Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $E(\bar{X}\bar{X^2})$

share|improve this question

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, egreg, Claude Leibovici, Davide Giraudo, Sami Ben Romdhane
If this question can be reworded to fit the rules in the help center, please edit the question.

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

1 Answer 1

up vote 0 down vote accepted

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 $$ E\bar{X}\bar{X}^2 = \frac{np(1-p+np)}{n^2} = \frac{p(1-p+np)}{n}. $$

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.