In deriving normal estimators using method of moments, why does the below equality hold?

$$ \frac{1}{n} \sum X_i^2 - \bar{X}^2 = \frac{1}{n} \sum (X_i - \bar{X})^2 $$

This is from Example 7.2.1 from Casella & Berger in which $X_1, \ldots, X_n$ are iid $\mathcal N(\theta, \sigma^2)$.


Just expand out the right side using FOIL to get $$\begin{align} \sum_{i=1}^n (X_i - \bar{X})^2 &= \sum_{i=1}^n (X_i^2 - 2X_i\bar{X} + \bar{X}^2)\\ &= \left(\sum_{i=1}^n X_i^2\right) -2\bar{X}\left(\sum_{i=1}^n X_i\right) + n\bar{X}^2\\ &= \left(\sum_{i=1}^n X_i^2\right) - 2\bar{X}\cdot n \bar{X} + n\bar{X}^2\\ &= \left(\sum_{i=1}^n X_i^2\right) - n\bar{X}^2 \end{align}$$

  • $\begingroup$ I was missing the step you have in line two to line three, e.g. writing the summation of the X_i's as n*X_bar. Makes sense now. Thanks! $\endgroup$ – Ravi Mar 2 '15 at 18:24

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