# Bernoulli random variables [closed]

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, HirshyDec 14 '15 at 18:30

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• What have you tried so far, and where did you get stuck? Add that to the question (not in a comment). – 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 $Var(X_1)$.
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 $\bar{X} = \frac{Y}{n}$, $$E[\bar{X}] = p = 0.5 \qquad \text{and} \qquad Var[\bar{X}]= \frac{p(1-p)}{n} = \frac{0.25}{n}$$