Distribution of sample variance of Bernoulli variables
I am facing the following problem, given $X_1 ... X_n$ a random sample of $Bernoulli(\theta)$ variables find the distribution of the sample variance $S^2 = \frac{1}{n} \sum_i(\bar{X} - X_i)^2$.
I have demonstrated that $S^2 = \bar{X} (1 - \bar{X})$ and i know $\bar{nX}$ has distribution $Binomial (n, \theta)$ but I have not been able to deduce the distribution of $S^2$.
$\endgroup$ 41 Answer
$\begingroup$You know the distribution of $\overline X$ and you’ve correctly determined that $S^2$ is a function of $\overline X$. Then the distribution of $S^2$ is just the composition of the two: The probability of a value of $S^2$ is the sum of the probabilities of the values of $\overline X$ that are its preimages; that is, for all integers $k$ from $0$ to $\left\lfloor\frac n2\right\rfloor$ we have:
$$ P\left(S^2=\frac kn\left(1-\frac kn\right)\right)=P\left(\overline X=\frac kn\lor\overline X=1-\frac kn\right)= \begin{cases} \binom nk\theta^k(1-\theta)^k&k=\frac n2\;,\\ \binom nk\left(\theta^k(1-\theta)^{n-k}+\theta^{n-k}(1-\theta)^k\right)&\text{otherwise}\;.\\ \end{cases} $$
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