CDF of Mixed distributions

Question

Suppose X and U are independent random variables with

$$P(X = k) =\frac{1}{N+1}\quad\quad\quad k=0,1,2,....N$$

and $U$ having a uniform distribution on $[0, 1]$. Let $Y = X + U$

a) Find the CDF of Y.

i know how to do transform continuous with continuous , but in mixed case i have no idea

Answer 1

$$\mathbb P\left(Y \le y\right) = \mathbb E\left[\mathbf 1_{Y \le y}\right] = \mathbb E\left[\mathbb E \left[\mathbf 1_{U \le y-X}\Big | X\right]\right] = \mathbb E\left[\mathbb E \left[\mathbf 1_{U \le y-X}\left(\mathbf 1_{y-X < 0} + \mathbf 1_{0 \le y-X < 1} + \mathbf 1_{y-X \ge 1}\right)\Big | X\right] \right]$$ since $$\mathbf 1_{U \le y-X}\mathbf 1_{y-X < 0} = 0,$$ $$\mathbf 1_{U \le y-X}\mathbf 1_{0 \le y-X < 1} = (y-X)\mathbf 1_{0 \le y-X < 1} = (y-X)\mathbf 1_{X \le y < X+1} = (y-X)\mathbf 1_{X=\lfloor y\rfloor} = (y-\lfloor y\rfloor)\mathbf 1_{X=\lfloor y\rfloor},$$ $$\mathbf 1_{U \le y-X}\mathbf 1_{y-X \ge 1} = \mathbf 1_{y \ge X+1} = \mathbf 1_{\lfloor y\rfloor \ge X+1}.$$ then

$$\mathbb P\left(Y \le y\right) = (y-\lfloor y\rfloor)\mathbb E\left[\mathbf 1_{X=\lfloor y \rfloor}\right] + \mathbb E\left[\mathbf 1_{\lfloor y\rfloor \ge X+1}\right] = (y - \lfloor y \rfloor) \mathbb P(X = \lfloor y \rfloor ) + \mathbb P (X \le \lfloor y\rfloor -1)$$

Now we will compute every term of the above sum $$\mathbb P(X \le \lfloor y \rfloor -1) = \left\{\begin{array}{cl}0 & \text{if}\;\; y < 1 \\ \frac{\lfloor y\rfloor}{N+1} & \text{if}\;\; 1 \le y < N+1 \\ 1 & \text{if}\;\; y \ge N+1\end{array}\right.$$

$$\mathbb P (X = \lfloor y \rfloor) = \left\{\begin{array}{cl}0 & \text{if}\;\; y < 0 \\ \frac{1}{N+1} & \text{if}\;\; 0 \le y < N+1 \\ 0 & \text{if}\;\; y \ge N+1\end{array}\right.$$

so $$\mathbb P(Y \le y) = \left\{\begin{array}{cl}0 & \text{if}\;\; y < 0 \\ \frac{y - \lfloor y \rfloor }{N+1}& \text{if}\;\; 0 \le y < 1 \\ \frac{y }{N+1} & \text{if}\;\; 1 \le y < N+1 \\ 1 & \text{if}\;\; y \ge N+1\end{array}\right.$$

Answer 2

Note that for $0\leqslant y\leqslant N+1$, distribution function of $Y$ is

$P(Y\leqslant y)=P(X+U\leqslant y)$

$\displaystyle\qquad\qquad\quad=\sum_{x=0}^{\lfloor y\rfloor}P(U\leqslant y-x\mid X=x)P(X=x)$

$\displaystyle\qquad\qquad\quad=\sum_{x=0}^{\lfloor y\rfloor}P(U\leqslant y-x)P(X=x)$

Can you proceed now? $Y$ should turn out to have another Uniform distribution.