Suppose $Z_i$ are independent Bernoulli random variables with differing probabilities $P_i$. Also suppose weights $W_i$ are positive and constant. Let's define the random variable $S$ which is the summation of each weighted $Z_i$ (i.e. $S=\sum_i W_iZ_i$).
The Poisson binomial distribution is a special case for this distribution, where $W_i=1$ for all $i$.
Does anyone have any idea of how to compute the skewness for the distribution of the random variable $S$?
I assume you're aware of basic properties of expectation:
$E(X+Y) = E(X)+E(Y)$
(the generalizations of these to $n$ variables is obvious by repeated application of the case with two terms)
$E(cX) = cE(X)$
I assume you're also aware that for independent random variables
$\text{Var}(X+Y) = \text{Var}(X)+\text{Var}(Y)$
and (for any random variable whose variance exists)
$\text{Var}(cX) = c^2\text{Var}(X)$
What you might not have seen is the extension to cumulants (also see here) of higher order than 2, which gives us (for independent random variables):
$\mu_k(X+Y) = \mu_k(X)+\mu_k(Y)$
Further, I assume it is immediately clear that
$\mu_k(cX) = c^k\mu_k(X)$
-- but if it is not, it should be easy for you to prove by elementary means.
These tools (specifically 3-6 or even just 5&6) should be sufficient to obtain the result you need in a couple of lines.
