Determining Which Mean Squared Error is Larger

Question

I have found the mean squared error of two estimators: $S^2$ and $S^{'2} = \frac{\sum(Y_i - \bar{Y})^2}{n}$, but I am unsure of a mathematical way to show which one is larger without just substituting arbitrary values of $n$.

The mean squared error for $S^2$ is $$\frac{2 \sigma^4}{n-1},$$ and the mean squared error for $S^{'2}$ is $$\sigma^4 \bigg( \frac{2}{n} - \frac{1}{n^2} \bigg).$$ How can I show that the mean squared error for $S^2$ is larger than that of $S^{'2}$ without simply substituting arbitrary values of $n$?

Answer 1

Assuming $n>1$, we have $$ -3n+1<0 \\ \implies 2n^2-3n+1 < 2n^2 \\ \implies (2n-1)(n-1) < 2n^2 \\ \implies \frac{2n-1}{n^2} < \frac{2}{n-1} \\ \implies \frac{2}{n} - \frac{1}{n^2} < \frac{2}{n-1}$$

Answer 2

Hint: You can expand $S^2$ by $n+1$: $$S^2=\sigma^4\cdot \frac{2n+2}{n^2-1}>\sigma^4\cdot\frac{2n-1}{n^2}=S^{'2} \quad \forall \ n>1$$