I have the following problem:

Let $Y_1, Y_2, \dots, Y_n$ be i.i.d. $\text{Uniform}(\theta, 1)$ random variables, and let an estimator be $\hat{\theta} = \min\{ Y_1, Y_2, \dots, Y_n \}$.

You may find the following information useful when answering this question:
Let $U_1, \dots, U_n$ be i.i.d. $\text{Uniform}(0, 1)$ random variables, and let $X = \min\{ U_1, \dots, U_n \}$, which has the density $f_X(x) = n(1 - x)^{n - 1}$.
(i) The first two moments are given by $E[X] = \dfrac{1}{n + 1}$ and $E[X^2] = \dfrac{2}{(n + 1)(n + 2)}$.
(ii) It is known that the random variable $X$ and $\dfrac{\hat{\theta} - \theta}{1 - \theta}$ have the same distribution. So this implies that $E[X] = E\left[ \dfrac{\hat{\theta} - \theta}{1 - \theta} \right]$ and $E[X^2] = E \left[ \dfrac{(\hat{\theta} - \theta)^2}{(1 - \theta)^2} \right]$.

Find the bias, variance, and MSE (mean squared error) of $\hat{\theta}$.

I am given the following description of the method of moments:

The main idea is based on expressing the population moments of the distribution of data in terms of its unknown parameter(s) and equating them to their corresponding sample moments. The parameter(s) are then estimated by the solutions of the resulting equations.

The solution begins by calculating the first two moments of $\hat{\theta}$:

$$\begin{align} E \left[ \dfrac{ \hat{\theta} - \theta }{ 1 - \theta } \right] = E[X] = \dfrac{1}{n + 1} \Rightarrow E \left[ \hat{\theta} \right] = \dfrac{ n \theta + 1 }{ n + 1 } \end{align}$$

$$\begin{align} E \left[ \left( \dfrac{\hat{\theta} - \theta }{ 1 - \theta } \right)^2 \right] = E \left[ X^2 \right] = \dfrac{2}{(n + 1)(n + 2)} &\Rightarrow \dfrac{ E\left[ \hat{\theta}^2 \right] - 2\theta E\left[ \hat{\theta} \right] + \theta^2 }{(1 - \theta)^2} = \dfrac{2}{(n + 1)(n + 2)} \\ &\Rightarrow E \left[ \hat{\theta}^2 \right] = \dfrac{n(n + 1) \theta^2 + 2n \theta + 2 }{ (n + 1)(n + 2) } \end{align}$$

How does the information in the problem statement, including (i) and (ii), and this solution align with the provided description of the method of moments? I'm a bit confused when trying to conceptually connect these two.

  • $\begingroup$ The method of moments wasn't used here; they simply equated equal equations and solved for $E[\hat \theta]$ and $E[\hat \theta^2]$ which will be needed to calculate bias, variance, and MSE. $\endgroup$
    – user308286
    Commented May 29, 2021 at 21:46
  • $\begingroup$ @Stacker ahh, ok, that makes sense then. Thanks for the clarification. $\endgroup$ Commented May 29, 2021 at 21:48

1 Answer 1


The method of moments wasn't used here; the beginning of the solution just equates equal expressions together to find $E[\hat \theta]$ and $E[\hat \theta^2]$, which will be needed to find the bias, variance, and MSE.

The method of moments is used for estimating a parameter. In this situation it would be used to estimate the parameter $\theta$. Here we are given that $\hat \theta=\min \{Y_1, Y_2, ..., Y_n\}$. This is the maximum likelihood estimator of $\theta$ if the density is $f(y)=\begin{cases}\frac{1}{1-\theta}&\theta\le y\le 1\\0&\text{otherwise}\end{cases}$ because it maximizes the likelihood $f(\textbf y|\theta)=\frac{1}{(1-\theta)^n}\textbf 1_{\min {Y_1,...,Y_n}\ge\theta}$, as $f$ is an increasing function of $\theta$. However if $f(y)=\begin{cases}\frac 1{1-\theta}&\theta<y<1\\0&\text{otherwise}\end{cases}$ this would not be the maximum likelihood estimator (no MLE would exist, as $\theta<\min \textbf Y$ and so can be made arbitrarily close to it but cannot be made equal to $\min \textbf Y$) and would just be any unnamed estimator. The method of moments estimator would be $E(Y)=m_1$ with $m_1$ being the first moment. So we would get $\frac{1+\theta}{2}=\bar Y\Rightarrow \hat\theta=2\bar Y-1$ as the method of moments estimator. Another estimator is a Bayes estimator, but this requires placing a prior on $\theta$ and also specifying a loss function.


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