Minimum of $\prod_{1\leq i \leq n} (1-p_i) \left[ \sum_{i=1}^n \frac{p_i}{1-p_i} + \sum_{i=1}^n \sum_{j=i+1}^n \frac{p_i p_j}{(1-p_i)(1-p_j)} \right]$

Let $$p_1, ..., p_n \in (0,q)$$ such that $$\sum_i p_i \geq q$$ for $$q\in[0,1]$$. I think that

$$\prod_{1\leq i \leq n} (1-p_i) \left[ \sum_{i=1}^n \frac{p_i}{1-p_i} + \sum_{i=1}^n \sum_{j=i+1}^n \frac{p_i p_j}{(1-p_i)(1-p_j)} \right]$$ Is minimized when $$p_i=\frac{q}{n}$$ for all $$i$$, but I don't know how to show it. Any hints? If not, then what is the minimizer?

I can show $$\prod_{1\leq i \leq n} (1-p_i) \left[ \sum_{i=1}^n \frac{p_i}{1-p_i} \right]$$ is minimized when $$p_i=\frac{q}{n}$$,

• I might be misreading your equations, but, trying it out for n=3, q=1, I get $\tfrac 23$ with your minimizer, but $p_i=0.99$ gives 0.03, and it seems like I can get as close to 0 as I want by getting closer to 1 Feb 18, 2023 at 3:23

Partial answer, for the case $$q=1$$, $$n>2$$, there seems to be no minimizer, but, you can get as close to 0 as you want by choosing $$p_i=1-\epsilon$$: $$\epsilon^n \left[ n\frac{1-\epsilon}{\epsilon} + \frac{n(n-1)}2 \frac{(1-\epsilon)^2}{\epsilon^2} \right]\\ =n \epsilon^{n-2} (1-\epsilon)\left(\epsilon+\frac{(n-1)(1-\epsilon)}{2}\right)$$ Since we assumed $$n>2$$, this scales down with $$\epsilon$$