# Proof Mean with a Smaller Sample Size is a Mean Preserving Spread

I am attempting the following exercise from Annie Liang's excellent lecture notes:

I have written an attempted proof (below), but I'm worried about the independence assumption I make. In particular, surely I cannot say that $$\mathbb{E}[\sum^n_{i = n'+1} Y_i | \overline{Y}_n) = \mu$$?

Attempt

WTS: $$\mathbb{E}[\overline{Y}_{n'} | \overline{Y}_n] = \overline{Y}_n$$

\begin{align} \overline{Y}_{n'} &= \frac{1}{n'} \sum_{i=1}^{n'} Y_i\\ &= \frac{1}{n'} \bigg[ \sum_{i=1}^{n} Y_i - \sum_{i=n'+1}^{n} Y_i \bigg] \\ &= \frac{1}{n'} \bigg[ n \overline{Y}_n - \sum_{i=n'+1}^{n} Y_i \bigg] \\ &= \frac{n}{n'} \overline{Y}_n - \frac{1}{n'} \bigg[ \sum_{i=n'+1}^{n} Y_i \bigg] \\ &= \overline{Y}_n + \dfrac{n - n'}{n'} \overline{Y}_n - \frac{1}{n'} \sum^n_{i = n'+1} Y_i \\ &= \overline{Y}_n + \dfrac{n - n'}{n'} \sum_{i=1}^n Y_i- \frac{1}{n'} \sum^n_{i = n'+1} Y_i \end{align}

I can then take expectations conditional $$\overline{Y}_n$$, letting $$\mathbb{E}(Y_n) = \mu$$. Note that independence $$\implies \sum^n_{i = n'+1} Y_i = (n - n') \mu$$. \begin{align} \mathbb{E}[\overline{Y}_{n'} | \overline{Y}_n] &= \overline{Y}_n + \frac{1}{n'}\frac{n-n'}{n} n \mu - \frac{1}{n'} \mu (n - n') \\ &= \overline{Y}_n + \mu \bigg[ \frac{n-n'}{n'} - \frac{n-n'}{n'} \bigg] \\ &= \overline{Y}_n \end{align}

• You cannot say $\mathbb{E}\left[\sum\limits^n_{i = n'+1} Y_i \mid \overline{Y}_n\right] = \mu$ for two reasons: the expectation of each term in the sum should be $\overline{Y}_n$ rather than $\mu$; and there are $(n-n')$ such terms rather than $1$. Commented May 17 at 20:31

The following fact is key. Since $$\bar{Y}_n=n^{-1}\sum_{i=1}^n Y_i$$, we can take expectations conditional on $$\bar{Y}_n$$ and use that $$\{Y_i\}$$ is an i.i.d. sequence to obtain $$\bar{Y}_n=\mathbb{E}[\bar{Y}_n|\bar{Y}_n]=n^{-1}\sum_{i=1}^n\mathbb{E}[Y_i|\bar{Y}_n]=\mathbb{E}[Y_1|\bar{Y}_n]$$
Using this, we find $$\mathbb{E}\left[\sum_{i=n'+1}^nY_i\Bigg|\bar{Y}_n\right]=\sum_{i=n'+1}^n \mathbb{E}[Y_i|\bar{Y}_n]=(n-n')\mathbb{E}[Y_1|\bar{Y}_n]=(n-n')\bar{Y}_n$$
Therefore, using your expression $$\bar{Y}_{n'}=\bar{Y}_n+\frac{n-n'}{n'}\bar{Y}_n-\frac{1}{n'}\sum_{i=n'+1}^nY_i$$ we can derive \begin{align*} \mathbb{E}[\bar{Y}_{n'}|\bar{Y}_n] &=\bar{Y}_n+\frac{n-n'}{n'}\bar{Y}_n -\frac{1}{n'}\mathbb{E}\left[\sum_{i=n'+1}^nY_i\Bigg|\bar{Y}_n\right] \\ &=\bar{Y}_n+\frac{n-n'}{n'}\bar{Y}_n-\frac{n-n'}{n'}\bar{Y}_n \\ &= \bar{Y}_n\end{align*}
We could have also gotten this without any manipulation of the sum by noting $$\mathbb{E}[\bar{Y}_{n'}|\bar{Y}_n]=\frac{1}{n'}\sum_{i=1}^{n'}\mathbb{E}[Y_i|\bar{Y}_n]\stackrel{i.i.d.}{=}\mathbb{E}[Y_1|\bar{Y}_n]=\bar{Y}_n$$