Interesting probability distribution of a mixed type random variable $Y$ Let $X$ and $U$ be independent random variables with:
$$P(X=k)=\frac1{N+1} \text{ for } k=0,1,2,\ldots,N$$
and $U$ having uniform $(0,1)$ distribution.
Let $Y=X+U$.
Find distribution function of $Y$.
I have tried to solve the problem by conditioning on value of $X$ and making use of total probability theorem.
I have got $P(Y\le y)=y-\frac N2$.  Is it correct? Please help.
 A: Suppose $k\in\{0,1,2,\ldots, N\}$ and $0\le a<b\le 1$.  Then
$$
\begin{align}
& \Pr(a<X+Y<b) = \Pr(X=k\ \&\ a<Y<b) = \frac 1 {N+1}\cdot(b-a) \\[10pt]
= {} & \frac{\text{length of the interval }(a,b)}{\text{length of the interval} (0,N+1)} \\[10pt]
= {} & \frac{\text{length of the interval }(k+a,k+b)}{\text{length of the interval} (0,N+1)}
\end{align}
$$
and that is what the probability would be if $X+Y$ has a continuous uniform distribution on $(0,N+1)$.
Now just prove that the distribution is determined by the probabilities assigned to intervals lying between two consecutive integers.
A: It is a uniform distribution on $[0, N+1]$. Just take $k\leq a<b\leq k+1$ and calculate $P(Y\in [a,b))$.
A: Let $Z$ have uniform distribution over $[0,N+1)$.
Define $X':=\lfloor Z\rfloor$ and $U':=Z-\lfloor Z\rfloor$.
It is not really difficult to show that $X'\simeq X$ and $U'\simeq U$ where $\simeq$ stands for having the same distribution.
For $u\in[0,1)$ and $k\in\{0,1,\dots,N\}$ we find:$$\Pr\left(U'\leq u\wedge X'=k\right)=\Pr\left(k\leq Z\leq k+u\right)=\frac{u}{N+1}=\Pr\left(U'\leq u\right)\left(X'=k\right)$$
showing that $U'$ and $X'$ are independent, just like $U$ and $X$ are independent.
Then $(X,U)\simeq(X',U')$ and consequently $Y=X+Y\simeq X'+Y'=Z$.
Proved is now that $Y$ has uniform distribution over $[0,N+1)$.
