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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 $\sigma_{\bar X}^2= \text{Var}[\bar X]$.

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

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  • $\begingroup$ 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
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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 $Var(X_1)$.

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

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