solution for $X1 + X2$ is:

• $f_X{_1}+f_X{_2}(z) = \int_{-\infty}^{\infty}f_X{_1}(x)f_X{_2}(z-x)\mathrm{d} x$
• Range of $X_1 + X_2$ is $[-1,1]$

For any $-0.5 < z \leq 0.5$:

$$f_X{_1} + f_X{_2} (z) = \int_{-\infty}^{\infty}f_X{_1}(x)f_X{_2}(z-x)dx \\ = \int_{-0.5}^{z} 1 dx\\ = z - (-0.5)\\ =z + 0.5$$

For any $0.5 < z < 1$:

$$f_X{_1} + f_X{_2} (z) = \int_{-\infty}^{\infty}f_X{_1}(x)f_X{_2}(z-x)dx \\ = \int_{z-0.5}^{0.5} 1 dx\\ = 0.5 - (z-0.5)\\ =1 - z$$

The conclusion is:

$$f_X{_1} + f_X{_2} (z) = \begin{cases}z+0.5 & -0.5 <z\leq 0.5 \\ 1-z & 0.5 < z < 1 \\ 0 & otherwise \end{cases}$$

My understanding for $X_1 + X_2 + X_3$ are:

• $f_({_X}{_1}_+{_X}{_2}_){_+}{_X}{_3}(z) = \int_{-\infty}^{\infty}f{_X}{_1}+_X{_2}(t)f_X{_3}(z-t)\mathrm{d} t$
• Range of $X_1 + X_2 + X_3$ is $[-1.5,1.5]$

Then my question for $X_1 + X_2 + X_3$ are:

1. What is the cases for $X_1 + X_2 + X_3$?
2. Can I use the same way of convolution of $2$ random variables integration to compute convolution of $3$ random variables?
3. May I know where I can find more information about convolution of $3$ independent random variable for uniform distribution?