Is negative probabilities allowed if the final distribution is valid? Illustrative example: 
The pgf for $X \sim \mathrm{Bin}(m,p_1)$ is $G_1(z)=[(1-p_1)+p_1z]^m$. If $m \sim \mathrm{Bin} (n,p_2)$ the final distribution has pgf of  $G_2(z)=[(1-p_1 p_2)+p_1 p_2z]^n$ representing $\mathrm{Bin}(n,p_1  p_2)$ distribution.
Mathematically the final answer is still a valid distribution for $s=p_1 p_2$ as long as $0<s <1$. This may be true even though $p_i<0$ or $p_i>1$ individually, as long as their product is positive and between 0 and 1.
For example, if $p_{\{1,2\}}=-\frac{1}{\sqrt 2}$ then both distributions alternate between positive and negative 'probabilities', yet their mixture is $s=\frac{1}{2}$; equivalent to flipping a coin $n$ times. 
This reminds me of improper priors where the probabilities are positive but the sum does not equal 100%. Here we may have a "distribution" where the sum is 100% but the probabilities may be negative.
If improper priors are allowed if the posterior is valid, are negative probabilities allowed if the final answer is a valid distribution?
 A: You can formulate this in terms of the following experiment: You flip two biased coins with probability of heads $p_1,p_2$ respectively $n$ times. You then ask: how many simultaneous occurrences of heads happened? This is answered by the binomial distribution with parameter $p_1p_2$. The fact of the matter is that in this line of reasoning, you're actively using the axioms of probability, which assume $0\leq p_1,p_2\leq 1$, along with assumptions on independence. However, you're right, if both $p$'s are negative, then $p_1p_2\geq 0$ might be a valid parameter for the binomial. A kind of analytic continuation for probability. 
Now you're saying that $p_1,p_2<0$ as a prior. This invalidates either prior as a probability distribution, which brings the entire experiment into question. Or, maybe your friend flipped the first coin $n$ times, and reported $p_1$ as the number of heads over the total as a negative number. This might be made valid if you were to extend the theory of the experiment to include a prior on your friends prior, which includes them drunkenly reporting a negative $p_1$ value. 
Here's the point. You are saying that $P:=p_1p_2$ is a kind of random variable, with some definition of priors on $p_1,p_2$. You're taking that random variable and plugging it into a binomial distribution, in hopes of getting a sensible answer. Whenever $0\leq P\leq 1$, you'll get such a sensible answer. Otherwise, you get nonsense. 
To resolve such issues, we usually say that the probability that $0\leq p_1,p_2\leq 1$ with probability 1. Any other option has probability 0. Conditioning on on events that have zero probability is a big no-no in introductory probability, and is formally defined later on using measure theory along with conditional expectation. 
