Vtyjkyuk

Say I have two biased coins with probabilities of coming up heads $p$ and $r$, where $pne r$.

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.

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:
$$p^n(frac1p - 1 + x)^n$$
$$prod p_i(frac1p_i - 1 + x)^n_i$$
As you can see, it can not be the same polynomial, unless $p_i$ equals $p$ for all $i$.

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: $$2q=p+rtag1$$

And $2nq(1-q)=np(1-p)+nr(1-r)$ or equivalently: $$2q-2q^2=p-p^2+r-r^2tag2$$

Substituting $(1)$ in $(2)$ we find:$$(p-r)^2=0$$

This contradicts that $pneq r$ so under that condition it is excluded that we are dealing with binomial distribution.