Suppose we have a random vector and a convex set such that

If you are doing things with convexity, then you may wonder whether

This is certainly true if only takes finitely many value in or is closed. In the first case, you just verify the definition of convexity and the second case, you may use the strong law of large numbers. But if you draw a picture and think for a while, you might wonder whether these conditions are needed as it looks like no matter what value takes, it can not go out of and the average should still belong to as long as is convex. In this post, we are going to show that it is indeed the case and we then have a theorem.

Theorem 1For any convex set , and for any random vector such that

its expectation is still in , i.e,

as long as the mean exists.

Skip the following remark if you don’t know or not familiar with measure theory.

Remark 1If you are a measure theoretic person, you might wonder whether should be Borel measurable. The answer isno. The set needs not to be Borel measurable. To make the point clear, suppose there is an underlying probability and is a random variable from this probability space to where is the borel sigma-algebra. Then we can either add the condition that the event or is understood as is a measurable event with respect to the completed measure space and we overload the notation to mean . The probability space is completed by the probability measure .

To have some preparation for the proof, recall the separating hyperplane theorem of convex set.

Theorem 2 (Separating Hyperplane theorem)Suppose and are convex sets in and , then there exists a nonzero such that

for all .

Also recall the following little facts about convexity.

- Any convex set in is always an interval.
- Any affine space of dimension in is of the form for some and .

We are now ready to prove Theorem 1.

*Proof of Theorem 1:* We may suppose that has nonempty interior. Since if it is not, we can take the affine plane containing with smallest dimension. Suppose is not in , then by separating hyperplane theorem, there exists a nonzero such that

Since almost surely, we should have

almost surely. Since , we see that with probability . Since intersection of the hyperplane of and is still convex, we see that that only takes value in a convex set in a dimensional affine space.

Repeat the above argument, we can decrease the dimension until . After a proper translation and rotation, we can say that takes its value in an interval in and we want to argue that the mean of is always in the interval.

This is almost trivial. Suppose the interval is bounded. If the interval is closed, then since taking expectation preserves order, i.e., , we should have its mean in the interval. If the interval is half open and half closed and if the means is not in the interval, then must be the open end of the interval since expectation preserves order, but this means that has full measure on the open end which contradicts the assumption that is in the interval with probability one. The case both open is handled in the same way. If the interval is unbounded one way, then the previous argument still works and if it is just , then for sure that . This completes the proof.

Nice!

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