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Which is a better estimator of $\mu$? the $2n$ is throwing me off.

Suppose we have a random sample of size $2n$ from a population denoted by X and E(X) = mu and Var(x) = sigma squared.

Let $$\bar{x}_1 = \frac{1}{2n}\sum_{ i }^{2n}Xi$$ and Let $$\bar{x}_2 = \frac{1}{n}\sum_{ i }^{n}X_i$$

be two estimators of $\mu$.

which is the better estimator of mu?

i know i must find the one with the smaller variance, but im not sure how to treat the 2n? do i square just the n or the two as well?

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  • $\begingroup$ One clarification: For the indexes in the sum, you write "from i to n" - do you mean "i goes from 1 to n" ? $\endgroup$ – Matti P. Sep 4 '19 at 6:19
  • $\begingroup$ Also, clearly $\bar{x}_1$ should be better, because $\bar{x}_2$ only looks at the first half of the sample; in other words the sample size is smaller in $\bar{x}_2$. $\endgroup$ – Matti P. Sep 4 '19 at 6:20
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Well, clearly in order to find variances of sample mean estimators $\bar x_1, \bar x_2$ you can use the following reasoning: $$\operatorname{Var}(\bar x_2) = \frac{1}{n^2} \operatorname{Var}\left(\sum_{i=1}^n X_i \right) = \frac{1}{n^2}\sum_{i=1}^n \operatorname{Var}(X_i) = \frac{1}{n^2}(n\sigma^2)=\frac{\sigma^2}{n}$$ $$\operatorname{Var}(\bar x_1) = \frac{1}{(2n)^2} \operatorname{Var}\left(\sum_{i=1}^{2n} X_i \right) = \frac{1}{(2n)^2}\sum_{i=1}^{2n} \operatorname{Var}(X_i) = \frac{1}{(2n)^2}(2n\sigma^2)=\frac{\sigma^2}{2n}$$ So, now you only have to compare them.

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  • $\begingroup$ Smaller variance for the larger sample is the main point. I mention confidence intervals in my Answer. $\endgroup$ – BruceET Sep 5 '19 at 3:04
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@Caran-d'Ache (+1) has pointed out that, for the full sample of $2n$ observations, $Var(\bar X_1) = \sigma^2/2n,$ while, for the 'half' sample, $Var(\bar X_2) = \sigma^2/n,$

If these are samples from a population with mean $\mu,$ then $E(\bar X_1) = E(\bar X_2) = \mu,$ so we say both means are unbiased estimators of $\mu.$

Generally speaking, when there is a choice between two two unbiased estimators of a parameter such as $\mu,$ we say that the one with the smaller variance is 'better'. By 'better' we mean more that smaller variance means that the less-variable estimator estimator is more likely to be near $\mu$ than the other.

Normal samples for illustration. For example, let $\mu = 70$ and $\sigma = 9.$ Let's suppose that the $2n = 2000.$ We use R to sample 2000 observations from $\mathsf{Norm}(\mu, \sigma)$ and to find the two sample means:

set.seed(904)
x = rnorm(2000, 70, 9)  
a1 = mean(x);  a2 = mean(x[1:1000])  
a1;  a2
[1] 69.90506$
[1] 69.77681

For larger sample, point estimate is better. Both $A_1 = \bar X_1 = 69.905$ and $A_2 = \bar X_2 = 69.777$ are close to the true mean $\mu = 70,$ but the one based on the larger sample is closer.

Confidence intervals. If $\sigma = 9$ is known, then a 95% confidence interval (CI) for $\mu$ based on the larger sample of 2000 observations is $\bar X_2 \pm 1.96\frac{9}{\sqrt{2000}},$ which computes to $(69.51,\, 70.30).$ In contrast, the CI $(69.22,\, 70.33)$ from the smaller sample is noticeably wider.

pm = c(-1,1)
a1 + pm*1.96 * 9/sqrt(2000)
[1] 69.51062 70.29950
a2 + pm*1.96 * 9/sqrt(1000)
[1] 69.21899 70.33464

Both CIs happen to include the true value $\mu = 70,$ but the CI from the larger sample is centered on a closer estimate and it is also shorter.

If $\sigma$ is unknown and has to be estimated by sample standard deviations, then the sample standard deviation $S_1$ from the larger sample is a better estimate than the sample SD $S_2$ from the smaller sample.

sd(x)
[1] 8.946693  # very close to mu = 9
sd(x[1:1000])
[1] 8.785786  # surprisingly smal)

Because samples are large, the CIs are about the same whether or not $\sigma$ is known. The 95% CI from the larger sample is $(69.51, 70.30)$ and the CI from the smaller sample is $(69.23, 70.32).$

General preference for larger samples. Finally, we can look at histograms of the the larger (top) and smaller samples. Each histogram is compared with the density of the population distribution. Neither histogram is an 'perfect' match to the population density, but the larger sample gives a much better match.

enter image description here

So not only does the larger sample give a better estimate of the population mean, the larger sample provides more information about the population than does the smaller one. So for almost any kind of statistical inference, the larger sample is better.

Note: In case it is of use, the R code to make the figure is shown below.

par(mfrow=c(2,1))
 hist(x, br=30, prob=T, ylim=c(0,0.045), col="skyblue2", main="Sample of 2000")
  curve(dnorm(x, 70, 9), add=T, lwd=2, col="red")
 hist(x[1:1000], br=30,  prob=T, col="skyblue2", main="Sample of 1000")
  curve(dnorm(x, 70, 9), add=T, lwd=2, col="red")
par(mfrow=c(1,1))
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  • $\begingroup$ perfectly instructive! $\endgroup$ – Caran-d'Ache Sep 5 '19 at 17:37

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