$X,U$ and $Y$ are random variables for which: $$ \forall i\in\{0,1,\dots,n-1\}: P(X=i)=\frac 1n,$$ $$ P(Y=1)=P(Y=-1)=\frac 12,$$ $$ U =_d U[0,1].$$ Show that $(U+X)Y = _d U[-n,n]$ if $U,X$ and $Y$ are mutually independent.
My attempt:
We note that $P\left( (U+X)Y\le z\right) = \frac 12\left(P(U+X\le z)+P(U+X\ge -z \right)$. We determine the distribution of $T=U+X$: $f_{T,X}(t,x)=f_U(t-x)f_X(x)$ (using $U\perp\!\!\!\perp X$). Since $X$ is discrete, we get the marginal distribution of $T$ by summing over all possible values that $X$ can take:
$$ f_{U+X}(t)=\sum_{x=0}^{n-1} f_U(t-x)P(X=x)=\frac 1n\sum_{x=0}^{n-1}f_U(t-x).$$ Inspired by the claim, we can note that $f_{U}(t-x)=1$ for $-n\le t \le n$. Therefore $$ f_{U+X}(t)=1\cdot I(-n\le t\le n).$$ Then, $$ P(U+X\le z)=\int_{-n}^zdt=z+n =P(U+X\ge -z).$$ We get $P((U+X)Y\le z)=F_{(U+X)Y}(z)=z+n$, $-n\le z\le n$.
This does not correspond to the CDF of a uniformly distributed RV on $[-n,n]$.
Where is my mistake?
Note that $$f_U(t-x) = I(0< t-x <1) = I(x<t <x+1).$$ Now, as we vary $x$ from $0$ to $n-1,$ for any $0<t<n,$ just one of these indicators has value $1$, the rest are zero. So, $$f_{U+X} (t) = \frac{1}{n}\sum_{x=0}^{n-1}I(x<t<x+1)=\frac{1}{n},\text{ for } 0<t<n.$$ This shows that $U+X\sim U[0, n].$ Now you can easily prove that $(U+X)Y\sim U[-n, n].$