# Sum binomial distribution not binominally distributed if the p is not equal for all

Suppose we have $X_1, ... X_n$ independent with $X_i \sim bin(n_i, p_i)$, prove that $Y := X_1 + ... + X_n$ is not binomially distributed.

My attempt: suppose $Y$ is binomially distributed, then it must be $bin(\sum_{i = 1}^nn_i, q)$ for some $q$ because it is possible that $Y = \sum_{i = 1}^nn_i$, but not $Y > \sum_{i = 1}^nn_i$. Then we have that $$E[Y] = \sum_{i = 1}^nn_iq$$ but also $$E[Y] = \sum_{i = 1}^nE[X_i] = \sum_{i = 1}^nn_iq_i$$ and $$Var(Y) = \sum_{i = 1}^nn_iq(1 - q)$$, but also, as the $X_i$ are iid: $$Var(Y) = \sum_{i = 1}^nn_iq_i(1 - q_i)$$

I'm trying to show that this implies that $q_i - q = 0$ for all $i$. From the expected values we know that $\sum_{i = 1}^nn_i(q - q_i) = 0$, but this doesn't imply that $q_i - q = 0$. I also tried substituting them in each other, but that didn't work out either. I'm not sure whether this will work, because it's possible that $Y$ satisfies these equations but is not binomially distributed.

I have reduced the information to the two equations $$\sum_{i = 1}^nn_i(q - q_i) = 0$$ and $$\sum_{i = 1}^nn_i(q^2 - q_i^2) = 0$$

• You'll need to impose some conditions on the $p_i$ because if $p_1=p_2=\cdots=p_n$, then the sum does have a binomial distribution.
– JimB
Commented Jan 3, 2018 at 18:43

This is an instance of Jensens inequality: you can view $$\sum \frac{n_i}{\overline{n}} q_i = \mathbb{E}_{\mu}[f]$$ where $\overline{n} = \sum_i n_i$. I leave it to you to figure out what $f$ and $\mu$ are.
Then your first equation gives you $$\mathbb{E}_{\mu}[f]=q$$ while the second reads $$\mathbb{E}_{\mu}[f^2]=q^2.$$
Now Jensens inequality is strict unless the function you are moving under the integral sign is affine. The only way a square can be affine is that the set of points it is defined on is a singleton, i.e. $q_i=q$ for any i.
• What does the $\mu$ in $\mathbb{E}_\mu[f]$ mean? Commented Jan 3, 2018 at 20:15