The question is:
Let $Y_1,....,Y_n$ be independent random variables each with the distribution $N\left ( \mu ,\sigma ^2 \right )$.
Let:
$\overline{Y} = \frac{1}{n}\sum_{i=1}^{n}Y_i$
$S^2 = \frac{1}{n-1}\sum_{i=1}^{n}\left ( Y_i - \overline{Y} \right )^2$
Show that: $S^2 = \frac{1}{n-1}\left [ \sum_{i=1}^{n}\left ( Y_i-\mu \right )^2 - n\left ( \overline{Y}-\mu \right )^2 \right ]$
This is as far as I've being able to get:
$S^2 = \frac{1}{n-1}\sum_{i=1}^{n}\left ( Y_i - \overline{Y} \right )^2$
$= \frac{1}{n-1}\left [ \sum_{i=1}^{n}\left ( Y_i-\overline{Y} \right )\sum_{j=1}^{n}\left ( Y_j -\overline{Y} \right ) \right ]$
I intend the first double summation to be for when $i=j$ and the second for when $i\neq j$...
$= \frac{1}{n-1}\sum_{i=1}^{n}\sum_{j=1}^{n}\left ( Y_i-\overline{Y} \right ) \left ( Y_j-\overline{Y} \right )+\frac{1}{n-1}\sum_{i=1}^{n}\sum_{j=1}^{n}\left ( Y_i-\overline{Y} \right ) \left ( Y_j-\overline{Y} \right )$
$= \frac{1}{n-1}\sum_{i=1}^{n}\left ( Y_i-\overline{Y} \right )^2 +\frac{1}{n-1}\sum_{i=1}^{n}\sum_{j=1}^{n}\left ( Y_i-\overline{Y} \right ) \left ( Y_j-\overline{Y} \right )$
Would anyone know how to finish the question off from here or any mistakes I've made thus far?
The question is exercise 1.4 from this textbook.
Thank you.