Bounding sum of weighted exponential I would like to upper bound $\sum_{i}^{n} P_{i} \cdot \exp(-\beta \cdot P_{i} \cdot n)$, where $P_{i}$ are probability measure such that $\sum_{i}^{n} P_{i} = 1$ and $\beta > 0$ is a constant. So basically it's a convex combination, which I could easily upper bound with Hölder's inequality that $$\sum_{i}^{n} P_{i} \cdot \exp(-\beta \cdot P_{i} \cdot n) \leq \max_{i} \exp(-\beta \cdot P_{i} \cdot n)$$
Is there any tighter upper bound for the sum like this ? Any idea or just reference would help !
 A: Since there is no other information available on the $P_i$ and on $\beta$, here is a modest approach where no one-size-fits-all bounds can be expected, rather bounds which fit special bodies.
First observe that OP's bound is actually tight if all $P_i = \frac{1}{n}$. Hence it only makes sense to look for improved bounds for a rather broad distribution of probabilities. In this case, denote with $P_{*}$  the smallest $P_i$,  we expect to have $P_*$ which is considerably lower than $\frac{1}{n}$.
Let us distinguish cases for $\beta$. Starting with very small $\beta$, we can use $e^{-x} < 1-x + x^2/2$ and obtain
$$
\sum_{i}^{n} P_{i} \cdot \exp(-\beta \cdot P_{i} \cdot n) \leq 1 -\beta \cdot n\sum_{i}^{n} P_{i}^2 + \frac{\beta^2 n^2}{2}\sum_{i}^{n} P_{i}^3
$$
By Cauchy-Schwarz, $n\sum_{i}^{n} P_{i}^2 \ge 1$, and the obvious $\sum_{i}^{n} P_{i}^3 \le 1$, so we have
$$
\sum_{i}^{n} P_{i} \cdot \exp(-\beta \cdot P_{i} \cdot n) \leq 1 -\beta + \frac{\beta^2 n^2}{2}
$$
Compare to OP's bound, and ignore terms of quadratic or higher order in $\beta$, we have indeed that
$$
 1 -\beta \leq 1 -\beta P_{*} n \simeq \exp(-\beta \cdot P_{*} \cdot n) = \max_{i} \exp(-\beta \cdot P_{i} \cdot n)
$$
since $ P_{*} < \frac{1}{n}$ was where we started. So this bound improves on OP's bound.
For not so small $\beta$, consider the following:
Consider the function $f(x) = x e^{- ax}$. It has a maximum of $\frac{e^{-1}}{a}$ at $x = \frac{1}{a}$. Let $a = \beta n$,  then if  $\beta > \frac{1}{n}$, this is a good upper bound. It is valid in any case, however depending on the $P_i$ if may not be attained.
So you have
$$
\sum_{i}^{n} P_{i} \cdot \exp(-\beta \cdot P_{i} \cdot n) \leq \sum_{i}^{n} \frac{e^{-1}}{\beta n} = \frac{e^{-1}}{\beta} 
$$
We compare to OP's bound and ask how good this is. We require
$$
\frac{e^{-1}}{\beta} = \frac{e^{-1} P_{*} \cdot n}{\beta P_{*} \cdot n} \leq\!\! ?  \; \exp(-\beta \cdot P_{*} \cdot n) = \max_{i} \exp(-\beta \cdot P_{i} \cdot n)
$$
Understand $q = \beta \cdot P_{*} \cdot n$ as dependent on $\beta$ , then we require
$$
e^{-1} P_{*} \cdot n \leq\!\! ? \; q e^{-q} \le e^{-1} 
$$
So this is favourable (since $P_{*} < \frac{1}{n}$)  if  $q$ of Order 1, i.e. $\beta \simeq  \frac{1}{ P_{*} \cdot n} > 1$.
If indeed $P_{*} << \frac{1}{n}$, then we have $q << 1$ and therefore approximately
$$
e^{-1} P_{*} \cdot n \leq\!\! ? \; q  = \beta \cdot P_{*} \cdot n
$$
i.e. $\frac{1}{P_{*} \cdot n} >> \beta > e^{-1}$ would then be the only requirement (where $\frac{1}{P_{*} \cdot n}$ is considered a very large number, so the left constraint is not severe).
From the discussion, it is also clear that the  ranges for $\beta$ obtained right now match the $\beta$-case which we just discuss.
