Define
\[\tilde{y} = \frac{1}{n}\left(\frac{1}{2} y_1 + \frac{3}{2} y_2 + + \frac{1}{2} y_3 + \dots + \frac{3}{2} y_n \right)\]
and assume \(n\) is an even number.
Show that this is unbiased.
Show that this is consistent.
Show that the sample average is more efficient that this.
The sample variance is
\[S^2 = \frac{1}{n-1}\sum_{i=1}^n (Y_i-\overline{Y})^2,\]
and we want to work out the unbiasedness of this object. That is, we want to take the \(\mathbb{E}\) operator over it. We will move the \(\mathbb{E}\) inside the sum (because it’s linear). So, let’s start with analysing one of the summands first:
\[\mathbb{E}\big[(Y_i-\overline{Y})^2\big]\]
Notice that this looks similar, but is not the same as
\[\mathbb{E}\big[(Y_i-\mu_Y)^2\big]\]
which is of course the population variance.
Let’s start with the following claim. I claim that this expression falls into that:
\[\mathbb{E}\big[(Y_i-\overline{Y})^2\big] = \operatorname{Var}(Y_i) + \operatorname{Var}(\overline{Y}) - 2\,\operatorname{Cov}(Y_i,\overline{Y}).\]
Why? Well, look at that. Remove the square and resolve the expectation:
\[\mathbb{E}[Y_i - \overline{Y}] = \mathbb{E}[Y_i] - \mathbb{E}[\overline{Y}] = \mu_Y - \mu_Y = 0.\]
That’s interesting, because it means that
\[\mathbb{E}\big[(Y_i-\overline{Y})^2\big] = \operatorname{Var}(Y_i - \overline{Y}).\]
why? you know why, that’s simply because
\[\operatorname{Var}(X) = E(X^2) - [E(X)]^2\] (fill in \((Y_i-\overline{Y}\) for \(X\) ).
Ok, next. We know this identity
\[\operatorname{Var}(A - B) = \operatorname{Var}(A) + \operatorname{Var}(B) - 2\,\operatorname{Cov}(A,B).\]
let’s substitute for A and B
\[ \mathbb{E}\big[(Y_i-\overline{Y})^2\big] = \operatorname{Var}(Y_i - \overline{Y})= \operatorname{Var}(Y_i) + \operatorname{Var}(\overline{Y}) - 2\,\operatorname{Cov}(Y_i,\overline{Y}).\]
Next, compute the individual terms, using the iid assumption:
\[\operatorname{Var}(Y_i) = \sigma^2, \qquad \operatorname{Var}(\overline{Y}) = \frac{\sigma^2}{n}.\]
What’s left is the covariance:
\[\operatorname{Cov}(Y_i, \overline{Y}) = \operatorname{Cov}\!\left(Y_i, \frac{1}{n}\sum_{j=1}^n Y_j\right) = \frac{1}{n}\operatorname{Cov}(Y_i, Y_i) + \frac{1}{n}\sum_{j\ne i}\operatorname{Cov}(Y_i, Y_j) = \frac{\sigma^2}{n}.\]
Put it all back together
\[\mathbb{E}\big[(Y_i-\overline{Y})^2\big] = \sigma^2 + \frac{\sigma^2}{n} - 2\cdot\frac{\sigma^2}{n} = \sigma^2\left(1 - \frac{1}{n}\right).\]
Summing over indices \(i=1,\dots,n\) as required in our initial expression, we get
\[\mathbb{E}\!\left[\sum_{i=1}^n (Y_i-\overline{Y})^2\right] = n\sigma^2\left(1 - \frac{1}{n}\right) = (n-1)\sigma^2.\]
and therefore we need to get rid of the \((n-1)\) there. How? Just divide through!
\[S^2 = \frac{1}{n-1}\sum_{i=1}^n (Y_i-\overline{Y})^2,\]
and we have the unbiased estimator
\[\mathbb{E}[S^2] = \sigma^2\]