Calculating variance of an estimator

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?

• Is the first one really correct? It seems to me that he claims $\text{Var}(\sum_i X_i) = n\text{Var}(X_1)$ (2nd =). Then, 3rd and 4th = are clearly wrong. – Antoine Jul 30 '18 at 8:38
• $X_i, ..., X_n$ are iid...forgot to mention. Then $Var(\sum_iX_i )= nVar(X_1)$, right? – user1607 Jul 30 '18 at 8:44
• I count lots of mistakes in this. E.g. do we really have $\frac{4n^2}9n=\frac4n$? – drhab Jul 30 '18 at 8:45
• @Antoine Sorry, but you are wrong. And probably the $X_i$ are meant to be independent (which should have been mentioned in the question). The variance of a sum of independent random variables equals the sum of variances. – drhab Jul 30 '18 at 8:49
• I appologize, i made mistakes when writing the questions and fixed them after they were poited out. – user1607 Jul 30 '18 at 8:56

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 $Var(X_1) = \frac{3}{4} \theta^2$.
• What is wrong with $Var(\frac{2n}{3}\sum_{i=1}^nX_i) = \frac{4n^2}{9} Var (\sum_{i=1}^n X_i) = \frac{4n^2}{9} Var(nX_1)= \frac{4n^2}{9} n^2 Var(X_1)$ ? – user1607 Jul 30 '18 at 9:05
• It is not true that $Var(\sum_{i} X_i) = Var(n X_1)$. For example if $n = 2$. Then your equation would read $Var(X_1 + X_2) = Var(2 X_1) = 4 Var(X_1)$ which is not correct. Instead you have $Var(X_1 + X_2) = Var(X_1) + Var(X_2) = 2 Var(X_1)$. – Matthias Klupsch Jul 30 '18 at 9:06
• thanks for the clarification. So then: $$Var(\hat{\theta})= \frac{4n^3}{9}Var(X_1)= \frac {4n^3}{9} \frac34 \theta^2 = \frac{n^3}{3}\theta^2$$ – user1607 Jul 30 '18 at 12:04