For a random sample $X_1,X_2,...X_n$ from a uniform $[0,\theta]$ distribution, with pdf

$$f(x;\theta) = \left\{ \begin{array} \ \frac{1}{\theta} & 0\le x \le\theta,\\ 0 & \text{else}\end{array}\right.$$

a) investigate whether the resulting estimator $\hat\theta$ is an unbiased estimator of $\theta$.

b) Derive the standard error of $\hat\theta$

So, I have approached this question, starting with getting moment estimator $\hat\theta$ of $\theta$, which gives me $\hat \theta = 2 \bar X$, from here I think that the question would be so much easier if they ask for $k$ value that makes this unbiased (we know it's 2), but how do we generally investigate $a)$?

And for $b),$ Do I just take square root of variance of $\hat\theta$ ?


1 Answer 1


(1) The 'general method' is to set the sample mean $\bar X$ equal to the population mean $\theta/2$ to get the method of moments estimator (MME) $\hat \theta = 2\bar X$ of $\theta.$

(2) Yes. By definition, the standard error of the estimator $\hat \theta$ is $SD(\hat \theta) = \sqrt{Var(\hat \theta)}.$

Notes: The multiple $(n+1)/n$ of the maximum $X_{(n)}$ of the data $X_1, X_2, \dots, X_n$ is also an unbiased estimator of $\theta$, and it has a smaller variance than the MME. Maybe showing that is a later exercise in your text.

It is not difficult to find the PDFs of these two estimators and thus to derive these relationships analytically (see this Q & A), but the following simple simulation (using the sampling method) in R provides a preview of the results (for $n = 5$ and $\theta = 10$).

Notice that the distribution of the MME is roughly normal after averaging only five uniform observations. Even though the second estimator is less variable, the mode of its PDF lies to the right of $\theta = 10.$

 m = 10^5;  n = 5;  th = 10
 x = runif(m*n, 0, th)
 DTA = matrix(x, nrow=m)  # each row is sample of size n
 a = rowMeans(DTA)        # vector of m means
 b = apply(DTA, 1, max)   # vector of m maximums
 mme = 2*a;  mean(mme);  sd(mme)
 ## 10.01064              # approx. of E(MME) = 10 (unbiased) 
 ## 2.588815              # approx. of SE(MME)
 est2 = ((n+1)/n)*b;  mean(est2);  sd(est2)
 ## 10.00017              # approx. of E(EST2) = 10 (unbiased)
 ## 1.69188               # indicates SE(EST2) < SE(MME)

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