Deriving variance of a linear estimator problem

Question

I have done parts A, B and C with no problems however part D is proving tricky:

var(yi) = var(xi + vi) = var(xi) + var(vi) + 2cov(xi,vi)

we know var(xi) = σ^2
and that var(vi) = w^2
and that cov(xi,vi) = 0

var(yi) = σ^2 + w^2

var(ybar) = Σ (yi/n) = (1/n)* Σ(var(yi)) = (1/n) * (n(σ^2 + w^2))

so I get that var(ybar) = σ^2 + w^2 because the n's cancel.

However the mark scheme seems to think that var(ybar) = (σ^2 + w^2)/n

Can anyone explain where I'm going wrong or even if the mark scheme is wrong? Thanks.

Answer 1

The error lies in the evaluation of $var(bar)$ - when you scale the random variable by factor $k$ (which is $1/n$ in your case), the variance is scaled by $k^2$ (i.e. $1/n^2$)
$$var(ybar) = \color{red}{var}(Σ (yi/n)) = (1/\color{red}{n^2})* Σ(var(yi))= (1/\color{red}{n^2})* (n(σ^2 + w^2)) = (σ^2 + w^2)/n$$