Respuesta :
Answer:
See the proof below.
Step-by-step explanation:
For this case we just need to apply properties of expected value. We know that the estimator is given by:
[tex]S^2_p= \frac{(n_1 -1) S^2_1 +(n_2 -1) S^2_2}{n_1 +n_2 -2}[/tex]
And we want to proof that [tex]E(S^2_p)= \sigma^2[/tex]
So we can begin with this:
[tex]E(S^2_p)= E(\frac{(n_1 -1) S^2_1 +(n_2 -1) S^2_2}{n_1 +n_2 -2})[/tex]
And we can distribute the expected value into the temrs like this:
[tex]E(S^2_p)= \frac{(n_1 -1) E(S^2_1) +(n_2 -1) E(S^2_2)}{n_1 +n_2 -2}[/tex]
And we know that the expected value for the estimator of the variance s is [tex]\sigma[/tex], or in other way [tex]E(s) = \sigma[/tex] so if we apply this property here we have:
[tex]E(S^2_p)= \frac{(n_1 -1 )\sigma^2_1 +(n_2 -1) \sigma^2_2}{n_1 +n_2 -2}[/tex]
And we know that [tex]\sigma^2_1 = \sigma^2_2 = \sigma^2[/tex] so using this we can take common factor like this:
[tex]E(S^2_p)= \frac{(n_1 -1) +(n_2 -1)}{n_1 +n_2 -2} \sigma^2 =\sigma^2[/tex]
And then we see that the pooled variance is an unbiased estimator for the population variance when we have two population with the same variance.
