Which estimator would be better in terms of Mean Square Error?

Let $$X_1,\ldots,X_n$$ be a a sample from a Uniform Distribution $$(0,\theta)$$ where $$\theta > 0$$ is an unknown parameter.

I have found the estimator based on the sample mean

$$\hat{\theta}=2\bar{X}$$

As well as the estimator based on the nth order statistic

$$\hat{\theta}=\frac{n+1}{n}X_{(n)}$$

Which would be a better estimator in the MSE sense ?

• MSE calculation is straightforward. What did you try? May 24, 2019 at 4:17
• @StubbornAtom I know the formula of the MSE, but I don't know what I'm suppose to do with it. Do I just subtract the estimator by the sample mean & estimator by the nth order statistic. $(\bar{X} - \hat{\theta})^2/n$ May 24, 2019 at 4:25
• What formula do you have? Because what you wrote does not make sense. Do you see that these are unbiased estimators of $\theta$ and so MSE is just variance? May 24, 2019 at 4:29
• I was using the formula of MSE, the sum of the difference squared divided by n. In that case, I got the variance to be $\frac{\theta^2}{12}$ for the sample mean and $\frac{(n)}{(n+1)^2}\theta^2$. Would that mean the sample mean is better then? May 24, 2019 at 4:37

Mean square error (MSE) of an estimator $$\hat\theta$$ for estimating $$\theta$$ is defined as $$\operatorname{MSE}_{\theta}(\hat\theta)=\operatorname E_{\theta}(\hat\theta-\theta)^2=\operatorname{Var}_{\theta}(\hat\theta)+(\text{bias}(\hat\theta))^2$$

Both your estimators are unbiased for $$\theta$$, so MSE here is just variance.

Now, $$\operatorname{Var}_{\theta}(2\overline X)=4\operatorname{Var}_{\theta}(\overline X)=4\times\frac{\theta^2}{12n}=\frac{\theta^2}{3n}$$

And keeping in mind that $$X_{(n)}/\theta\sim \mathsf{Beta}(n,1)$$ we have,

\begin{align} \operatorname{Var}_{\theta}\left[\frac{n+1}{n}X_{(n)}\right]&=\left(\frac{n+1}{n}\right)^2\operatorname{Var}_{\theta}(X_{(n)}) \\&=\left(\frac{n+1}{n}\right)^2\times\frac{n\theta^2}{(n+1)^2(n+2)} \\&=\frac{\theta^2}{n(n+2)} \end{align}

It is clear that

$$\frac{\theta^2}{3n}-\frac{\theta^2}{n(n+2)}=\frac{\theta^2}{n}\left[\frac{1}{3}-\frac{1}{n+2}\right]> 0\quad,\forall \,n> 1$$

This proves $$T=\frac{n+1}{n}X_{(n)}$$ has the smaller MSE, which is not surprising since it is known that $$T$$ is the uniformly minimum variance unbiased estimator of $$\theta$$.

As pointed out by @Henry, it is worth mentioning that $$X_{(n)}$$ is a sufficient statistic for $$\theta$$ (and hence responsible for data condensation without losing information about the parameter) whereas $$\overline X$$ is not. So even without any calculation we expect the estimator based on $$X_{(n)}$$ to have a smaller mean square error.

• Can you explain why there is a $\frac{1}{n}$ in the sample mean and if so, why there is a $\frac{1}{n+2}$ on the nth order statistic ? May 24, 2019 at 7:17
• Because that's what you get from calculation. Variance of sample mean is $\sigma^2/n$ and that of the order statistic follows from the variance of beta distribution (or you can write down the density of $X_{(n)}$ and find the variance directly). May 24, 2019 at 7:33
• +1 Also worth noting that $X_{(n)}$ is a sufficient statistic here while $\bar X$ is not May 24, 2019 at 7:43
• Thank you StubbornAtom but if you want to compare $X_{(n)}$ with $\bar{X}_{n}$ not $\frac{n+1}{n}X_{(n)}$, how can you show the same conclusion using your answer ?
– user347910
Feb 25, 2020 at 13:24
• @CechMS Find the bias $E(X_{(n)}-\theta)$ and use the formula right at the beginning for the MSE. The variance is already shown. Feb 25, 2020 at 14:22