Given that $ \operatorname{Var}(x)=\frac{3}{4}\theta^2$, I want t find the variance of estimator $\hat{\theta_1} = \frac{2n}{3}\sum_{i=1}^nX_i$. EDIT: $X_1,…,X_n$ are independent and identically distributed (iid)
I proceed as follows:
$$ \operatorname{Var}(\hat{\theta}) = \operatorname{Var}(\frac{2n}{3}\sum_{i=1}^nX_i)=\frac49 \frac1n \operatorname{Var}(X_1)= \frac49 \frac1n \frac34 \theta^2 = \frac{1}{3n}\theta^2 $$
or
$$ \operatorname{Var}(\hat{\theta}) = \operatorname{Var}(\frac{2n}{3}\sum_{i=1}^nX_i)= \operatorname{Var}(\frac{2n}{3}nX_1)= \operatorname{Var}(\frac{2}{3}X_1)=\frac49 \frac34\theta^2 = \frac13\theta^2 $$
Our professor provided us with the solution, so I know the first approach is the correct one, however i do not understand what is wrong with the second approach?
Best Answer
Both approaches seem to be problematic in my view with the second being worse than the first because there is no reason why your equalities should hold. The first approach only has the problem of lacking correct calculations with rational numbers as far as I can tell.
The identity $$Var(\frac{2n}{3}\sum_{i=1}^nX_i) = \frac{4n^2}{9}n Var(X_1)$$ is correct (if we assume that the $X_1, \dots, X_n$ are independent and identically distributed), since $Var(\alpha X) = \alpha^2 Var(X)$ for any constant $\alpha$ and $Var(X + Y) = Var(X) + Var(Y)$ for independent $X$ and $Y$.
However, $$\frac{4n^2}{9}n Var(X_1)=\frac{4 n^3}{9}Var(X_1) $$ and not $\frac{4 }{n}Var(X_1)$ or $\frac{4 }{9 n}Var(X_1)$ (how are the $n$ supposed to cancel?)
You should now try to plug in $Var(X_1) = \frac{3}{4} \theta^2$.