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Let $\mathbf{p}=(p_1,\dots,p_m)$ be a vector in $[0,1]^m$ and let $\mathbf{X}=(X_1,\dots,X_m)$ be a vector of independently-distributed binomial random variables such that $X_i\sim \text{Binom}(n,p_i)$. Further, let $h(\mathbf{X})$ be a convex function on $[0,n]^m$.

Question: Is the real-valued function $g(\mathbf{p})=\mathbb{E}_\mathbf{X}[h(\mathbf{X})\ |\ \mathbf{p}]$ convex on $[0,1]^m$?

Notes:

  1. This question is a follow-up from a previous question, from which we know that in the univariate case, i.e., $m=1$, $g(p)$ is convex.
  2. I have done some numerical testing and it appears that $g(\mathbf{p})$ is convex
  3. The function $g(\mathbf{p})$ is identical to one representation of the multivariate Berstein polynomial. I haven't seen this representation much in the literature (see another MO question). However, there is another more common representation, for which some convexity results exist.
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  • $\begingroup$ from your notation, is $g$ a vector valued function? then Noah's argument from the previous question immediately yields the answer. I presume that what you really want is a real-valued $g$ that should be convex on $[0,1]^m$? $\endgroup$
    – Suvrit
    Commented Mar 16, 2013 at 1:57

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The answer is no, a counterexample is for $n=1$, $m=2$: let $h$ be zero on (0,0), (0,1), (1,0), and let $h(1,1)=1$. It is rather clear that you can get it for a convex $h$.

Then, $g(p_1,p_2)=p_1 p_2$, which is not a convex function on the unit square.

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    $\begingroup$ Thanks! I agree that the answer is no. Another example of a convex $h$ that makes $g$ non-convex is $h(X_1,X_2)=(1-X_1)+(1-X_2)+1$. $\endgroup$
    – Hugh Medal
    Commented Mar 18, 2013 at 11:17
  • $\begingroup$ @Hugh The function $h$ defined by $h(x_1,x_2)=(1-x_1)+(1-x_2)+1$ is linear hence there is no counterexample there. If you mean $h(x_1,x_2)=(1-x_1)(1-x_2)+1$, then this is Victor's example modulo an irrelevant affine part. $\endgroup$
    – Did
    Commented Mar 26, 2013 at 17:14

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