Estimator Properties and Analysis

The problem deals with independent and identically distributed (i.i.d.) random variables \(X_i\) defined as \(X_i = \theta + \epsilon_i\), where \(\theta\) is an unknown parameter to be estimated, and \(\epsilon_i\) are i.i.d. random errors. The probability density function (PDF) of \(\epsilon_i\) is given by \(g(\epsilon) = |\epsilon|\) for \(-1 < \epsilon < 1\).

We are given two estimators for \(\theta\):

1. \(T_1 = \frac{1}{n} \sum_{i=1}^n X_i\), which is the sample mean of \(X_i\).
2. \(T_2 = X_{\left(\left[\frac{3 n}{4}\right]+1\right)}\), which is the sample 75th percentile of \(X_i\).

We need to evaluate the correctness of statements regarding the consistency, asymptotic normality, and asymptotic variance of these estimators.

Distribution of Error Term ε

First, let's analyze the distribution of the error term \(\epsilon\). The PDF is \(g(\epsilon) = |\epsilon|\) for \(-1 < \epsilon < 1\). The PDF is symmetric around 0.

• Mean of \(\epsilon\): \(E[\epsilon] = \int_{-1}^1 \epsilon g(\epsilon) d\epsilon = \int_{-1}^1 \epsilon |\epsilon| d\epsilon\). Since \(\epsilon |\epsilon|\) is an odd function and the integration interval is symmetric around 0, \(E[\epsilon] = 0\).
• Variance of \(\epsilon\): \(Var(\epsilon) = E[\epsilon^2] - (E[\epsilon])^2 = E[\epsilon^2] - 0^2 = E[\epsilon^2]\). \(E[\epsilon^2] = \int_{-1}^1 \epsilon^2 g(\epsilon) d\epsilon = \int_{-1}^1 \epsilon^2 |\epsilon| d\epsilon = \int_{-1}^0 \epsilon^2 (-\epsilon) d\epsilon + \int_0^1 \epsilon^2 (\epsilon) d\epsilon = \int_{-1}^0 -\epsilon^3 d\epsilon + \int_0^1 \epsilon^3 d\epsilon = \left[-\frac{\epsilon^4}{4}\right]_{-1}^0 + \left[\frac{\epsilon^4}{4}\right]_0^1 = (0 - (-\frac{(-1)^4}{4})) + (\frac{1^4}{4} - 0) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\). So, \(Var(\epsilon) = \frac{1}{2}\).
• CDF of \(\epsilon\): \(F_\epsilon(\epsilon) = \int_{-1}^\epsilon g(u) du = \int_{-1}^\epsilon |u| du\).
  • For \(-1 < \epsilon \le 0\), \(F_\epsilon(\epsilon) = \int_{-1}^\epsilon -u du = [-\frac{u^2}{2}]_{-1}^\epsilon = -\frac{\epsilon^2}{2} - (-\frac{1}{2}) = \frac{1 - \epsilon^2}{2}\).
  • For \(0 < \epsilon < 1\), \(F_\epsilon(\epsilon) = \int_{-1}^0 -u du + \int_0^\epsilon u du = \frac{1}{2} + [\frac{u^2}{2}]_0^\epsilon = \frac{1 + \epsilon^2}{2}\).
  So, \(F_\epsilon(\epsilon) = \begin{cases} \frac{1 - \epsilon^2}{2} & -1 < \epsilon \le 0 \\ \frac{1 + \epsilon^2}{2} & 0 < \epsilon < 1 \end{cases}\).
• 75th percentile of \(\epsilon\): We need the value \(q_{0.75}\) such that \(F_\epsilon(q_{0.75}) = 0.75\). Since \(0.75 > 0.5\), \(q_{0.75}\) is positive. \(\frac{1 + q_{0.75}^2}{2} = 0.75 \implies 1 + q_{0.75}^2 = 1.5 \implies q_{0.75}^2 = 0.5 \implies q_{0.75} = \sqrt{0.5} = \frac{1}{\sqrt{2}}\) (since \(q_{0.75} > 0\)).

Estimator T1 (Sample Mean) Analysis

\(T_1 = \frac{1}{n} \sum_{i=1}^n X_i = \frac{1}{n} \sum_{i=1}^n (\theta + \epsilon_i) = \frac{1}{n} (n\theta + \sum_{i=1}^n \epsilon_i) = \theta + \frac{1}{n} \sum_{i=1}^n \epsilon_i\).

• Expectation: \(E[T_1] = E[\theta + \frac{1}{n} \sum \epsilon_i] = \theta + \frac{1}{n} \sum E[\epsilon_i] = \theta + \frac{1}{n} \sum 0 = \theta\). \(T_1\) is an unbiased estimator for \(\theta\).
• Variance: \(Var(T_1) = Var(\theta + \frac{1}{n} \sum \epsilon_i) = Var(\frac{1}{n} \sum \epsilon_i)\). Since \(\epsilon_i\) are i.i.d., \(Var(T_1) = \frac{1}{n^2} \sum Var(\epsilon_i) = \frac{1}{n^2} n Var(\epsilon) = \frac{1}{n} Var(\epsilon) = \frac{1}{n} \cdot \frac{1}{2} = \frac{1}{2n}\).
• Consistency: An unbiased estimator is consistent if its variance approaches 0 as \(n \to \infty\). Here, \(Var(T_1) = \frac{1}{2n} \to 0\) as \(n \to \infty\). Thus, \(T_1\) is a consistent estimator for \(\theta\).
• Asymptotic Normality: By the Central Limit Theorem (CLT), for i.i.d. random variables \(\epsilon_i\) with finite mean (0) and variance (1/2), the sample mean \(\frac{1}{n} \sum \epsilon_i\) is asymptotically normal. Specifically, \(\sqrt{n} (\frac{1}{n} \sum \epsilon_i - E[\epsilon_i]) \xrightarrow{d} N(0, Var(\epsilon))\). Substituting, \(\sqrt{n} (T_1 - \theta - 0) \xrightarrow{d} N(0, 1/2)\). This means \(\sqrt{n} (T_1 - \theta) \xrightarrow{d} N(0, 1/2)\). Therefore, \(T_1\) is asymptotically normal for \(\theta\).
• Asymptotic Variance: The asymptotic variance of \(T_1\) is given by \(\frac{1}{n} \cdot Var(\epsilon) = \frac{1}{2n}\).

