Convolution of a Binomial and Uniform Distribution I am given that $X$ is a random variable with a Binomial distribution with parameters $(n,p)$ and that $Y$ is a random variable with a Uniform distribution on $(0,1)$. We assume independence. I want to find the distribution of the sum of $X$ and $Y$.
First, I define $Z:=X+Y$ and I want to find $F_z(z)=P(Z\le z)$.
Now, I understand how to go about this problem if $X$ and $Y$ are both discrete or both continuous, however in this case $X$ is discrete while $Y$ is continuous. For example, if I had two continuous distributions then:
$P(Z\le z) = P(X+Y\le z)=\int_{-\infty}^{+ \infty}f_X(z-x) f_Y(y) dy$
Do I need to transform the pmf of $X$ into a continuous function and if so how can I do this? This is supposedly an easy question so perhaps there is a very straightforward way.
 A: You can calculate the distribution without thinking about convolutions at all: Note that if you know the value of $Z$, say $Z=z$, then with probability $1$, $X=\lfloor z\rfloor$ (the greatest integer $\le z$) and $Y=Z-X$. So the probability density on the interval $(k,k+1)$ will just be $\binom{n}{k}p^k(1-p)^{n-k}$.
If you do wish to think of convolution, do a formal calculation with delta functions: The distribution of $X$ is given by
$$ f_X(x)=\sum_{k=0}^n \binom{n}{k}p^k(1-p)^{n-k} \delta(x-k), $$
and that of $Y$ by $f_Y(y)=[0<y<1]$ (where the bracket is the Iverson bracket), hence
$$
\begin{aligned}
  f_X*f_Y(z)&=\int_{-\infty}^{\infty} f_X(x)f_Y(z-x)\,dx \\
   &= \int_{z-1}^{z} f_X(x) \,dx \\
   &= \int_{z-1}^{z} \sum_{k=0}^n \binom{n}{k}p^k(1-p)^{n-k} \delta(x-k) \,dx \\
   &= \sum_{k=0}^n \binom{n}{k}p^k(1-p)^{n-k} \int_{z-1}^{z} \delta(x-k) \,dx \\
   &= \sum_{k=0}^n \binom{n}{k}p^k(1-p)^{n-k} \bigl[z-1<k<z\bigr] \\
   &= \sum_{k=0}^n \binom{n}{k}p^k(1-p)^{n-k} \bigl[k<z<k+1\bigr]
\end{aligned}
$$
(again using the Iverson bracket at the end). That final expression is just a restatement of what I said in the first paragraph.
