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I am a mathematics undergraduate student and I am reading some statistics notes for interest. Recently I have encountered some problem in this question (hopefully not a too naive one):

Given is a random sample $X_1,X_2,\ldots,X_n$ from a distribution with finite variance $\sigma^2$. We estimate the expectation of the distribution with the sample mean $\bar{X_n}$ on $n$ observations and the same expectation using the sample mean $\bar{X_{5n}}$ on $5$ times $n$ observations. What is the relative efficiency $Var(\bar{X_n})/Var(\bar{X_{5n}})$ with respective to $\bar{X_n}$?

My intuition leads me to the answer $1/5$ but I can't really prove it. What I have done is as follow:

$\dfrac{Var(\bar{X_n})}{Var(\bar{X_{5n}})}=\dfrac{E[\bar{X_n^2}]-E[\bar{X_n}]^2}{E[\bar{X_{5n}^2}]-E[\bar{X_{5n}}]^2}=\dfrac{E[\bar{X_n^2}]-\mu^2}{E[\bar{X_{5n}^2}]-\mu^2}$

But I just stop here and can't proceed. Could I have some help? Hints or full answer are appreciated.

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1 Answer 1

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If the $X_i$ are independent (and probably they are in this context) then:$$\text{Var}\overline{X}_n=\text{Var}\left(\frac1{n}\sum_{i=1}^nX_i\right)=\frac1{n^2}\text{Var}\left(\sum_{i=1}^nX_i\right)=\frac1{n^2}\sum_{i=1}^n\text{Var}X_i=\frac{\sigma^2}{n}$$

The third equation is based on independence.

Likewise we find $\text{Var}\overline{X}_{5n}=\frac{\sigma^2}{5n}$ so that: $$\frac{\text{Var}\overline{X}_n}{\text{Var}\overline{X}_{5n}}=5$$

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  • $\begingroup$ Oh I see...so using the identity $Var X=E(X^2)-E(X)^2$ is not a good choice this time. Thank you for your answer! $\endgroup$ Aug 11, 2017 at 9:55

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