# 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.

• 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. – Harald Hanche-Olsen Feb 3 at 14:34
• Yes, edited the original post, thanks. – R. Rayl Feb 3 at 14:35
• Related : math.stackexchange.com/q/1113762 – Jean Marie Feb 3 at 14:43
• 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 ? – Jean Marie Feb 3 at 14:46
• 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. – R. Rayl Feb 3 at 14:51

## 1 Answer

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 (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 (again using the Iverson bracket at the end). That final expression is just a restatement of what I said in the first paragraph.

• 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? – R. Rayl Feb 3 at 15:03
• Yes it is. It avoids difficult-to-read information in lower case as indices. – Jean Marie Feb 3 at 15:46