Distribution of the sum of binomial random variables

Question

What is the distribution of the sum of random variables given by $$n = n_1 + n_2 + \cdots + n_k $$ where each $n_i$ is binomially distributed random variable define by $B(n_i,p)$.

I'm not sure how to approach this problem. I believe that the random variable with mean $\eta = \eta_1 + \cdots + \eta_n$ and variance $\sigma^2 = \sigma_1^2 + \cdots +\sigma_n^2$ The central limit theorem states that under certain general conditions, the distribution $F(x)$ of $X$ approaches a normal distribution with the same mean and variance. So does this apply in this same situation? Thanks for your help in solving this, I really appreciate it!

Update: The random variables can be treated as independent, which should help make the answer significantly easier

Answer 1

You did not state that these $k$ random variables are independent, and without that there are many different distributions that could arise in this way.

However, notice this:

The distribution $\mathrm{Bin}(n_i,p)$ is that of the number of successes in $n_i$ independent trials, with probability $p$ of success on each trial.

If the $k$ random variables are independent, then their sum is the number of successes in $n_1+\cdots+n_k$ independent trials with probablity $p$ of success on each trial. That tells you what its distribution is.

If you don't know the fact in the indented statement above, then the problem can still be solved by other means, but it's a lot more work.

Answer 2

I may be misinterpreting the question, but it should just be another binomial with parameters $\sum_{i=1}^k n_i$ and $p$.

Given the way it's written, I'm assuming it's the same $p$ for each one and that you're using $n_i$ to refer to both the number of items drawn for each binomial RV and the variable itself.

To see this, note that we can write a $B(n_i, p)$ random variable as $\sum_{j=1}%{n_i} X_j$ where $X_j$ is 1 with probability $p$ and 0 with probability $1-p$. As long as $p$ is the same for each one, you're just summing up a bunch of sums and get $\sum_{j=1}^n X_j$ which is $B(n, p)$.

Answer 3

The moment generating function of a Binomial(n,p) random variable is $(1-p+pe^t)^n$. The moment generating function of a sum of independent random variables is the product of the corresponding moment generating functions, which in this case is $\prod_{i=1}^k (1-p + pe^t)^{n_i} = (1-p+pe^t)^{\sum_i n_i}$, which is a Binomial$(\sum_i n_i , p)$ r.v.

Aside: You should use different letters for the parameter and the RV itself.