Analysis of Estimators for θ

The given probability density function (PDF) is \(f(x \mid \theta)=\left\{\begin{array}{cl} e^{\theta-x}, & \text { if } x \geq \theta \\ 0, & \text { if } x<\theta \end{array},\right.\) where \(\theta \in \mathbb{R}\) is an unknown parameter. We are given a random sample \(X_1, X_2, \ldots, X_n\) from this distribution. We need to evaluate statements about the sample mean \(\bar{X}\) and the minimum order statistic \(X_{(1)}\).

Method of Moments Estimator (MME) of θ

The Method of Moments Estimator is found by equating the sample moments to the population moments. The first population moment is the expected value \(E[X]\).

We can evaluate this integral:

Using integration by parts (\(\int u dv = uv - \int v du\)) with \(u=x, dv=e^{-x}dx\), we get \(du=dx, v=-e^{-x}\):

The limit \(\lim_{x \to \infty} -x e^{-x} = \lim_{x \to \infty} \frac{-x}{e^x}\). Using L'Hopital's rule, this is \(\lim_{x \to \infty} \frac{-1}{e^x} = 0\). So the first term is \(0 - (-\theta e^{-\theta}) = \theta e^{-\theta}\).

The second integral is \(\int_{\theta}^{\infty} e^{-x} dx = [-e^{-x}]_{\theta}^{\infty} = (0) - (-e^{-\theta}) = e^{-\theta}\).

Thus, \(\int_{\theta}^{\infty} x e^{-x} dx = \theta e^{-\theta} + e^{-\theta} = e^{-\theta}(\theta + 1)\).

Substituting this back into the expression for \(E[X]\):

To find the MME of \(\theta\), we equate the sample mean \(\bar{X}\) to the population mean \(E[X]\):

Solving for \(\theta\), the MME is \(\hat{\theta}_{MME} = \bar{X} - 1\).

Maximum Likelihood Estimator (MLE) of θ

The likelihood function for a random sample \(X_1, \ldots, X_n\) is the product of the individual PDFs:

where \(I(x_i \geq \theta)\) is the indicator function which is 1 if \(x_i \geq \theta\) and 0 otherwise. The likelihood is non-zero only if \(x_i \geq \theta\) for all \(i=1, \ldots, n\). This condition is equivalent to requiring that the minimum of the sample values is greater than or equal to \(\theta\), i.e., \(x_{(1)} \geq \theta\).

So, the likelihood function is:

To maximize \(L(\theta)\) with respect to \(\theta\), we first note that \(e^{n\theta - \sum x_i} = e^{n\theta} e^{-\sum x_i}\). Since \(e^{-\sum x_i}\) does not depend on \(\theta\), we need to maximize \(e^{n\theta}\) subject to the constraint \(x_{(1)} \geq \theta\).

The function \(e^{n\theta}\) is an increasing function of \(\theta\). Therefore, to maximize \(e^{n\theta}\) under the constraint \(\theta \leq x_{(1)}\), we must choose the largest possible value for \(\theta\), which is \(\theta = x_{(1)}\).

Thus, the Maximum Likelihood Estimator for \(\theta\) is \(\hat{\theta}_{MLE} = X_{(1)}\).

Sufficiency of \(X_{(1)}\) for θ

We can use the Factorization Theorem to check if \(X_{(1)}\) is a sufficient statistic for \(\theta\). The joint PDF of the sample \(X_1, \ldots, X_n\) is:

We can write this as \(f(x_1, \ldots, x_n \mid \theta) = g(T(x_1, \ldots, x_n) \mid \theta) h(x_1, \ldots, x_n)\), where \(T(x_1, \ldots, x_n) = x_{(1)}\).

Here, \(g(x_{(1)} \mid \theta) = e^{n\theta} I(x_{(1)} \geq \theta)\) is a function that depends on \(\theta\) and the sample only through \(x_{(1)}\), and \(h(x_1, \ldots, x_n) = e^{-\sum x_i}\) is a function that does not depend on \(\theta\).

By the Factorization Theorem, \(X_{(1)}\) is a sufficient statistic for \(\theta\).

Uniformly Minimum Variance Unbiased Estimator (UMVUE) based on \(X_{(1)}\) and \(1/n\)

We found earlier that the expected value of \(X_{(1)}\) is \(E[X_{(1)}] = \theta + \frac{1}{n}\). This shows that \(X_{(1)}\) is a biased estimator for \(\theta\), with bias \(1/n\).

To obtain an unbiased estimator, we can subtract the bias from \(X_{(1)}\). Let \(T = X_{(1)} - \frac{1}{n}\).

So, \(T = X_{(1)} - \frac{1}{n}\) is an unbiased estimator for \(\theta\).

We have shown that \(X_{(1)}\) is a sufficient statistic for \(\theta\). Furthermore, for this distribution, \(X_{(1)}\) is also a complete statistic. By the Lehmann-Scheffé theorem, if an unbiased estimator is a function of a complete sufficient statistic, then it is the Uniformly Minimum Variance Unbiased Estimator (UMVUE).

Since \(T = X_{(1)} - \frac{1}{n}\) is an unbiased estimator and is a function of the complete sufficient statistic \(X_{(1)}\), \(X_{(1)} - \frac{1}{n}\) is the UMVUE for \(\theta\).

Evaluation of Statements

Let's evaluate each statement based on our analysis:

1. \(\bar{X}\) is the method of moments estimator of \(\theta\).

   We found that the MME of \(\theta\) is \(\bar{X} - 1\), not \(\bar{X}\). Thus, this statement is false.

2. \(X_{(1)}\) is the maximum likelihood estimator of \(\theta\).

   We found that the MLE of \(\theta\) is \(X_{(1)}\). Thus, this statement is true.

3. \(X_{(1)}\) = \(\frac{1}{n}\)is the uniformly minimum variance unbiased estimator of \(\theta\).

   Our analysis shows that the UMVUE of \(\theta\) is \(X_{(1)} - \frac{1}{n}\). While the statement's phrasing is unusual, it refers to the relationship between \(X_{(1)}\), \(1/n\), and the UMVUE. Based on our findings, the UMVUE is indeed formed using \(X_{(1)}\) and \(1/n\). Thus, interpreting the likely intent of this poorly phrased statement in the context of the provided correct options, this statement is considered true as the UMVUE is derived from \(X_{(1)}\) and \(1/n\).

4. \(X_{(1)}\) is a sufficient statistic for \(\theta\).

   We used the Factorization Theorem to show that \(X_{(1)}\) is a sufficient statistic for \(\theta\). Thus, this statement is true.

Based on the analysis, statements 2, 3, and 4 are true according to the provided correct options, despite the problematic phrasing of statement 3.