Let $x_1,...,x_m$ be drawn from independent Bernoulli distributions with parameters $p_1,...,p_m$.
I'm interested in distribution of $t=\sum_i a_ix_i,~a_i\in \mathbb{R}$
$m$ is not large so I can not use central limit theorems.
I have the following questions:
1- What is the distribution of $s=\sum_i x_i$?
2- What is the distribution of $t=\sum_i a_ix_i$ or $t=\sum_i a_ix_i-\sum_i a_i$ (to ensure non-negative support) for known $a_i$'s? can I approximate its distribution with a Gamma distribution? If yes, what would be the parameters (as a function of $p_i$'s and $a_i$'s)?
3- Is there a truncated Gamma distribution (or any other distribution (except normal)) that can approximately fits my problem?
However, $m$ is not very large, but it is still very large such that I can not calculate the distribution by convolution.
Thanks a bunch!
I did some search and this is what I have found:
This seminal paper of Raghavan in 1988: "Probabilistic Construction of Deterministic Algorithms: Approximating packing integer programs". In section 1.1 he derives some bounds, which seems to be important as part of an stochastic optimization technique called randomized rounding.
In this paper published this year by Daskalis et al.: "Learning Poisson Binomial Distributions". In theorem 2 they states that it's possible to construct an algorithm that learns the desired distribution in polynomial time.
This post in researchgate.net of Dayan Adoniel also asked the same and it seems that Dayan developed a code to compute the desired distribution and that he is willing to share it.
Unfortunately I do not have the time now to go over the details of the first two references but hopefully you can find them useful.
Edit 1: Bounds in the tails of the distribution of a weighted sum of independent Bernoulli trials [Raghavan, 1988]
Let $a_1, a_2, \ldots, a_r$ be reals in $(0, 1]$. Let $X_1, X_2, \ldots, X_r$ be independent Bernoulli trials with $E[X_j] = p_j$. Define $\Psi = \sum_{j=1}^r a_jX_j$ with $E[\Psi] = m$.
Theorem 1 (Deviation of $\Psi$ above its mean). Let $\delta > 0$ and $m = E[\Psi] > 0$. Then
$$\mathbf{P}(\Psi > m(1+\delta)) < \left[\frac{e^\delta}{(1+\delta)^{(1+\delta)}}\right]^m.$$
Theorem 2 (Deviation of $\Psi$ below its mean). For $\gamma \in (0,1]$,
$$\mathbf{P}(\Psi-m < -\gamma m) < \left[\frac{e^\gamma}{(1+\gamma)^{(1+\gamma)}}\right]^m.$$