I am trying to solve a problem which consists of:
$X_1, X_2 \overset{\text{i.i.d}}{\sim} U([{\mu -1, \mu + 1}])$. Find the density of $Y=X_1+X_2$.
I have read the related questions to this topic, but it still does not help me find a solution. Here is what I have tried so far:
Let: $$y = x_1 + x_2 \Rightarrow y - x_1 = x_2$$
$$f_{X_1}(x_1) = {1 \over 2} = f_{X_2}(x_2) \rightarrow \int_{\mu -1}^{\mu+1}{1\over 2}dx_1 = {1\over 2}x_1 \bigg|_{\mu -1}^{\mu+1} = {1\over 2}(\mu + 1 - \mu - 1) = {2\over 2} = 1$$
Thus the joint density is:
$$f_{X_1, X_2} (x_1, x_2) = {1\over 2}{1\over 2} = {1\over 4}$$
Now substituting $x_2 = y - x_1$:
$$f_{X_1, X_2} (x_1, y - x_1) = {1\over 2}{1\over 2} = {1\over 4}$$
and to get the marginal distribution of $y$:
$$f_{Y}(y) = \int_{-\infty}^{\infty} f_{Y}(y - x_1)f_{X_1}(x_1)dx_1 $$
Now comes the part where I have difficulties. The bounds of $y = x_1 + x_2$ are:
$$2\mu-2 \lt y \lt 2\mu +2$$
and the bounds for $x_1$ are:
$$\mu-1 \lt x_1 \lt \mu +1$$
So I have to split the bounds of $y$ into three parts:
$$ \begin{cases} 2\mu-2 + (\mu+1) \lt y,& \text{ for } 2\mu -2 \lt y \lt \mu-1 \\ \mu - 1 \lt y,& \text{ for } \mu -1 \lt y \lt \mu+1 \\ y - (\mu+1)\lt \mu+1, & \text{ for } \mu +1 \lt y \lt 2\mu+2 \end{cases} $$
which would result in:
$$ f_{Y}(y) = \begin{cases} \int_{2\mu-2 + (\mu+1)}^{y} {1\over 2} dx_1 \\ \int_{\mu - 1}^{y} {1\over 2} dx_1\\ \int_{y - \mu+1}^{\mu+1} {1\over 2} dx_1 \end{cases} $$
Is that correct?
I suggest you an easier (In my opinion) way to get the result.
$X_1$ and $X_2$ can be interpreted as two copies of Lebesgue measure multiplied by an opportune costant (In this case $\frac{1}{2}$) in $[\mu-1,\mu+1]$.
Then, using the product of measure ($X_1$ and $X_2$ are indipendet) you can easly get that $\mathbb{P}(X_1 \in A, X_2 \in B)=\mathscr{L^1}(A)\mathscr{L^1}(B)$ where $\mathscr{L}^1(A)=\mathbb{P}(X_1 \in A)$ for any measurable set of $[\mu-1,\mu+1]$ and the same for $\mathscr{L^1}(B)$.
Then $F_Y(y)=\mathbb{P}(X_1+X_2 \leq y)=\mathbb{P}(X_1 \leq y - X_2)$ can easly get computing the area of $\{(t,h) \in \mathbb{R}^2:h \leq y-t\} \cap [\mu-1,\mu+1] \times [\mu-1,\mu+1]$ which is easy because there are only two possible figure: a triangle and a trapeze. Then is it sufficient to derivate (where it is possible) the function $F_Y(y)$.