Expectation of a sample variance

Question

Let $Y_1$ and $Y_2$ be two independent realizations of the random variable $Y$ such that $E(Y_1)=E(Y_2)=E(Y)$ and $Var(Y_1)=Var(Y_2)=Var(Y)$.

With $Y_1$ and $Y_2$ as our sample, the sample variance is written as $\sum (Y_i-\overline Y)^2$.

I want to write the sample variance in terms of $Y_1$ and $Y_2$ only, is it correct to state that $\overline Y=E(Y_1)=E(Y_2)=E(Y)$ and then get as sample variance: $$[Y_1-E(Y_1)]^2+[Y_2-E(Y_2)]^2$$

If so, I am asked to show that $E(\text{sample variance})=Var(Y)$ but I am only able to show that it is equal to $2Var(Y)$.

Answer 1

No, $\bar Y = (Y_1 + Y_2)/2$ is the sample mean, which is a statistic, not an expectation. So the sample variance is $$\left(Y_1 - \frac{Y_1 + Y_2} 2 \right)^2 + \left(Y_2 - \frac{Y_1 + Y_2} 2 \right)^2 = \left(\frac{Y_2 - Y_1}{2}\right)^2 + \left(\frac{Y_1 - Y_2}{2}\right)^2 = \frac{(Y_1 - Y_2)^2}{2}.$$ So what is the expectation of this statistic? You would expand $$(Y_1 - Y_2)^2 = Y_1^2 - 2Y_1 Y_2 + Y_2^2,$$ employ the linearity of expectation, and under the independence assumption, $$\operatorname{E}[Y_1 Y_2] \overset{\text{ind}}{=} \operatorname{E}[Y_1]\operatorname{E}[Y_2].$$ The details have been left as an exercise.