Simplify a weighted average with enumerations as weights

109 Views Asked by At

Could you help me simplify the following expression: $$\forall n_0 \ge 0, \forall m_1,m_2 \gt 0, \forall k_1 \in [[0,n_0m_1]],\forall k_2 \in [[0,n_0m_2]],$$ $$\overline{k_0}_{(n_0,m_1,m_2)}(k_1,k_2) = \frac{\sum_{k_0=0}^{n_0} w_{k_0} k_0}{\sum_{k_0=0}^{n_0} w_{k_0}},$$ $$w_{k_0} = \binom{m_1-1}{k_1-k_0}_{n_0+1} \binom{m_2-1}{k_2-k_0}_{n_0+1}$$ with the notations: $$\biggl(\binom{n}{k}\biggr) = \binom{n+k-1}{k}$$ $$\biggl(\binom{-n}{k}\biggr) = (-1)^k \binom{n}{k}$$ $$\binom{n}{k}_{m} = \sum_{i=0}^n \biggl(\binom{-n}{i}\biggr) \biggl(\binom{n}{k-im}\biggr)$$ (actually, $\bigl(\binom{n}{k}\bigr)$ is the number of multisets and $\binom{n}{k}_{m}$ is a kind of generalized Pascal's triangle, with $\binom{n}{k}_{2} = \binom{n}{k}$ but giving different triangles with for example m=3 or m=4)

I tried to use falling and rising factorials but it didn't led me as far as I'd want to.
I also read about Dixon's identity and MacMahon Master theorem but I don't see how to use them.


My goal would be to compute the corresponding function $p_0$ (with $[x]$ being some rounding function): $$\forall p_1,p_2 \in [0,1], p_0(p_1,p_2)=\lim_{n_0,m_1,m_2 \to \infty} \frac{\overline{k_0}_{n_0,m_1,m_2}([p_1 n_0 m_1],[p_2 n_0 m_2])}{n_0}$$

I computed it for some large values of $n_0,m_1,m_2$ and it seems to converge.


To add some background, this is my current attempt to answer this question, which I will edit soon to explain how I found this expression.

To summarize, it's an attempt to compute a good a priori probability for $P(S|A\&B)$, with A and B being independent and knowing the values of $P(S|A)$ and $P(S|B)$ (and maybe $P(S)$)