# How to show $(n+1) \max(X_1, X_2, \ldots , X_n)/n$ is a method of moments estimator

If $X_i \sim$ uniform$(0,\theta)$, how can I show that the estimator $(n+1) \max(X_1, X_2, \ldots , X_n)/n$ is a method of moments estimator for $\theta$? For example, we know the first sample moment is $\sum_{i=1}^{n} X_i / n$ and that the first population moment is $\theta / 2$. Thus, the estimator we get from the method of moments is $2 \bar{X}$. I don't know how to get $(n+1) X_{(n)} / n$ by the method of moments.

• What is the expected value of $X_{(n)}$? – dsaxton Mar 8 '16 at 23:57
• One does not estimate the values of a random variable. Are you asking for the distribution of the maximum? For prediction of the maximum? Are you considering actual data or are you asking a theoretical question about random variables? – whuber Mar 9 '16 at 0:26
• @dsaxton The expected value of X(n) should be n*theta/(n+1) – purod Mar 9 '16 at 1:37
• @whuber What I want to ask is that if theta is not known, we want to find an unbiased estimator for it. Using other method, I get estimate of (n+1)*x(n)/n. However, I am asked to get this estimator with method of moment. I know that mean = theta/2 then an estimator for theta by method of moment should be 2*mean. I don't know how to get estimator (n+1)*theta/n with method of moment – purod Mar 9 '16 at 1:40
• It's a trivial method of moments estimator based on a sample of size one from a "stretched" beta distribution over $(0, \theta)$ (the distribution of the maximal order statistic). This isn't really the idea behind method of moments though, generally you work with the underlying distribution and not the distribution of a sample-based statistic. – dsaxton Mar 9 '16 at 1:46

Let $X_1, \dots X_n \sim U(0, \theta)$. Let $X_{(n)} = \max\{X_1, \dots, X_n\}.$ The traditional way to answer a question "What is the method of moments estimator?" is to do what you did in your statement of the question.
Define $Y = X_{(n)}$. Now you will find the method of moments estimator for $\theta$ that characterizes the distribution of $Y$ (not the distribution of $X_1$).
You already have $E(Y) = \dfrac{n \theta}{n+1}$, and that gives you the desired result.