Solution to Exercise 6.12.

  Showing that $\hat{\sigma}^2$ is and unbiased estimator of $\sigma^2$ means showing that $E[\hat{\sigma}^2$ is $\sigma^2$. Now we can split the fraction in the definition of $\hat{\sigma}^2$ as in

$$\hat{\sigma}^2 \; = \; \frac{n_1 - 1}{n_1 + n_2 -2} S_1^2 \, + \, \frac{n_2 - 1}{n_1 + n_2 -2} S_2^2$$

Using the linearity property of expectation and that each S_i² is unbiased and unbiased estimator of $\sigma^2$we have

$$E[ \hat{\sigma}^2 ] \; = \; \frac{n_1 - 1}{n_1 + n_2 -2} E[S_1^2] \, + \, \frac{n_2 - 1}{n_1 + n_2 -2} E[S_2^2]$$

$$\; = \; \frac{n_1 - 1}{n_1 + n_2 -2} \sigma^2 \, + \, \frac{n_2 - 1}{n_1 + n_2 -2} \sigma^2$$

$$\; = \; \left[ \frac{n_1 - 1}{n_1 + n_2 -2} \, + \, \frac{n_2 - 1}{n_1 + n_2 -2} \right] \sigma^2$$

$$\; = \; \frac{n_1-1+n_2-1}{n_1 + n_2 -2} \sigma^2 \; = \; \sigma^2$$

as desired.

Note: One can see that any statistic of the form

$$T \; = \; a S_1^2 + (1-a) S_2^2 ,$$

where a is a conastant will be an unbiased estimator of $\sigma^2$.