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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

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3 Answers 3

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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.

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  • $\begingroup$ The question does not state that they are independent. I agree with you that it changes it a lot. However, if it wasn't independent, but the same probability $p$ existed for each $n$, then would it still be binomial as you suggested? $\endgroup$ Commented Oct 8, 2016 at 22:58
  • $\begingroup$ @BillyThorton : No. For example, suppose $Y_1, Y_2, Y_3, Y_4 \sim \text {i.i.d Bernoulli}(p)$, i.e. each is the number of successes in one trial, and $X_1,X_2,X_3$ are respectively $Y_1,Y_2,Y_3$ and $X_4,X_5,X_6$ are respectively $Y_2,Y_3,Y_4$. Then the sum $X_1+X_2 + X_3 + X)4 + X_5 + X_6$ would be $(Y_1+Y_2+Y_3) + (Y_2+Y_3+Y_4) = Y_1+2Y_1+2Y_3 + Y_4$. Working out the probability distribution of that sum should be a simple exercise and you get a random variable whose expected value is $6p$ and whose variance is $10p(1-p)$. To show that that cannot be a binomial distribution$,\ldots\qquad$ $\endgroup$ Commented Oct 8, 2016 at 23:09
  • $\begingroup$ $\ldots\,$ suppose it is $\text{Bin}(n,r)$. Then $6p=nr$ and $10p(1-p) = nr(1-r)$. Solve that for $r$ and $n$ and you will get that for some values of $p$, $n$ is not even an integer. $\qquad$ $\endgroup$ Commented Oct 8, 2016 at 23:11
  • $\begingroup$ I feel like the result of this comment is that it either has to be independent (and therefore a binomial distribution) or else it becomes an almost unsolvable mess? $\endgroup$ Commented Oct 8, 2016 at 23:28
  • $\begingroup$ @BillyThorton : Maybe "unsolvable" is a bit exaggerated, but something like that. $\endgroup$ Commented Oct 8, 2016 at 23:34
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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 $B(n, p)$.

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  • $\begingroup$ I just saw Michael Hardy's post above, and he is right. I assumed they are independent. Otherwise you need to know something about the dependence to solve it. $\endgroup$ Commented Oct 8, 2016 at 22:56
  • $\begingroup$ What do you mean by knowing something about the dependence? If we assume we don't know does what you say still apply? $\endgroup$ Commented Oct 8, 2016 at 23:05
  • $\begingroup$ @BillyThorton: Sorry for the necro. If still interested, it's dealt with within the concept of the Copula. $\endgroup$
    – MSIS
    Commented Jan 21 at 4:30
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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.

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