3
$\begingroup$

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.
$\endgroup$
1
  • $\begingroup$ 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
    Mar 16, 2013 at 1:57

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ 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
    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
    Mar 26, 2013 at 17:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.