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.

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.
