# Simple proof of Prékopa's Theorem: log-concavity is preserved by marginalization

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.

• 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. – alpoge May 23 '19 at 21:40
• @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. – Sascha May 23 '19 at 21:50

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 $$D_{zz} f > 0$$.