Good UPPER bounds for $\log(\sum_{i=1}^n p_ie^{z_i})-\sum_{i=1}^np_iz_i$ where $(p_i)_i$ is a probability vector Let $x=(z_1,\ldots,z_n)$ be real vector and $(p_1,\ldots,p_n)$ be a probability vector.
Question
$\log(\sum_{i=1}^n p_ie^{z_i})-\sum_{i=1}^np_iz_i \le ???$
Observation
This paper allows us to upper-bound things like $\Delta_f(z,p):=f(\sum_{i}z_i p_i) - \sum_i f(p_iz_i)$, thus providing a kind of reversed Jensen's inequality.
Indeed, it was shown that

If $f$ is concave, $a:=\min_i z_i$ and $b := \max_i z_i$, then
  $$
\Delta_f(z,p) \le \max_{p,q \ge 0,\;p+q=1} pf(b)+qf(a) - f(pb+qa).
$$

I could probably use this with $f=\log$ to bound $\Delta_{\log} (z,p)$.
 A: One answer - subgaussian variables generalize this property.
Let $\mu = \sum_i p_i z_i$, then the distribution is considered $\sigma^2$-subgaussian if for all $\lambda \in \mathbb{R}$, 
$$ \log\left(\sum_i p_i e^{\lambda(z_i - \mu)}\right) \leq \frac{\lambda^2 \sigma^2}{2} $$
i.e.
$$ \log\left(e^{-\lambda \mu} \sum_i p_i e^{\lambda z_i}\right) \leq \frac{\lambda^2 \sigma^2}{2} $$
i.e.
$$ \log\left(\sum_i p_i e^{\lambda z_i}\right) - \lambda \mu \leq \frac{\lambda^2 \sigma^2}{2} . $$
Your expression is the case $\lambda=1$.
In a sense this is not an answer, just a definition for when and how your inequality can be satisfied, but hopefully useful because we know ways to show variables are subgaussian:


*

*Any random variable bounded in $[a,b]$ is $\left(\frac{b-a}{2}\right)^2$-subgaussian.

*If $X$ is $\sigma^2$-subgaussian, then $cX$ is $c^2\sigma^2$-subgaussian.

*If $X$ and $Y$ are independent and $\sigma_1^2,\sigma_2^2$ subgaussian, then $X+Y$ is $\sigma_1^2 + \sigma_2^2$ subgaussian.

