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How can you go from $4V(\bar X)$ to $\displaystyle \frac{4}{n}V(X_1)$? I understand the rest of the steps...
$\endgroup$ 01 Answer
$\begingroup$Assuming that $X_1,\ldots,X_n$ are independent and identically distributed (this is used explicitly in equality 3 and 4) we have$$ \begin{align} \mathrm{Var}(\bar{X})&=\mathrm{Var}\left(\frac{1}{n}\sum_{i=1}^nX_i\right)=\frac{1}{n^2}\mathrm{Var}\left(\sum_{i=1}^n X_i\right)=\frac{1}{n^2}\sum_{i=1}^n\mathrm{Var}(X_i)\\ &=\frac{1}{n^2}n\mathrm{Var}(X_1)=\frac{1}{n}\mathrm{Var}(X_1). \end{align} $$
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