# Show that $E[\mu_i^2|X_i]=\sigma^2$ and $Var(\mu_i|X_i)=\sigma^2$

I'm a little stuck on this review problem so help would be greatly appreciated!

Q: We have the regression model $Y_i=\beta_0+\beta_1X_i+\mu_i$ and we assume that the expected errors are $0$. We also know that the asymptotic variance of $\hat{\beta_1}$ is: $$\frac{Var(\mu_i(X_i-E(X_i))}{Var(X_i)}$$

EDITED: Now I have to Show that IF $E[\mu_i^2|X_i]=\sigma^2$ THEN $Var(\mu_i|X_i)=\sigma^2$.

My thoughts: What does knowing the asymptotic variance have to do with anything in this problem? Can I simply solve the following equation?

$$Var(\mu_i^2|X_i)=E[\mu_i^2|X_i] - E([\mu_i|X_i])^2$$

But how?

Step by step help would be greatly appreciated. Many thanks in advance!

• what is the definition of $\sigma^2$? Feb 2, 2015 at 18:52
• $\sigma^2$ is the variance, I'm not sure what you mean. Apologies, I added the homework tag although this is not homework. Feb 2, 2015 at 18:56
• You haven't supplied enough information to answer this question. You are asking us to show that $\sigma^2$ is a particular variance. But how is $\sigma^2$ defined in the first place?
– whuber
Feb 2, 2015 at 19:52
• Can you edit your title and question body to reflect your changed question? Feb 2, 2015 at 22:09
• "we assume that the expected errors are 0" - is this in conditional expectations? can you write it as an equation? Feb 2, 2015 at 22:11

$Var(μ_i|X_i)=E[μ^2_i|X_i]−E([μ_i|X_i])^2$
and that the that the expected errors are 0 (i.e. $E([μ_i|X_i]=0$. Then your result follows immediately
• This is a good answer (+1) but essentially does someone's homework. It goes some way beyond the "hints and guidance" called for in the self-study tag wiki for routine bookwork problems. I think it would have been better (at least initially) to replace your text following the first equation with something like "can you say anything about the term involving the expected error $E([\mu_i|X_i])$?" Feb 2, 2015 at 22:12
• Thanks for the reference to the wiki on self-study. The variance term should refer to $var(\mu_i)$ and not $var(\mu_i^2)$. Feb 4, 2015 at 15:49