 # Convex combination with binomial probabilities. Suppose that we have some z_k>0, k={0,1,⋯,n}. I want to compare weighted averages of z_k's when the weights are defined by binomial probabilities. puntdald8 2022-09-13 Answered
Convex combination with binomial probabilities
Suppose that we have some zk>0, k={0,1,⋯,n}. I want to compare weighted averages of zk's when the weights are defined by binomial probabilities.
More specifically, for p and q, where p,q∈(0,1), and for some λ∈(0,1), let x=λp+(1−λ)q.
In this case, should the following be true for all λ∈(0,1)?
$max\left\{\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right){p}^{k}\left(1-p{\right)}^{n-k}{z}_{k},\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right){q}^{k}\left(1-q{\right)}^{n-k}{z}_{k}\right\}\ge \sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right){x}^{k}\left(1-x{\right)}^{n-k}{z}_{k}.$
I could see that it's true for $n=2$ but I can't show that it still holds for any $n>2$.
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Step 1
No, this is not the case, not even for $n=2$. You excluded 0 and 1, but by continuity we can still use them for a simple counterexample:
Step 2
For $p=0$, $q=1$, the left-hand side is zero whereas the right-hand side is positive for any $\lambda \in \left(0,1\right)$.

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