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 is the expectatio $ \langle F \rangle_z(y) := \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 above probability measure.
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.
