The Lumber Room

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Convex, continuous, Jensen’s

with 3 comments

On the principle that things I’ve forgotten at least thrice are worth remembering…

A convex function (need a mnemonic?) is f such that
\displaystyle f(\lambda x + (1-\lambda)y) \le \lambda f(x) + (1-\lambda)f(y) for any λ in [0,1].

Midpoint convex (almost) implies convex
A natural question is whether we need all λ in [0,1]; what about just ½? Functions which satisfy
\displaystyle f(\frac{x+y}{2}) \le \frac{f(x)+f(y)}{2}
are called midpoint convex.

It turns out that when things are “nice” as in most frequently encountered cases, a midpoint convex function is convex. (A counterexample is this crazy non-measurable nowhere-continuous bounded-in-no-interval function: since R is a vectorspace over Q, there is a basis (that includes π, say), and any real number x has a unique representation as a finite rational sum of basis elements. Define f(x) to be the coefficient of π in the representation of x. This function is in fact linear over Q.)
For midpoint convex to mean convex, any of the following conditions will do:

  • f is continuous [Proof: prove the inequality for λ with denominators that are powers of 2, then use denseness in R.]
  • f is continuous at some point (any one point will do) [Proof: see next section]
  • f is (Lebesgue) measurable
  • f is bounded on some measurable set with positive measure

Convex functions are continuous
In fact, a stronger theorem is true: If a midpoint convex function is discontinuous at a single point, then it is unbounded on every interval and (hence) discontinuous everywhere.
Proof: Suppose wlog that f is discontinuous at 0, that f(0)=0, and that a sequence x_n \rightarrow 0 has f(x_n) \rightarrow m for some positive m. Prove that \liminf f(2^k x_n) \ge 2^k m, so f is unbounded near 0. Then use this to show unbounded everywhere.

Note that convex functions are bounded on every interval, so by above theorem they are continuous.

Convex functions are differentiable almost everywhere
In fact a convex function has left and right derivatives at all points, these are monotonically increasing, and there are only countably many points where they don’t coincide.

All the above is from pages 10 to 16 of this book I had found through googling; if anyone’s reading this, please tell me if you know better proofs.

A proof of Jensen’s inequality
Here, Jensen’s inequality is the statement that for a convex function f,
\displaystyle \mathbb{E}\left[f(X)\right] \geq f\left(\mathbb{E}\left[X\right]\right) .

An equivalent definition of a convex function is that at any point we can find a line through the point that lies entirely below the function. (Why?) In other words, for any point a, there is a “slope” m such that
\displaystyle f(x) \ge f(a) + m(x-a) for all x.
With this definition, Jensen’s inequality has an absurdly short proof: take a to be the mean \mathbb{E}\left[X\right], then take expectation.

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Written by S

Tue, 2010-06-29 at 02:32:05 +05:30

Posted in mathematics

3 Responses

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  1. Your use of “almost always” is misleading. As far as real valued functions go, if there is one counterexample, most things are counterexamples. The set of midpoint convex functions is a linear subspace of the set of convex functions. Any proper subspace of a vector space over an infinite field is almost *nothing* in the whole space (for instance, a line is almost nothing in the plane, almost all continuous functions are nowhere differentiable, etc.).

    vipulnaik

    Thu, 2010-07-01 at 09:55:14 +05:30

    • Ah yes. Good point, thanks: I’ve changed it to “in most frequently encountered cases”.
      Since I knew that “almost all”, “almost everywhere” and “almost sure” have technical meanings; I should have avoided using “almost always” in a nontechnical sense. :-)

      S

      Thu, 2010-07-01 at 12:11:56 +05:30

  2. OK, too many of my interviewers have told me “Jensen’s inequality is the reason all of us make money here in quant finance” and I can’t yet seem to figure out how. Damn.

    Mohan

    Wed, 2010-07-28 at 15:23:10 +05:30


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