The following problem is from the book "Probability and Statistics" which is part of the Schaum's outline series. It can be found on page 71 and is problem number 2.74. It is under the section Convolutions. Therefore, I want to solve it using a convolution.
Problem:
Let $X$ and $Y$ be identically distributed independent random variables with density function
$$ f(x) = \begin{cases}
1, & \text{for } 0 \leq t \leq 1 \\
0, & \text{otherwise }
\end{cases} $$
Find the density function of $X + Y$ and check your answer.
Answer:
Let $u = x + y$. Let $g$ be the density function we seek. This gives us:
\begin{align*}
g(u) &= \int_{-\infty}^{\infty} f(v)f(u-v) \, dv
\end{align*}
Now for $v > 1$ we have $f(v) = 0$ so we can rewrite the integral.
\begin{align*}
g(u) &= \int_{-\infty}^{1} f(v)f(u-v) \, dv
\end{align*}
Now for $v < 0$ we have $f(v) = 0$ so we can rewrite the integral.
\begin{align*}
g(u) &= \int_{0}^{1} f(v)f(u-v) \, dv \\
g(u) &= \int_{0}^{1} f(u-v) \, dv
\end{align*}
Now, I am stuck. How do I complete the problem? Am I right so far?
Here is an updated solution based on comments from the group. We want: $$ 0 \leq u - v \leq 1 $$ Now we solve for $v$ $$ -u \leq -v \leq 1 - v $$ $$ u -1 \leq v \leq u $$ Now we have the following integral:
\begin{align*} g(u) &= \int_{u-1}^{u} 1 \, dv\\ g(u) &= dv \Big|_{v = u - 1}^{v = u} \\ g(u) &= u - (u - 1) \\ g(u) &= 1 \end{align*} Where did I go wrong?
We have $f(t)= \mathbf 1_{0\leq t\leq 1}$, so therefore $$\begin{align}g(u)&=\int_\Bbb R f(v)~f(u-v)~\mathrm d v\\&=\int_\Bbb R\mathbf 1_{0\leq v\leq 1}~\mathbf 1_{0\leq u-v\leq 1}~\mathrm d v\\&=\int_\Bbb R \mathbf 1_{0\leq u\leq 2}~\mathbf 1_{\max(0,u-1)\leq v\leq\min(1,u)}~\mathrm d v\\&=\int_{\max(0,u-1)}^{\min(1,u)}\mathbf 1_{0\leq u\leq 2}~\mathrm d v\end{align}$$