# If $x_i$ is from a random sample is $Var(\bar x \mid x_i)=0$?

If $x_i$ is from a random sample, is the conditional variance of the mean (or the sum of squares, really any statistic based on $x$) just treated as a constant? I saw this in a OLS variance of a parameter proof. I'm just looking for some rational as to why it is true.

I think this would also work for expected values So for example: $$E(\bar x \mid x_i)=\bar x$$ is true as well. Whats the reasoning being these things?

I believe it can also be assumed that $x_i$ is identically distributed random sample.

• You start by $x_i$ being "from" a random sample, and in the last sentence, it appears that you treat for some reason $x_i$ as being the whole sample. What is it? Sep 12, 2015 at 21:47

When you know all your $X_i$ then you know your $\bar{X}$, hence $E(\bar{X}|X_1,X_2,...,X_n)=\bar{X}$ as your sample mean is no longer a random variable but a constant now.

Thus, $Var(\bar{X}|X_1,X_2,...,X_n)=E(\bar{X}^2|X_1,...,X_n)-(E(\bar{X}|X_1,...,X_n))^2=\bar{X}^2-\bar{X}^2=0$

• Although I certainly believe the last sentence, I don't see how it follows from the rest of your answer.
– user88319
Aug 29, 2015 at 4:36
• I do not understand your comment. Sorry, please rephrase it. Aug 29, 2015 at 7:32
• Apologies if I was too terse. You don´t mention in your answer that $E[\overline{X}^2|X_1, \cdots, X_n] = \overline{X}^2$, so your last sentence seemed unjustified to me.
– user88319
Aug 30, 2015 at 5:36

Whenever you see $$\bar{X}\mid \mathbf{x}$$

or some other form of it, it means to fix the value of $\mathbf{x}$.

So, in your case, we seek $$\text{Var}\left[\bar{X}\mid X_1, \cdots, X_n\right]\text{.}$$ But here's the thing. By definition, $$\bar{X} = \dfrac{1}{n}\sum\limits_{i=1}^{n}X_i$$ so thus $\bar{X}$ is constant because $X_1, \cdots, X_n$ are fixed (or constant). Since the variance of a constant is $0$, the result follows.