Distribution of a sum of a product of Bernoulli vectors divided by the sum of the first Bernoulli vector

I'm trying to understand the distribution of a normalized sum of a product of Bernoulli random variables. Specifically, I have two vectors $$M_1$$ and $$M_2$$ of length n. Each element of each vector is a Bernoulli random variable with varying success probability $$p_i$$. We can assume for now that each element is independent from all other elements and the vectors are independent from each other. With this assumption the sum of elements of vector $$M_1$$ is $$\sum_i M_{1i}$$ and follows a Poisson binomial distribution with expected value $$\sum_i p_{1i}$$. The element-wise product of both vectors $$M_1 M_2$$ consists again of Bernoulli random variables, this time with success probabilities $$p_{1i}p_{2i}$$ for each element. The sum of this vector of products $$\sum_i M_{1i}M_{2i}$$ (which is also the inner product of the two vectors) is then again distributed with a Poisson binomial distribution with expected value $$\sum_i p_{1i}p_{2i}$$. What I am interested in now is the distribution of $$Z=\frac{\sum_i M_{1i}M_{2i}}{\sum_i M_{1i}}$$. Basically this is the number of common successes between the two vectors divided by the number of successes in the first vector. This problem seems simple enough to possibly have a closed-form expression, but I don't know what to look for. Assumptions of n being large are acceptable, but exact solutions are preferred. Any pointers to either a solution or a problem of a similar form in an application would be greatly appreciated.

• If by "closed-form" you mean an expression that asymptotically in $n$ is $o(n)$ in length, you will necessarily be disappointed because it depends (at a minimum) on all the $p_{2i}$ separately. Approximations can be developed by making assumptions about the average relationships among the $p_{i1}$ and $p_{2i},$ so if you have such information, please share it.
– whuber
Commented Jan 28 at 19:25
• thank you for the clarification. I'm fine if the expression depends on all the p_{2i}, especially if like in the Poisson Binomial Distribution it would result in a straightforward expected value and variance. Even better of course would be if it converges to a well-known distribution. On the average relationship between p_{1i} and p_{2i} : I don't think there is anything else but to say that they are conditionally independent from each other. But would specifying P(p_{2i}|p_{1i}) be helpful? I might be able to make an assumption based on what my data shows Commented Jan 31 at 15:52
• Convergence results likely depend on details of what happens to the $p_{ji}$ as $i$ grows large.
– whuber
Commented Jan 31 at 15:54
• p_{ji} would not change in distribution as i gets large, in the sense of probabilities becoming smaller or larger, if that's what you mean. Commented Jan 31 at 15:58
• Are you supposing these values are themselves random variables? If so, that would be useful information to include -- along with any assumptions about those distributions.
– whuber
Commented Jan 31 at 16:25

Here is my attempt at the distribution; more elegant/compact solutions quite possibly exist however.

Let $$A:=\sum_i M_{1i}$$ and $$B:=\sum_i M_{1i}M_{2i}$$. Also, suppose for some given $$A$$, let $$M_1^{A,k}$$ correspond to a specific $$M_1$$ vector whose sum is $$A$$. So, there are $${n\choose A}$$ different $$M_1^{A,k}$$ vectors for a given value of $$A$$.

The joint PMF $$P(A=a,B=b)$$ where $$b\leq a\leq n$$ is \begin{align*} P(A,B) &= P(A)P(B|A)\\ &= P(A)\sum_kP(B|M_1^{A,k})P(M_1^{A,k}|A)\\ &= \sum_kP(B|M_1^{A,k})P(M_1^{A,k}) \end{align*} Now, $$P(B|M_1^{A,k})$$ is a Poisson-binominal PMF. Specifically, suppose $$\{j\}$$ is the set of indices of the "successful" components of $$M_1^{A,k}$$. We then look at $$M_2$$ and focus on its $$j$$ components as well; $$P(B|M_1^{A,k})$$ is simply the probability that $$B$$ of those $$M_{2j}$$ components are successful.

Now, we look at $$P(M_1^{A,k})$$, which is simply the probability of having the $$j$$ components be the successful ones, ie $$P(M_1^{A,k})=\prod_{i\in\{j\}}p_{1i}\cdot \prod_{i\not\in\{j\}}(1-p_{1i})$$

Now that we have expressions for our summands in the joint PMF $$P(A,B)$$, we turn to $$Z=\frac{B}{A}$$. Clearly $$Z$$ is discrete and rational-valued. Its PMF will be $$P(Z=z)=\sum_{(a,b)\;st.\;b/a=z,\; b\leq a\leq n }P(A=a,B=b)$$

Technically that's it, and it's "closed-form" given all the sums/products are finite. But it's ugly sorry!

• Thank you for your answer, this is really helpful with notation and with the conditional probability idea, will +1 once I have enough reputation. I won't accept it yet hoping that someone will come up with a smart assumption that can lead to a nice closed-form solution or at least approximation. Could you think of a similar problem to this which does have a known distribution as the solution? Maybe I could take some inspiration from there Commented Jan 31 at 16:49
• @NilsR I imagine things will simply greatly if further assumptions or restrictions are placed on the $p_{ji}$ values; otherwise, I fail to see how we can avoid long sums or products. The other thing of course is that $Z$ is rational, so its PMF will likely exihibit lots of "unsmoothness", which is not conducive to the existence of a compact PMF (which is likely to be continuous if extended to the reals). Commented Jan 31 at 23:33