"Integrating" a concave function over the Poisson distribution

Question

Assume that $f(x)$ is a concave function and $P_\lambda(k)=\frac{\lambda^k}{k!}e^{-\lambda}$ is the Poisson distribution.

Now I "integrate" $f(x)$ over this distribution, like:

$$g(\lambda) := \sum_k f(k) P_\lambda(k)$$

What can be said about $g$ as a function of $\lambda$?

Is it still concave?

Answer 1

Note that $P'_k(\lambda)=P_{k-1}(\lambda)-P_k(\lambda)$ and hence $$P_k''(\lambda) = P_{k-2}(\lambda)-2P_{k-1}(\lambda)+P_k(\lambda).$$ Then $$g''(\lambda)=\sum_k f(k)\left((P_{k-2}(\lambda)-2P_{k-1}(\lambda)+P_k(\lambda)\right)=\sum_k\left(f(k{+}2)-2f(k{+}1)+f(k)\right)P_k(\lambda).$$ But $f(k{+}2)-2f(k{+}1)+f(k)\le0$ because $f$ is concave.