2

How did they get from equation (3) to equation (4)?

$$S^2 = \frac{1}{n} \sum (X_i - \bar{X})^2 \tag{0}$$

$$E[S^2] = E\Big[\frac{1}{n} \sum (X_i - \bar{X})^2 \Big]\tag{1}$$

$$E[S^2] = E\Bigg[\frac{1}{n} \sum \limits_{i=1}^{n}\Big[~[(X_i - \mu)-(\bar{X}-\mu)]^2~\Bigg]\tag{2}$$

$$E[S^2] = \Bigg[ \frac{1}{n} \sum \limits_{i=1}^{n} \Big[~(X_i-\mu)^2-2(X_i-\mu)(\bar{X}-\mu)+(\bar{X}-\mu)^2~\Big] ~\Bigg]\tag{3}$$

$$E[S^2] = E\Bigg[~\frac{1}{n} \Big[~\sum \limits_{i=1}^{n} (X_i - \mu)^2 - n(\bar{X} - \mu)^2 \Big]~\Bigg]\tag{4}$$

$$E[S^2] = \frac{1}{n} \sum \limits_{i=1}^{n} E[X_i-\mu)^2] - E[(\bar{X}-\mu)^2]\tag{5}$$

$$E[S^2] = \sigma^2 - \sigma_X^2\tag{6}$$

$$E[S^2] = \sigma^2 - \frac{1}{n}\sigma^2\tag{7}$$

$$E[S^2] = \frac{n-1}{n}\sigma^2\tag{8}$$

Equation (8) shows that $S^2$ is a biased estimator of $\sigma^2$

Leaky Capacitor
  • 621

1 Answers1

5

I will spell out the steps, hope this helps. $$E[S^2] = \Bigg[ \frac{1}{n} \sum \limits_{i=1}^{n} \Big[~(X_i-\mu)^2-2(X_i-\mu)(\bar{X}-\mu)+(\bar{X}-\mu)^2~\Big] ~\Bigg]\\ =\frac{1}{n}\sum\limits_{i=1}^n\Bigg[(X_i-\mu)^2\Bigg] -2\frac{1}{n}\Bigg[\sum\limits_{i=1}^n(X_i-\mu)(\bar{X}-\mu)\Bigg] + \frac{1}{n}\sum\limits_{i=1}^n\Bigg[(\bar{X}-\mu)^2\Bigg]\\ =\frac{1}{n}\sum\limits_{i=1}^n\Bigg[(X_i-\mu)^2\Bigg] -2(\bar{X}-\mu)\frac{1}{n}\Bigg[\sum\limits_{i=1}^n(X_i-\mu)\Bigg] + \frac{1}{n}\sum\limits_{i=1}^n\Bigg[(\bar{X}-\mu)^2\Bigg]\\ =\frac{1}{n}\sum\limits_{i=1}^n\Bigg[(X_i-\mu)^2\Bigg] -2(\bar{X}-\mu)(\bar{X}-\mu) + \Bigg[(\bar{X}-\mu)^2\Bigg]\\ =\frac{1}{n}\sum\limits_{i=1}^n\Bigg[(X_i-\mu)^2\Bigg] -(\bar{X}-\mu)^2\\ =\frac{1}{n}\sum\limits_{i=1}^n\Bigg[(X_i-\mu)^2 - n(\bar{X}-\mu)^2\Bigg]$$

Jürgen Sukumaran
  • 7,645