Let $X_1,X_2,X_3,\dotsc$ be an i.i.d. sequence of Bernoulli random variables where $X_i=0$ with probability $0.5$ and $X_i=1$ with probability $0.5$. For a fixed value of $N$, define $$\bar X = \frac{X_1+X_2+X_3+\dotsb +X_N}{N}$$

Find $\mu_{\bar X} = E[\bar X]$ and $\sigma_{\bar X}^2= \text{Var}[\bar X]$.


closed as off-topic by Davide Giraudo, colormegone, Pragabhava, Mark Viola, Hirshy Dec 14 '15 at 18:30

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." – Davide Giraudo, colormegone, Pragabhava, Mark Viola, Hirshy
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ What have you tried so far, and where did you get stuck? Add that to the question (not in a comment). $\endgroup$ – drhab Dec 14 '15 at 9:36
  • $\begingroup$ Just reiterating, please check here for help with formatting your questions. $\endgroup$ – Em. Dec 14 '15 at 9:50

The expectation $E$ has the property $$E\left[\frac1NX_1+\frac1NX_2+\ldots+\frac1NX_N\right]=\frac1NE[X_1]+\ldots+\frac1NE[X_N]$$ which is known as the linearity of expectation. This works regardless of whether the random variables are independent. But if they are independent (as in your case) then this also works for the variance (but be careful, you have to square the scalar) as follows $$Var\left(\frac1NX_1+\frac1NX_2+\ldots+\frac1NX_N\right)=\frac1{N^2}Var(X_1)+\ldots+\frac1{N^2}Var(X_N)$$ Now in your case things simplify even more because the random variables are identically distributed, so that $$E[X_1]=E[X_2]=\ldots=E[X_N] \quad \text{and}\quad Var(X_1)=Var(X_2)=\ldots=Var(X_N)$$ Therefore, you may write $$E\left[\frac{X_1+X_2+\ldots+X_N}{N}\right]=\frac{1}{N}N\cdot E[X_1]=E[X_1]$$ and $$Var\left(\frac{X_1+X_2+\ldots+X_N}{N}\right)=\frac{1}{N^2}N\cdot Var(X_1)=\frac1NVar(X_1)$$ So, all this reduces to finding $E[X_1]$ and $Var(X_1)$.

  • $\begingroup$ Thank you this helps a lot. I couldn't quite understand Bernoulli random variables but this help me have a better understanding $\endgroup$ – Brian Byrne Dec 14 '15 at 10:11
  • 1
    $\begingroup$ You are welcome. But actually this does not work only for Bernoulli random variables. It explains more about the properties of expectation and variance in general rather than the properties of the sum of Bernoulli's. The other answer focuses on that. $\endgroup$ – Jimmy R. Dec 14 '15 at 10:13

Let $X_i \sim Ber(p)$, then $Y = \sum_{i=1}^n X_i \sim Bin(n,p)$.

So, $$E[Y] = np \qquad \text{and} \qquad Var[Y]= np(1-p) $$ as $\bar{X} = \frac{Y}{n}$, $$E[\bar{X}] = p = 0.5 \qquad \text{and} \qquad Var[\bar{X}]= \frac{p(1-p)}{n} = \frac{0.25}{n} $$


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