# Simplify a weighted average with enumerations as weights

Could you help me simplify the following expression: $$\forall n_0 \ge 0, \forall m_1,m_2 \gt 0, \forall i_1 \in [[0,n_0m_1]],\forall i_2 \in [[0,n_0m_2]],$$ $$m_{n_0,m_1,m_2}(i_1,i_2) = \frac{\sum_{k=0}^{n_0} w_k k}{\sum_{k=0}^{n_0} w_k n_0},$$ $$w_k = w(i_1-k,m_1-1,n_0+1) w(i_2-k,m_2-1,n_0+1),$$ $$w(i,n,m) = \sum_{k=0}^n (-1)^k \binom{n}{k} \biggl(\binom{n}{i-km}\biggr)$$ (with the convention that $$0!=1$$ and $$\forall k \lt 0, \frac{1}{k!} = 0$$, and the notations $$m_j = n_j +1$$ and $$\bigl(\binom{n}{k}\bigr) = \binom{n+k-1}{k}$$)

which gives: $$m_{n_0,m_1,m_2}(i_1,i_2) = \frac{\sum_{k_0=0}^{n_0} \sum_{k_1=0}^{m_1-1} (-1)^{k_1} \binom{m_1-1}{k_1} \bigl(\binom{m_1-1}{i_1-k_0-k_1(n_0+1)}\bigr) \sum_{k_2=0}^{m_2-1} (-1)^{k_2} \binom{m_2-1}{k_2} \bigl(\binom{m_2-1}{i_2-k_0-k_2(n_0+1)}\bigr) k_0}{\sum_{k_0=0}^{n_0} \sum_{k_1=0}^{m_1-1} (-1)^{k_1} \binom{m_1-1}{k_1} \bigl(\binom{m_1-1}{i_1-k_0-k_1(n_0+1)}\bigr) \sum_{k_2=0}^{m_2-1} (-1)^{k_2} \binom{m_2-1}{k_2}\bigl(\binom{m_2-1}{i_2-k_0-k_2(n_0+1)}\bigr) n_0}$$

The goal would be to have a more compact expression, especially for the weights of the average ($$w(i,n,m)$$), to get rid of the sums, and to make it easier to compute the corresponding function: $$\forall x_1,x_2 \in [0,1], m(x_1,x_2)=\lim_{n_0,m_1,m_2 \to \infty, } m_{n_0,m_1,m_2}(\lfloor x_1 n_0m_1 \rfloor,\lfloor x_2 n_0m_2 \rfloor)$$

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.

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

• Hm, the first sum $m_{n_0, m_1, m_2}(i_1, i_2) = \frac{i}{n_0}\frac{\sum w_k}{\sum w_k} = \frac{i}{n_0}$, or am I missing something? – pisoir Dec 16 '18 at 16:24
• Nope it was a stupid mistake, thanks! I corrected it. – CidTori Dec 16 '18 at 18:00