# Distribution function of the sum of poisson and uniform random variable.

Merry Christmas to everybody. I am working on the following problem.

Let $X$ and $Y$ be independent Poisson($\lambda$), respectively Uniform$(0,1)$ random variables. Find the distribution function of the random variable $Z := X+Y$.

I have the solution here but I don't understand it completely.

\begin{align*} F_Z(a) &= \mathbb P(X+Y \le a) = \sum_{i = 0}^{\lfloor a\rfloor-1} \mathbb P(X = i) + \mathbb P(X = \lfloor a\rfloor, Y \le a-\lfloor a\rfloor) \\ &= \sum_{i = 0}^{\lfloor a\rfloor-1} \mathbb P(X = i) + \mathbb P(X = \lfloor a\rfloor) \cdot \mathbb P(Y \le a-\lfloor a\rfloor) = \sum_{i = 0}^{\lfloor a\rfloor-1} \frac{\lambda^i}{i!} e^{-\lambda} + \frac{\lambda^{\lfloor a\rfloor}}{\lfloor a\rfloor!} e^{-\lambda} \cdot (a-\lfloor a\rfloor). \end{align*}

What I don't understand is:

\begin{align*} \mathbb P(X+Y \le a) = \sum_{i = 0}^{\lfloor a\rfloor-1} \mathbb P(X = i) + \mathbb P(X = \lfloor a\rfloor, Y \le a-\lfloor a\rfloor) \end{align*}

$\lfloor \cdot \rfloor$ denotes the floor function.

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Let us write the probability $\Pr(X+Y\leqslant a)$ as $$\begin{eqnarray} \Pr(X+Y \leqslant a) &=& \sum_{m=0}^\infty \Pr(X+Y\leqslant a \mid X=m) \Pr(X=m) \\ &=& \sum_{m=0}^\infty \Pr(m+Y\leqslant a) \Pr(X=m) \\ &=& \sum_{m=0}^\infty \Pr(Y\leqslant a-m) \Pr(X=m) \end{eqnarray}$$ Because with probability $0\leqslant Y \leqslant 1$ with probability 1, $\Pr(Y \leqslant a - m)$ equals zero for those $m$, that $a - m \leqslant 0$, i.e. for all $m \geqslant \lfloor a \rfloor + 1$ and equals to one for those $m$ that $a-m > 1$, i.e. $m < \lfloor a\rfloor$.
Hence $$\begin{eqnarray} \Pr(X+Y \leqslant a) &=& \sum_{m=0}^{\lfloor a \rfloor -1} \Pr(X=m) + \Pr(X=\lfloor a \rfloor) \underbrace{\Pr(Y \leqslant a - \lfloor a \rfloor )}_{a - \lfloor a \rfloor = \{a\}} \end{eqnarray}$$