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usul
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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.)

usul
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