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I am using a book, where the following exercise appears:

Two estimators for a random sample of size $n$ (Bernoulli population):

$T_1=\frac{\sum\limits_{i=1}^n X_i + 2X_n}{n+2}$

$T_2=\frac{\sum\limits_{i=1}^{n-2} X_i + 2X_n}{n+2}$

I want to compare them in terms of their relative efficiency.

I know that:

$VAR[T_2]=VAR\bigg[\frac{\sum\limits_{i=1}^{n-2} X_i + 2X_{n}}{n+2}\bigg] \\ =\frac{1}{(n+2)^2}.\{VAR[X_1+X_2+...+X_{n-2}]+VAR[2X_n]\} \\ $

however I am not sure how to continue... the variance of a Bernoulli is $p.q$, should I replace each $X_i$ with $p.q$?

The book suggests:

(1.1) $= \frac{1}{(n+2)^2}.[(n+2).pq+4pq] \\ = \frac{1}{(n+2)^2}.(n+2).pq \\ = \frac{pq}{n+2}$

however, I do not understand these three steps (1.1).

The same for the estimator $T_1$:

$VAR[T_1]=VAR\bigg[\frac{\sum\limits_{i=1}^{n} X_i + 2X_n}{n+2}\bigg] \\ =\frac{1}{(n+2)^2}.\{VAR[X_1+X_2+...+X_n+2X_n]\}$

the book suggests:

(2.1) $=\frac{1}{(n+2)^2}.\{VAR[X_1+X_2+...+X_{n-1}+3X_n]\} \\ = \frac{n+8}{(n+2)^2}.pq$

I do not understand these steps (2.1).

Sorry for the very basic question.

Sinval
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  • Use the property of variance $\operatorname{Var}(c_1X_1+c_2X_2)=c_1^2\operatorname{Var}(X_1)+c_2^2\operatorname{Var}(X_2)$ if $c_1,c_2$ are constants and $X_1,X_2$ are independent random variables. – Momo Nov 30 '21 at 02:05
  • Can you show me how in $T_1$ or $T_2$? – Sinval Nov 30 '21 at 07:17
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    $\operatorname{Var}(T_1)=\operatorname{Var}\left(\frac{\sum_{i=1}^n X_i + 2X_n}{n+2}\right)=\frac{1}{(n+2)^2}\operatorname{Var}\left(\sum_{i=1}^n X_i + 2X_n\right)\=\frac{1}{(n+2)^2}\operatorname{Var}(X_1+X_2+\ldots+X_{n-1}+X_n+2X_n)\=\frac{1}{(n+2)^2}\operatorname{Var}(X_1+X_2+\ldots+X_{n-1}+3X_n)\=\frac{1}{(n+2)^2}(\operatorname{Var}(X_1)+\operatorname{Var}(X_2)+\ldots+\operatorname{Var}(X_{n-1})+\operatorname{Var}(3X_n))\=\frac{1}{(n+2)^2}(\operatorname{Var}(X_1)+\operatorname{Var}(X_2)+\ldots+\operatorname{Var}(X_{n-1})+9\operatorname{Var}(X_n))\=\frac{1}{(n+2)^2}((n-1)pq+9pq)$ – Momo Nov 30 '21 at 12:24
  • Thank you so much. And now I detected a typo in the book (in the first line of (1.1), it should be $\frac{1}{(n+2)^2}.[(n-2)pq+4pq]$ instead of $\frac{1}{(n+2)^2}.[(n+2)pq+4pq]$ – Sinval Dec 01 '21 at 19:57

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