I am stuck on how to sum this distributions together.
Question: Let $\xi$ be a discrete random variable such that $P(\xi = -1) =\frac{1}{2}$ and $P(\xi = 1)=\frac{1}{2}$. Let $\eta \sim U [-1, 1 ]$ (so, it has continuous uniform distribution). Let $\xi$ and $\eta$ be independent. Find the probability distribution function of the random variable $\xi + \eta.$
My Attempt:. $$F_{\xi + \eta} (X) = P(\xi + \eta \leq X)$$
$$\implies P\left( (\xi = -1) \cap (\eta \leq X) \right) + P((\xi =1)\cap (\eta \leq X-1))$$
$$= \frac{1}{2} P(\eta \leq X) + \frac{1}{2} P(\eta \leq X-1)$$
$$ F_{\xi + \eta}(x) = \left\{ \begin{array}{rcl}
-1 ,
& x\leq-1 \\
x , & x\in [-1,0] \\
1 , & x\geq 1
\end{array}\right. + \frac{1}{2} \left\{ \begin{array}{rcl}
-1 ,
& x\leq-1 \\
(x-1) , & x\in [0,1] \\
1 , & x\geq 1
\end{array}\right.$$ At this point I am stuck. I guess some thing is wrong with my approach. Any help or hint is appreciated. Thanks in Advance!
Best Answer
Let's go from here $\frac{1}{2} P(\eta-1 \leq X) + \frac{1}{2} P(\eta+1 \leq X)$
Hint: Notice that $\eta-1 \sim U[-2,0]$ and $\eta+1\sim U[0,2]$
Spoiler: With those changes you get
$F_{\xi + \eta}(x) = \frac{1}{2}\left\{ \begin{array}{rcl} 0 , & x\leq-2 \\ 1-\frac{x}{2} , & x\in [-2,0] \\ 1 , & x\geq 0 \end{array}\right. + \frac{1}{2} \left\{ \begin{array}{rcl} 0 , & x\leq0 \\ \frac{x}{2} , & x\in [0,2] \\ 1 , & x\geq 2 \end{array}\right.$
$=\begin{cases}0 & x\leq-2\\ \frac{1}{2}-\frac{x}{4} & x\in[-2,0]\\ \frac{1}{2}+\frac{x}{4} & x\in[0,2]\\ 1 &x\geq 2 \end{cases} $ by simply summing over each interval