Mean and Variance for a sum of independent weighted bernoulli random variables with different probabilities of success

Suppose $Z_i$ are independent Bernoulli random variables with differing probabilities $P_i$. Also suppose weights $W_i$ are positive and constant.

Can you tell me the mean and variance for the random variable $S$ which is the summation of each weighted $Z_i$ (i.e. $W_iZ_i$). Furthermore is there a simple distribution for this random variable (similar to the Poisson Binomial distribution)?

• What exactly are the weights supposed to mean? In what sense are you "summing" these variables? Are you literally forming the random variable $\sum_i w_i Z_i$ (as you write), or are you perhaps trying to form a mixture of their distributions, or maybe something else? And where you write that the weights are "constant" does that mean they are all equal to each other?
– whuber
Oct 10, 2014 at 16:26

I understand $Z_i$ takes value 1 with probability $P_i$ and $0$ otherwise.

Computing the mean of the weighted sum is simple: by linearity of the $\mathrm E$ operator,

$$\mathrm E \sum_i W_i Z_i = \sum_i W_i \mathrm E Z_i = \sum_i W_i P_i.$$

As for the variance, since the variables are independent you can just sum their variances. So:

$$\mathrm{Var} \sum_i W_i Z_i = \sum_i W_i^2 \mathrm{Var}Z_i = \sum_i W_i^2 P_i(1-P_i).$$

The distribution of $\sum_i W_i Z_i$ does not have a name, to my knowledge.

• To make it clearer that this answers (quite correctly!) one possible interpretation of the question, consider pointing out that the weighted sum of $n$ variables you are working with will not describe a distribution supported on the integers $\{0,1,\ldots,n\}$, which is the support of a Poisson-Binomial distribution (unless all the weights are equal to $1$). This contrast casts some doubt on whether your interpretation, albeit a natural one, truly is the intended one.
– whuber
Oct 10, 2014 at 20:33