# Finding $E\Bigl(\overline{Y^2}\Bigm|\overline{Y\vphantom{Y^2}}\Bigr)$ by Basu's theorem?

Suppose $Y_1,\ldots,Y_n$ are a random sample of normal distribution $\mathcal{N}(\mu,1)$. If $\overline{Y^2}=\displaystyle\frac{1}{n}\sum_{i=1}^n Y_i^2$, how can I find $E\Bigl(\overline{Y^2}\Bigm|\overline{Y\vphantom{Y^2}}\Bigr)$ by Basu's theorem?

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Crossposted: stats.stackexchange.com/q/35846/2970 – cardinal Sep 8 '12 at 19:09
– cardinal Sep 8 '12 at 19:15

## 1 Answer

By Basu's theorem $\frac{1}{n-1}\sum_{i=1}^n (Y_i- \bar{Y})^2$ is independent of $\bar{Y}$. Hence $$\mathbb{E} \left(\frac{1}{n-1}\sum_{i=1}^n (Y_i- \bar{Y})^2 \mid \bar{Y} \right) = \mathbb{E} \left(\frac{1}{n-1}\sum_{i=1}^n (Y_i- \bar{Y})^2 \right)$$ Develop both sides. You expression will appear on the left-hand side.

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This will actually work for $N(\mu,\sigma^2)$ if you consider the family $\{N(\mu,\sigma^2)\,:\,\mu\in\mathbb{R}\}$ with $\sigma^2$ fixed. For that family, the sample mean is sufficient and the residual sum of squares is ancillary. – Michael Hardy Sep 7 '12 at 17:14
Using Basu for this is serious overkill (I realize that's what the question is asking for, though). It should really only be seen as a simple example to understand what Basu's theorem says and not as a serious means of proof of the desired fact. :-) – cardinal Sep 8 '12 at 19:11

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@cardinal : serious overkill is too soft to describe this ;) – vanna Sep 8 '12 at 19:42
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