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.

  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.