Bounding the tail of sum of discrete distributions (via sub-gaussianity) I have the following problem: we have a sequence of random variables $Z_1, ..., Z_n$ which are summed up; let's denote $X$ to be their sum. We observe a number $\epsilon$ that is sampled from $X$ and we wish to somehow describe/bound the probability $P(X \geq \epsilon)$; that is, we want to compute the extremeness (quantile) of $\epsilon$. We are interested only in the right tail of $X$.
Each of $Z_i$ is a discrete distribution with probabilities $p_i=[p_1,...p_k], k \in N$ and corresponding values given as $x=[x_1,...,x_k]=[-\log(p_1),...-\log(p_k)]$ (we assume p > 0). The values of $p_i$ (which also give us the $x_i$) are known. The distributions $Z_1,...Z_n$ are independent (but not necessary identically distributed).
Since we know all $p_i$, we can compute the means and variances of $Z_i$. (The mean of $Z_i$ happens to be the entropy of $Z_i$, by the way how we defined the values $x_i$.) By summing the means and variances of $Z_i$, we get the mean $\mu$ and variance $\sigma^2$ of $X$.
To bound the tail probability of $X$, we can use Chebyshev inequality: we have $P(X \geq \epsilon) \leq \frac{\sigma^2}{(\epsilon - \mu)^2}$. However, the bound is very loose and I am looking for a better one.
Also, in the limit for $n \to \infty$, the $X$ will be normally distributed by CLT; however, I will mostly deal with $n$ between 10-30, where the normal approximation is not so good.
For these reasons, I am trying to utilize the subgaussianity of $Z_i$. We say that random variable $Z$ is b-subgaussian, if
$$
\mathbb{E}\left[ \exp\lambda (Z - \mathbb{E}[Z])\right] \leq \exp \left( \frac{\lambda^2b^2}{2} \right) \quad \forall \lambda \in \mathbb{R}
$$
The subgaussianity is preserved through summation (if $Z_1$ is $b_1$-subgaussian and $Z_2$ is $b_2$-subgaussian, then $Z_1 + Z_2$ is $\sqrt{b_1^2 + b_2^2}$ subgaussian.) Therefore, if we compute $b_i,...b_n$ such that $Z_1,...,Z_n$ are $b_i,...b_n$-subgaussian, we can obtain $b$ such that $X$ is b-subgaussian.
The subgaussian property gives us a much stronger bound on the tail of $X$ compared to the Chebyshev inequality: if $X$ is $b-$subgaussian, then
$$
P(X \geq \epsilon) \leq \exp \left( \frac{-(\epsilon-\mu)^2}{2b^2}\right)
$$
That is, lower values of $b$ are better (give us tighter bound).
Any distribution with values bounded by interval $[c,d]$ is $b-$subgaussian for $b \geq \frac{(d-c)^2}{4}$ by Hoeffding's lemma. However, this bound is again too lose for my usecase.
That is, I would like to ask:

*

*Is there some algorithm, that given the discrete probability distribution $p=[p_1,...p_k]$ and corresponding values $x$ (in my case, I have $x=-log(p)$) computes $b$ such that the distribution is $b-$subgaussian for reasonably low $b$?

*Alternatively, is there some algorithm, that given my values of $p$ and $x$ and given some $b$ decides whether the distribution is $b-$subgaussian? (I can then use it to find the lowest $b$ by binary search).

*After all, I try to bound the tail probability of $X$ - maybe the subgaussianity approach is not a good idea? Are there any better / reasonable ways how to obtain a tight bound?

P.S. Another way how to estimate the probability in the tail is by Monte-Carlo sampling from $X$ (which I can, since I know all $Z_i$). However, this will be computationally too expensive for me. In my use case, I know all possible values of $Z$ in advance (there may be about 100 possible values of vectors $p$, and I know them in advance) and I can precompute even computationally expensive stuff. Then, I am given the actual sequence of $Z$s (the sequence will be about 10-30 of $Z$s selected from the possible values of $Z$ that were given in advance), and I need to compute the bound on the tail of $X$ as fast as possible.
References:
I got the introduction to subgaussianity from Lattimore, Tor, and Csaba Szepesvári. Bandit algorithms (2020), chapter 5. The book is available here.
Another resourse I went through is this lecture notes (it mention the usage Hoeffidng's lemma).
edit: this material states an equivalent condition for subgaussianity, that can be checked algorithmically (Theorem 2.1, page 6 of the .pdf). I am currently trying to understand the proof and somehow construct an algorithm that decide if distribution is $\sigma$-gaussian for given $\sigma$.

This answer aims at your first question: how to obtain a reasonably low value of $b$ given $p$. Ideally, this actually provides the best value of $b$.
Denote $\mathbb{E}[Z] = H(Z)$. On the left side of the inequality for b-subgaussian definition, we have
\begin{align}
\mathbb{E}\left[e^{\lambda(Z-H(Z))}\right] 
&= e^{-\lambda H(Z)}\left(\sum_{i=1}^k p_ie^{-\lambda \log(p_i)} \right)\\
&= e^{-\lambda H(Z)}\left(\sum_{i=1}^k p_i^{1-\lambda} \right)\\
&= e^{-\lambda H(Z)}\phi(\lambda) \quad,
\end{align}
where $\phi(\lambda) = \sum_{i=1}^k p_i^{1-\lambda}$. Now notice that
$$ e^{-\lambda H(Z)}\phi(\lambda) \leq e^{\frac{\lambda^2b^2}{2}}, \,\forall \lambda \, \in \, \mathbb{R} \iff -\lambda H(Z) + \log(\phi(\lambda)) \leq \frac{\lambda^2b^2}{2}, \, \forall \lambda \, \in \, \mathbb{R} \quad. $$
For $\lambda = 0$, the equation is trivially satisfied. For $\lambda \neq 0$, define
$$ f(\lambda) = \frac{-\lambda H(Z) + \log(\phi(\lambda))}{\lambda^2} \quad.$$
Note $f$ is smooth, $\underset{\lambda \rightarrow +\infty}{\lim} f(\lambda) = \underset{\lambda \rightarrow -\infty}{\lim} f(\lambda) = 0$. Moreover, you can actually show that (use L'Hôpital twice) $\underset{\lambda \rightarrow 0}{\lim} f(\lambda) $ exists and is well defined. Therefore, this function is bounded above. Let $M = \underset{\lambda \, \in \, \mathbb{R}}{\sup} f(\lambda) > 0$, the best $b$ is then
$$ b = \sqrt{2M}$$.
Given $p$, $H(Z)$ is a constant, and then just use numerical methods to obtain $M$.
Below is an image of $f$ generated using R for $p = (0.1, 0.2, 0.3, 0.4)$ from $-100$ to $100$ in steps of $0.001$ (skipping 0).

