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Suppose $X_1, \dots, X_n$ are iid binomial random variables with parameters $k$ and $p$. So $$P(X_k = k|k,p) = \binom{k}{x}p^{x}(1-p)^{k-x}$$

Here $k$ and $p$ are unknown and we want to find point estimators for them. Why is the second population moment $kp(1-p)+k^{2}p^{2}$? That is, we have:

$$\bar{X} = kp$$ $$\frac{1}{n} \sum X_{i}^{2} = kp(1-p)+k^{2}p^{2}$$

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Do you mean $P(X_k=x|k,p)$? –  Thomas Andrews Jan 30 '12 at 12:47
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You seem to have $k$ and $i$ mixed up: you may want something like $P(X_i = x_i|k,p)$ on the left hand side of your first formula. I also suspect your last two formulae are expectations on the left hand side, i.e. $E\left[ \frac{1}{n} \sum X_i \right]$ and $E\left[ \frac{1}{n} \sum X_i^2 \right]$ –  Henry Jan 30 '12 at 12:48

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You need to know or be able to work out that the expectation of a sum is the sum of the expectations, that the sum of variances of independent random variables is the sum of the variances, and that the variance (or second central moment) is the second moment minus the square of the first moment.

We can consider each $X_i$ to be distributed as $\sum_{j=0}^k Y_{i,j}$ where each $Y_{i,j}$ is an independent Bernoulli random variable taking the value $1$ with probability $p$ and $0$ with probability $1-p$, and $E[Y_{i,j}]=p$. Similarly $E[Y_{i,j}^2]=p$ and $E[\left(Y_{i,j}-E[Y_{i,j}]\right)^2]=E[Y_{i,j}^2]-E[Y_{i,j}]^2 =p - p^2 = p(1-p)$.

So $E[X_{i}]=\sum_{j=0}^k E[Y_{i,j}]=kp$ and $E[\left(X_{i}-E[X_{i}]\right)^2]=\sum_{j=0}^k E[\left(Y_{i,j}-E[Y_{i,j}]\right)^2]=kp(1-p).$ But $E[X_{i}^2]=E[\left(X_{i}-E[X_{i}]\right)^2] + E[X_{i}]^2 = kp(1-p) + k^2 p^2$, which is really the result you want.

If you need it spelt out $E\left[ \frac{1}{n} \sum_{i=1}^n X_i^2 \right] = \frac{1}{n} E\left[ \sum_{i=1}^n \left(kp(1-p) + k^2 p^2\right) \right] = kp(1-p) + k^2 p^2$.

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