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.

  • 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, 2019 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, 2019 at 21:50

1 Answer 1


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.


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