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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.

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    $\begingroup$ Nitpick: I think you meant “I want to find the distribution of the sum of 𝑋 and 𝑌.” The distributions get convolved; the variables are added. $\endgroup$ – Harald Hanche-Olsen Feb 3 at 14:34
  • $\begingroup$ Yes, edited the original post, thanks. $\endgroup$ – R. Rayl Feb 3 at 14:35
  • $\begingroup$ Related : math.stackexchange.com/q/1113762 $\endgroup$ – Jean Marie Feb 3 at 14:43
  • $\begingroup$ I think you can be as well inspired by this answer math.stackexchange.com/q/1169353 that rightly promotes the idea to use Dirac's $$\delta$s ; but are you familiar with these notations ? $\endgroup$ – Jean Marie Feb 3 at 14:46
  • $\begingroup$ Ok, I see, thank you. I'm not familiar with that at all, but very happy to learn. My only concern is that since I have not encountered this before, I doubt this is the intended way to solve the problem. $\endgroup$ – R. Rayl Feb 3 at 14:51
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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.

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  • $\begingroup$ Thank you, this has really made it clear to me. Would I be right in saying that the Iverson bracket is equivalent to an indicator function on the event inside the bracket? $\endgroup$ – R. Rayl Feb 3 at 15:03
  • $\begingroup$ Yes it is. It avoids difficult-to-read information in lower case as indices. $\endgroup$ – Jean Marie Feb 3 at 15:46

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