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suppose $X_1, X_2,\ldots,X_n$ be a random sample of $B(1,p)$ distribution.if $$\bar{X}=\frac{1}{n}\sum_{i=1}^n X_i, \bar{X^2}=\frac{1}{n}\sum_{i=1}^n X_i^2$$ how can I calculate $E(\bar{X}\bar{X^2})$

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closed as off-topic by heropup, egreg, Claude Leibovici, Davide Giraudo, Sami Ben Romdhane Apr 4 at 11:26

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What have you tried? Given your posting history, you should at least have been able to attempt something. –  heropup Apr 4 at 9:07

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I don't know if I'm missing something essential here... As $X_i$ follows a $B(1,p)$ distribution, $X_i$ only takes the values zero and one. Therefore, $X_i^2 = X_i$ and so $$ \bar{X}^2 = \frac{1}{n}\sum_{i=1}^n X_i^2 = \frac{1}{n}\sum_{i=1}^n X_i. $$ Therefore, we obtain $$ E\bar{X}\bar{X}^2 = E(\bar{X})^2 = E\left(\frac{1}{n}\sum_{i=1}^n X_i\right)^2 = \frac{1}{n^2} E\left(\sum_{i=1}^n X_i\right)^2. $$ As each $X_i$ is $B(1,p)$ and independent of each other, $Y = \sum_{i=1}^n X_i$ follows a $B(n,p)$ distribution. Therefore, it has mean $np$ and variance $np(1-p)$. The second moment is therefore $$ E\left(\sum_{i=1}^n X_i\right)^2 =EY^2 = VY + E^2Y = np(1-p) + (np)^2 = np(1-p+np), $$ and so $$ E\bar{X}\bar{X}^2 = \frac{np(1-p+np)}{n^2} = \frac{p(1-p+np)}{n}. $$

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