Let $\{Y_1, \ldots, Y_n\}$ be a random sample with $E(Y_i)= \mu_Y$ and $\operatorname{var}(Y_i) = \sigma_Y^2$. Show that $Y = \sum_{i=1}^n \frac{Y_i}{n-2}$ is a biased estimator of the mean?
This is what I have so far.
$$E \left( \sum_{i=1}^n \frac{Y_i}{n-2}\right) = \frac{1}{n-2}E \left( \sum_{i=1}^n Y_i \right) = \frac{1}{n-2} \sum_{i=1}^n E(Y_i) = \frac{n\mu_Y}{n-2}$$
This is as far as I have simplified the mean. I am just confused how/ if this shows that $Y$ is a biased estimator.
You wan't to show $E\left[ \sum\frac{Y_i}{n-2} \right] \neq \mu_Y$. But $E\left[ \sum\frac{Y_i}{n-2} \right] = \frac{n}{n-2}\mu_Y$ which is different for $\mu_y$ for all $n$.
However, note that $\lim_{n\to\infty}E \left[ \sum\frac{Y_i}{n-2} \right] =\mu_Y$. Thus, $\sum\frac{Y_i}{n-2}$ is asymptotically unbiased. That is, for big $n$, it is almost an unbiased estimator for the mean.