Let $X_1, X_2, ..., X_n$ be a random iid sample from a population with mean $\theta$.

Now I am wondering about the intuition behind $E(X_1| \overline X ) = \overline X$, the sample mean.

If we just consider $X_1$ (or any $X_i$ for that matter) we have that $E(X_1) = \theta$ as the expected value of any random observation from a population will be the population mean.

Now given that we know $\overline X$ how come that changes what we expect to get for $X_1$? $X_1$ is still the same random observation from the population as before...but it seems that knowing the sample mean 'overrides' what we expect to get for an observation...is this correct?

  • 2
    $\begingroup$ Hint: What is $E(X_1+X_2+\cdots+X_n\,|\,\bar X)$? How do $E(X_i\,|\,\bar X)$ and $E(X_j\,|\,\bar X)$ differ? (I suspect your intuition is really concerning $E(X_i\,|\,X_j,\bar X)$.) $\endgroup$
    – whuber
    Commented Jun 27, 2014 at 18:57
  • $\begingroup$ stats.stackexchange.com/q/374989/119261 $\endgroup$ Commented Feb 12, 2021 at 16:26

1 Answer 1


Using whuber's hint:

$\begin{align*} \left.\text{E}\left[ \bar{X} \right| \bar{X}\right] &= \bar{X}\\ \left.\text{E}\left[\dfrac{1}{n}\sum_{i=1}^n X_i \right|\bar{X}\right] &= \bar{X}\\ \dfrac{1}{n}\sum_{i=1}^n \text{E}[ X_i |\bar{X}] &=\bar{X} \\ \dfrac{1}{n}n\text{E}[ X_1 |\bar{X}] &= \bar{X}\\ \text{E}[ X_1 |\bar{X}] &= \bar{X} \end{align*}$

  • 7
    $\begingroup$ I guess the hint was meant for the OP. $\endgroup$ Commented Jun 27, 2014 at 21:04

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