4
$\begingroup$

The following result is well-known:

Suppose that $H(x,y)$ is a log-concave distribution for $(x,y) \in \mathbb R^{m \times n}$ so that by definition we have $$H \left( (1 - \lambda)(x_1,y_1) + \lambda (x_2,y_2) \right) \geq H(x_1,y_1)^{1 - \lambda} H(x_2,y_2)^{\lambda},$$ and let $M(y)$ denote the marginal distribution obtained by integrating over $x$ $$M(y) = \int_{\mathbb{R}^m} H(x,y) \, dx.$$ Let $y_1$ $y_2 \in \mathbb R^n$ and $\lambda \in (0,1)$ be given. Then the Prékopa–Leindler inequality applies. It can be written in terms of $M$ as $$M((1-\lambda) y_1 + \lambda y_2) \geq M(y_1)^{1-\lambda} M(y_2)^\lambda$$ which is the log-concavity for $M$.

Now, I wanted to understand this for a very simple example where $f: \mathbb R^2 \rightarrow \mathbb R:$

$$e^{-g(y)} = \int_{\mathbb R} e^{-f(y,z)} \ dz.$$

Then, I want to prove that $g''\ge 0$ if $f$ satisfies $D^2f > 0$ globally as a matrix. We assume for simplicity that $f$ is such that the above integral is well-defined.

It is easy to see that

$$g''(y) = \langle D_{yy}f \rangle_z - \operatorname{ Var}_z (D_{y}f)$$

where $\langle \cdot \rangle_z$ is the expected value $$ \langle F \rangle_z(y) := \frac{\int_{\mathbb R} F(y,z) e^{-f(y,z)} \ dz}{ \int_{\mathbb R} e^{-f(y,z)} \ dz} $$ and $\operatorname{ Var}_z$ is the variance with respect to the probability measure with density $p(z) \propto e^{-f(y,z)}$.

However, it is not at all clear to me from this representation why $g''\ge 0$ holds.

Is there a pedestrian way to see this from the above expression for the second derivative?

I am looking for a more "Calculus" based derivation (using the 2nd derivative) of this inequality than the usual convex-combinatorial arguments.

$\endgroup$
  • 2
    $\begingroup$ The proof in Tao’s notes is super pedestrian, no? terrytao.wordpress.com/tag/prekopa-leindler-inequality (You can ignore the tensor product argument since your case is exactly the case he reduces to.) Sorry that this doesn’t literally answer your question, but I feel it’s a very good explanation worth sharing. $\endgroup$ – alpoge May 23 '19 at 21:40
  • 1
    $\begingroup$ @alpoge I indeed want to have a proof that relies less on convex-combinatorial arguments, like in Tao's notes, but one that uses more calculus. $\endgroup$ – Sascha May 23 '19 at 21:50
6
$\begingroup$

By the Brascamp–Lieb concentration inequality, we have $$ \operatorname{ Var}_z (D_{y}f) \le \langle (D_{zz}f)^{-1} (D_{zy} f)^2 \rangle_z \;, $$ and hence, $$ g''(y) \ge \langle (D_{zz}f^{-1}) (D_{zz} f D_{yy} f - (D_{z y} f)^2 ) \rangle_z = \langle (D_{zz}f^{-1}) \det D^2 f \rangle_z > 0 $$ since $D^2 f$ is globally positive definite (by assumption), which implies that $\det D^2 f > 0$ and $D_{zz} f > 0$.

A much more general version of this result can be found in Theorem 4.2 of Brascamp and Lieb's original paper (cited below), where they regard this result as a sharpened version of Prékopa's theorem. See Section 4 of:

Brascamp, Herm Jan; Lieb, Elliott H., On extensions of the Brunn-Minkowski and Prekopa-Leindler theorems, including inequalities for log concave functions, and with an application to the diffusion equation, J. Funct. Anal. 22, 366-389 (1976). ZBL0334.26009.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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