How to improve estimation if $\sigma^2$ is known I'd like to ask this rather theoretical question about estimating a linear model with known $\sigma^2$ for the error term. I read that if the variance of the error terms was known, it would've been possible to know the population slopes and intercepts. Can you give me a hint on how to insert this information in the estimator? Trying to minimize $(\varepsilon'\varepsilon - \sigma^2)^2$ it's kind of messy. Is it a good direction to try $\operatorname{Var}[y] = \operatorname{Var}[x\beta] + \sigma^2$ assuming $x$ and $\varepsilon$ are independent?
 A: Perhaps there's a misunderstanding of the Gauss-Markov theorem taking place here. An assumption of the Gauss-Markov theorem is that $\mathrm{Var}\left(\epsilon_i \right) = \sigma^2 $. The best-linear unbiased estimator under the Gauss-Markov assumptions is $\hat{\mathbf{b}} = (X'X)^{-1}X'\mathbf{y}$. Observe that $\sigma^2$ doesn't enter the formula for the estimator not because it was assumed you didn't know $\sigma^2$. Rather, $\sigma^2$ doesn't enter the formula because it doesn't help!
If you drop the requirement that the estimator is unbiased (or otherwise deviate from the assumptions or objective of the Gauss-Markov theorem), you can get estimators that do depend on $\sigma^2$, such as the James-Stein estimator.
A: Building on @dsaxton's comment, here are two scatterplots with the same errors $u$, drawn from $N(0,\sigma^2)$, $\sigma^2=4$. The first has slope $\beta=4$, the second $\beta=6$. 
How do you want to use knowledge of $\sigma^2$ to pin down $\beta$ if two populations with the same $\sigma^2$ have different $\beta$?

sigma <- 2
beta1 <- 4
beta2 <- 6
u <- rnorm(100,sd=sigma)
x <- runif(100)
y1 <- beta1*x + u
y2 <- beta2*x + u

par(mfrow=c(1,2))
plot(x,y1,ylim=c(-4,8))
plot(x,y2,ylim=c(-4,8))

A: Thanks for your answers. I cannot find again the web site where it said something like "we do not know sigma, otherwise we could find the population intercept and slope".
I found it intriguing, I thought maybe there is some trick (by improving the estimator I meant reducing the variance of the estimator).
Thanks a lot for your time, you're very nice people,
Chris
