Pdf of sum of two uniform random variables on $\left[-\frac{1}{2},\frac{1}{2}\right]$

Question

Let $X$ and $Y$ be independent RVs and uniformly distributed on $\left[-\frac{1}{2},\frac{1}{2}\right]$, what is the pdf of $X+Y$? Everything I can find is on the interval $[0,1]$ and I don't know how to choose the integration limits for this non generic case.

Answer 1

We must find the p.d.f. of $X$, $Y$, and joint p.d.f. of $X$ and $Y$ as below. \begin{align*} f_X(x)=\dfrac{1}{\frac{1}{2}-(-\frac{1}{2})}=1 \text{, for } -\frac{1}{2}\leq x\leq \frac{1}{2} \text{ and } 0 \text{ otherwise.} \end{align*} \begin{align*} f_Y(y)=\dfrac{1}{\frac{1}{2}-(-\frac{1}{2})}=1 \text{, for } -\frac{1}{2}\leq y\leq \frac{1}{2} \text{ and } 0 \text{ otherwise.} \end{align*}

\begin{align*} f_{X,Y}(x,y)=f(x)f(y)=1\cdot 1=1 \text{, for } -\frac{1}{2}\leq x\leq \frac{1}{2}, -\frac{1}{2}\leq y\leq \frac{1}{2} \text{ and } 0 \text{ otherwise.} \end{align*}

Let $U=X+Y$ and $V=X$.

We can use transformation method to get p.d.f. of $X+Y$.

We find the inverse of transformation, as below.

$X=V$ and $Y=U-X=U-V$.

We have a new domain of $U$ and $V$, i.e. $-\dfrac{1}{2}\leq V\leq \dfrac{1}{2}$ and $-\dfrac{1}{2}\leq U-V\leq \dfrac{1}{2}$

We find the jacobian as below.

\begin{align*} \vert J\vert &= \begin{vmatrix} \dfrac{\partial X}{\partial U}&\dfrac{\partial X}{\partial V}\\ \dfrac{\partial Y}{\partial U}&\dfrac{\partial Y}{\partial V}\\ \end{vmatrix} = \begin{vmatrix} 0&1\\ 1&-1\\ \end{vmatrix} =\vert-1\vert =1. \end{align*}

Now, we have the joint probability of $U$ and $V$ as below

\begin{align*} f_{U,V}(u,v)&=f_{X,Y}(x,y)\cdot \vert J\vert\\ &= f_{X,Y}(v,u-v) \cdot 1\\ &= 1. \end{align*}

for $-\dfrac{1}{2}\leq V\leq \dfrac{1}{2}$ and $-\dfrac{1}{2}\leq U-V\leq \dfrac{1}{2}$ and $0$ otherwise.

To find the p.d.f. of $X+Y$, we find the marginal pdf of $U$ from the joint p.d.f. $f_{U,V}(u,v)$. We have p.d.f. of $U=X+Y$ as below

\begin{align*} f_{U}(u)&=\int\limits_{u-\frac{1}{2}}^{u+\frac{1}{2}} 1 dv\\ &=\left[v\right]_{u-\frac{1}{2}}^{u+\frac{1}{2}}\\ &= \left(u+\frac{1}{2}\right)-\left(u-\frac{1}{2}\right)\\ &=1 \end{align*}

for $-\dfrac{1}{2}\leq U \leq \dfrac{1}{2}$ and $0$ otherwise.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{-1/2}^{1/2} \bracks{-\,{1 \over 2} < z - x < {1 \over 2}}\dd x} \\[5mm] = &\ \int_{-1/2}^{1/2} \bracks{z - {1 \over 2} < x < z + {1 \over 2}} \dd x \\[5mm] = &\ \bracks{-\,{1 \over 2} < z + {1 \over 2} < {1 \over 2}} \int_{-1/2}^{z + 1/2}\dd x \\[2mm] + &\ \bracks{-\,{1 \over 2} < z - {1 \over 2} < {1 \over 2}} \int_{z - 1/2}^{1/2}\dd x \\[5mm] = &\ \bracks{-1 < z < 0}\pars{z + 1} + \bracks{0 < z < 1}\pars{1 - z} \\[5mm] = &\ \bbx{\large\bracks{\vphantom{A^{A^{A}}} \verts{z} < 1} \pars{\vphantom{A^{A}}1 - \verts{z}}} \\ & \end{align}