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

Question

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.

Answer 1

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

Answer 2

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.