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

CC BY-SA 4.0
2024-05-17T10:34:59.160Z
2024-05-17T10:34:59.160Z
Joseph Basford
<p>I am attempting the following exercise from Annie Liang's excellent <a href="https://www.anniehliang.com/lecture-notes" rel="noreferrer">lecture notes</a>:
<a href="https://i.sstatic.net/XIpNnvic.png" rel="noreferrer"><img src="https://i.sstatic.net/XIpNnvic.png" alt="enter image description here" /></a></p>
<p>I have written an attempted proof (below), but I'm worried about the independence assumption I make. In particular, surely I cannot say that <span class="math-container">$\mathbb{E}[\sum^n_{i = n'+1} Y_i | \overline{Y}_n) = \mu$</span>?</p>
<p><strong>Attempt</strong></p>
<p>WTS: <span class="math-container">$\mathbb{E}[\overline{Y}_{n'} | \overline{Y}_n] = \overline{Y}_n$</span></p>
<p><span class="math-container">$$\begin{align}
\overline{Y}_{n'} &amp;= \frac{1}{n'} \sum_{i=1}^{n'} Y_i\\
&amp;= \frac{1}{n'} \bigg[ \sum_{i=1}^{n} Y_i - \sum_{i=n'+1}^{n} Y_i \bigg] \\
&amp;= \frac{1}{n'} \bigg[ n \overline{Y}_n - \sum_{i=n'+1}^{n} Y_i \bigg] \\
&amp;= \frac{n}{n'} \overline{Y}_n - \frac{1}{n'} \bigg[ \sum_{i=n'+1}^{n} Y_i \bigg] \\
&amp;= \overline{Y}_n + \dfrac{n - n'}{n'} \overline{Y}_n - \frac{1}{n'} \sum^n_{i = n'+1} Y_i \\
&amp;= \overline{Y}_n + \dfrac{n - n'}{n'} \sum_{i=1}^n Y_i- \frac{1}{n'} \sum^n_{i = n'+1} Y_i
\end{align}$$</span></p>
<p>I can then take expectations conditional <span class="math-container">$\overline{Y}_n$</span>, letting <span class="math-container">$\mathbb{E}(Y_n) = \mu$</span>.
Note that independence <span class="math-container">$\implies \sum^n_{i = n'+1} Y_i = (n - n') \mu$</span>.
<span class="math-container">$$\begin{align}
\mathbb{E}[\overline{Y}_{n'} | \overline{Y}_n] &amp;= \overline{Y}_n + \frac{1}{n'}\frac{n-n'}{n} n \mu - \frac{1}{n'} \mu (n - n') \\
&amp;= \overline{Y}_n + \mu \bigg[ \frac{n-n'}{n'} - \frac{n-n'}{n'} \bigg] \\
&amp;= \overline{Y}_n
\end{align}$$</span></p>

CC BY-SA 4.0
2024-05-17T20:31:57.363Z
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 $1$.
Henry
<microeconomics><statistics>
Proof Mean with a Smaller Sample Size is a Mean Preserving Spread
matthewoulton
# Proof Mean with a Smaller Sample Size is a Mean Preserving Spread

(Asked by: username_0 on 2024-05-17)

I am attempting the following exercise from Annie Liang's excellent [lecture notes](https://www.anniehliang.com/lecture-notes):
[![enter image description here](https://i.sstatic.net/XIpNnvic.png)](https://i.sstatic.net/XIpNnvic.png)

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 $\implies \sum^n\_{i = n'+1} Y\_i = (n - n') \mu$.
$$\begin{align}
\mathbb{E}[\overline{Y}\_{n'} | \overline{Y}\_n] &= \overline{Y}\_n + \frac{1}{n'}\frac{n-n'}{n} n \mu - \frac{1}{n'} \mu (n - n') \\
&= \overline{Y}\_n + \mu \bigg[ \frac{n-n'}{n'} - \frac{n-n'}{n'} \bigg] \\
&= \overline{Y}\_n
\end{align}$$

**Tags:** microeconomics, statistics
> **Comments:**
> **username_1**: 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 $1$.  

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## 1 answers

### Answer by username_2 on 2024-05-17. Score: 3

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 $$\mathbb{E}[\bar{Y}\_{n'}|\bar{Y}\_n]=\frac{1}{n'}\sum\_{i=1}^{n'}\mathbb{E}[Y\_i|\bar{Y}\_n]\stackrel{i.i.d.}{=}\mathbb{E}[Y\_1|\bar{Y}\_n]=\bar{Y}\_n$$


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