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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?

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  • $\begingroup$ 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. $\endgroup$ – Antoine Jul 30 '18 at 8:38
  • $\begingroup$ $X_i, ..., X_n$ are iid...forgot to mention. Then $Var(\sum_iX_i )= nVar(X_1)$, right? $\endgroup$ – user1607 Jul 30 '18 at 8:44
  • $\begingroup$ I count lots of mistakes in this. E.g. do we really have $\frac{4n^2}9n=\frac4n$? $\endgroup$ – drhab Jul 30 '18 at 8:45
  • $\begingroup$ @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. $\endgroup$ – drhab Jul 30 '18 at 8:49
  • $\begingroup$ I appologize, i made mistakes when writing the questions and fixed them after they were poited out. $\endgroup$ – user1607 Jul 30 '18 at 8:56
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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 $Var(X_1) = \frac{3}{4} \theta^2$.

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  • $\begingroup$ 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) $ ? $\endgroup$ – user1607 Jul 30 '18 at 9:05
  • $\begingroup$ 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 $Var(X_1 + X_2) = Var(X_1) + Var(X_2) = 2 Var(X_1)$. $\endgroup$ – Matthias Klupsch Jul 30 '18 at 9:06
  • $\begingroup$ thanks for the clarification. So then: $$ Var(\hat{\theta})= \frac{4n^3}{9}Var(X_1)= \frac {4n^3}{9} \frac34 \theta^2 = \frac{n^3}{3}\theta^2$$ $\endgroup$ – user1607 Jul 30 '18 at 12:04
  • $\begingroup$ Yes that's correct. $\endgroup$ – Matthias Klupsch Jul 30 '18 at 12:22

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