2

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?

cardinal
  • 7,420
hadisanji
  • 999

1 Answers1

3

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.

Michael Hardy
  • 1
vanna
  • 1,514
  • 8
  • 21
  • 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 07 '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 08 '12 at 19:11
  • 1
    @cardinal : serious overkill is too soft to describe this ;) – vanna Sep 08 '12 at 19:42