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 $S$?

  • $\begingroup$ This is pretty straightforward (by the look of it a routine bookwork-style question). Is this for some class? (If not, how does it arise?) $\endgroup$
    – Glen_b
    Commented May 23, 2016 at 23:12

1 Answer 1


I assume you're aware of basic properties of expectation:

  1. $E(X+Y) = E(X)+E(Y)$

    (the generalizations of these to $n$ variables is obvious by repeated application of the case with two terms)

  2. $E(cX) = cE(X)$

    I assume you're also aware that for independent random variables

  3. $\text{Var}(X+Y) = \text{Var}(X)+\text{Var}(Y)$

    and (for any random variable whose variance exists)

  4. $\text{Var}(cX) = c^2\text{Var}(X)$

    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):

  5. $\mu_k(X+Y) = \mu_k(X)+\mu_k(Y)$

    Further, I assume it is immediately clear that

  6. $\mu_k(cX) = c^k\mu_k(X)$

    -- 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.


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