Question

I. Suppose population 1 has mean μί with variance σ2 and population 2 has mean μ2 with the same variance σ2. Let s and s denote the sample variances from two samples with size ni and n2 from the corresponding populations, respectively. Show that the pooled estimator 1i+(2-1)si pooled ni + n2 -2 is an unbiased estimator of σ2.

Answer 1

Given the population 1 with mean $\mu _{1}$ and population 2 with mean $\mu _{2}$ with the same variance $\sigma ^{2}$.

Let $X_{1},X_{2},..,X_{n1}$ and $Y_{1},Y_{2},..,Y_{n2}$ be two random samples from two populations with common variance $\sigma ^{2}$.

Let $s_{1}^{2}$ and $s_{2}^{2}$ denote the sample variances from two samples with size $n_{1}$ and $n_{2}$ from the corresponding populations.

TO PROVE:

The pooled estimator $s_{pooled}^{2}=[(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}]/(n_{1}+n_{2}-2)$ is an unbiased estimator of $\sigma ^{2}$.

PROOF:

In order to prove that the pooled estimator is an unbiased estimator of $\sigma ^{2}$, we first prove that the two variances are unbiased estimators of $\sigma ^{2}$.

Let $s_{1}^2=(1/(n_{1}-1))\sum_{i=1}^{n_{1}}(X_{i}-X_{bar})^{2}$ be the variance of sample 1.

We know that $(n_{1}-1)s_{1}^{2}/\sigma ^{2}\sim \chi ^{2}$ distribution with $(n_{1}-1)$ degrees of freedom.

Using the fact that the expected value of a $\chi ^{2}$ random variable with $\nu$ degrees of freedom is $\nu$,

$E[(n_{1}-1)s_{1}^{2}/\sigma ^{2}]=n_{1}-1$

$\Rightarrow ((n_{1}-1)/\sigma ^{2})E(s_{1}^{2})=n_{1}-1$

$\Rightarrow E(s_{1}^{2})=\sigma ^{2}$

Thus $s_{1}^2$ is an unbiased estimator of $\sigma ^{2}$.   $\rightarrow (1)$

In a similar manner,

Let $s_{2}^2=(1/(n_{2}-1))\sum_{i=1}^{n_{2}}(X_{i}-X_{bar})^{2}$ be the variance of sample 2.

We know that $(n_{2}-1)s_{2}^{2}/\sigma ^{2}\sim \chi ^{2}$ distribution with $(n_{2}-1)$ degrees of freedom.

Using the fact that the expected value of a $\chi ^{2}$ random variable with $\nu$ degrees of freedom is $\nu$,

  $E[(n_{2}-1)s_{2}^{2}/\sigma ^{2}]=n_{2}-1$

$\Rightarrow ((n_{2}-1)/\sigma ^{2})E(s_{2}^{2})=n_{2}-1$

$\Rightarrow E(s_{2}^{2})=\sigma ^{2}$

Thus $s_{2}^2$ is an unbiased estimator of $\sigma ^{2}$.   $\rightarrow (2)$

Now let us consider the pooled variance $s_{pooled}^{2}=[(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}]/(n_{1}+n_{2}-2)$

From $(1), (2)$ and  linearity of expected value:

$E[s_{pooled}^{2}]=E[[(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}]/(n_{1}+n_{2}-2)]$

  $=[(n_{1}-1)/(n_{1}+n_{2}-2)]E[s_{1}^{2}]+[(n_{2}-1)/(n_{1}+n_{2}-2)]E[s_{2}^{2}]$

  $=[(n_{1}-1)/(n_{1}+n_{2}-2)]\sigma ^{2}+[(n_{2}-1)/(n_{1}+n_{2}-2)]\sigma ^{2}$

  $=[[(n_{1}-1)+(n_{2}-1)]/(n_{1}+n_{2}-2)]\sigma ^{2}$

  $=[(n_{1}+n_{2}-2)/(n_{1}+n_{2}-2)]\sigma ^{2}$

$E[s_{pooled}^2]=\sigma ^{2}$

This shows that $s_{pooled}^2$ is an unbiased estimator of $\sigma ^{2}$.