In one of my classes, we stated that if $X_i$ are independent Bernoulli random variables with p proportion of success, then the distribution of the sum $\sum X_i$ is Binomial(n,p). I already proved this, but it got me thinking: if there is a distribution for the sum of these n Bernoulli random variables, is there a special distribution of the sum of n Binomial random variables?
I know there is a relationship between the Binomial distribution and the Binomial coefficient, and moreover that there is a relation between the Binomial coefficient and the Multinomial coefficient (i.e. the Binomial coefficient is a special case of the Multinomial coefficient). I know there is also such a thing as the Multinomial distribution, so my suspicion is that this distribution of the sum of Binomial random variables is somehow Multinomial. But, that's well beyond the scope of my class, so I was curious if there was some intuitive explanation about why it is or is not the Multinomial distribution.
Thanks so much!
Consider the sequence $(X_k)_{k=1}^m$ where $\forall k\in\Bbb N\cap [1;m] : X_k\sim\mathcal{Bin}(n,p)$ .
Each $X_k$ is a count of successes among $n$ independent Bernoulli distributions with identical success rate $p$.
Therefore $\sum\limits_{k=1}^m X_k$ is the count of successes among $m$ times $n$ independent Bernoulli distributions with identical success rate $p$.
Thus the sum on $m$ random variables with distribution $\mathcal{Bin}(n,p)$, is a random variable with a distribution of $\mathcal{Bin}(nm,p)$.
Remark: it is somewhat more complicated when the success rate is not identical.