Say I have two biased coins with probabilities of coming up heads $p$ and $r$, where $p\ne r$.
If I toss each of the coins $n$ times, the number of heads that show up is Poisson Binomial distributed.
But how do I prove that it is not Binomial? I tried using MGFs but the problem gets complex really fast.
Thank you for your help.
On
If it would be binomial then it must have parameters $2n$ and $q$.
Then the expectation should be $2nq$ and variance should be $2nq(1-q)$.
That leads to $2nq=\mathbb EX=np+nr$ or equivalently: $$2q=p+r\tag1$$
And $2nq(1-q)=np(1-p)+nr(1-r)$ or equivalently: $$2q-2q^2=p-p^2+r-r^2\tag2$$
Substituting $(1)$ in $(2)$ we find:$$(p-r)^2=0$$
This contradicts that $p\neq r$ so under that condition it is excluded that we are dealing with binomial distribution.
But MGF gives you an anwser immediately. Binomial distribution's MGF is $$(1 - p + pe^t)^n$$ But for your distribution it looks like this $$\prod(1 - p_i + p_ie^t)^{n_i}$$ Think about this thing as you're thinking about polynomials, replace $e^t$ term with $x$ and check what happens: $$p^n(\frac{1}{p} - 1 + x)^n$$ $$\prod p_i(\frac{1}{p_i} - 1 + x)^{n_i}$$ As you can see, it can not be the same polynomial, unless $p_i$ equals $p$ for all $i$.