Estimator T2 (Sample 75th Percentile) Analysis

\(T_2 = X_{\left(\left[\frac{3 n}{4}\right]+1\right)}\) is the sample 75th percentile of \(X_i\).

• Population 75th Percentile of \(X_i\): The distribution of \(X_i\) is \(F_X(x) = P(X_i \le x) = P(\theta + \epsilon_i \le x) = P(\epsilon_i \le x - \theta) = F_\epsilon(x - \theta)\). The 75th percentile of \(X_i\), denoted \(\xi_{0.75}\), is the value such that \(F_X(\xi_{0.75}) = 0.75\). This means \(F_\epsilon(\xi_{0.75} - \theta) = 0.75\). From our analysis of \(\epsilon\), the 75th percentile of \(\epsilon\) is \(\frac{1}{\sqrt{2}}\). Therefore, \(\xi_{0.75} - \theta = \frac{1}{\sqrt{2}}\), which implies \(\xi_{0.75} = \theta + \frac{1}{\sqrt{2}}\).
• Consistency: The sample \(p\)-th percentile is a consistent estimator for the population \(p\)-th percentile under mild conditions (which hold here). Thus, \(T_2\) is a consistent estimator for \(\theta + \frac{1}{\sqrt{2}}\). This implies \(T_2 - \frac{1}{\sqrt{2}}\) is a consistent estimator for \(\theta\).
• Asymptotic Normality and Variance: The sample \(p\)-th percentile \(X_{([np]+1)}\) is asymptotically normal with mean equal to the true \(p\)-th percentile \(\xi_p\) and asymptotic variance \(\frac{p(1-p)}{n [f_X(\xi_p)]^2}\), provided \(f_X(\xi_p) > 0\). * Here \(p = 0.75\), \(\xi_{0.75} = \theta + \frac{1}{\sqrt{2}}\). * The PDF of \(X\) is \(f_X(x) = \frac{d}{dx} F_X(x) = \frac{d}{dx} F_\epsilon(x - \theta) = g(x - \theta) = |x - \theta|\) for \(\theta - 1 < x < \theta + 1\). * We need \(f_X(\xi_{0.75}) = f_X(\theta + \frac{1}{\sqrt{2}}) = |(\theta + \frac{1}{\sqrt{2}}) - \theta| = |\frac{1}{\sqrt{2}}| = \frac{1}{\sqrt{2}}\). Since \(\frac{1}{\sqrt{2}} \approx 0.707\), which is between -1 and 1, and \(f_X(\xi_{0.75}) = \frac{1}{\sqrt{2}} > 0\), the asymptotic normality result applies. * The asymptotic variance of \(T_2\) (as an estimator for \(\xi_{0.75} = \theta + \frac{1}{\sqrt{2}}\)) is \(\frac{p(1-p)}{n [f_X(\xi_p)]^2} = \frac{0.75(1-0.75)}{n (\frac{1}{\sqrt{2}})^2} = \frac{\frac{3}{4} \cdot \frac{1}{4}}{n \cdot \frac{1}{2}} = \frac{\frac{3}{16}}{\frac{n}{2}} = \frac{3}{16} \cdot \frac{2}{n} = \frac{3}{8n}\). * The sample quantile \(T_2\) is asymptotically normal with mean \(\theta + \frac{1}{\sqrt{2}}\) and asymptotic variance \(\frac{3}{8n}\). * Now consider \(T_2 - \frac{1}{\sqrt{2}}\). As \(T_2\) is asymptotically normal for \(\theta + \frac{1}{\sqrt{2}}\), it follows that \(T_2 - \frac{1}{\sqrt{2}}\) is asymptotically normal for \((\theta + \frac{1}{\sqrt{2}}) - \frac{1}{\sqrt{2}} = \theta\). The asymptotic variance is \(Var(T_2 - \frac{1}{\sqrt{2}}) = Var(T_2) = \frac{3}{8n}\).

Option Analysis

Let's examine each statement based on our analysis:

1. T 1 is a consistent and asymptotically normal estimator of θ: Our analysis shows \(T_1\) is consistent and asymptotically normal for \(\theta\). This statement is correct.
2. \(T_2-\frac{1}{\sqrt{2}}\)is a consistent and asymptotically normal estimator of θ: Our analysis shows \(T_2 - \frac{1}{\sqrt{2}}\) is consistent and asymptotically normal for \(\theta\). This statement is correct.
3. The asymptotic variance of T 1 is \(\frac{1}{2 n}\): Our analysis shows the asymptotic variance of \(T_1\) is \(\frac{1}{2n}\). This statement is correct.
4. The asymptotic variance of T 2 is \(\frac{3}{8 n}\): Our analysis shows the asymptotic variance of the sample 75th percentile \(T_2\) is \(\frac{3}{8n}\). This statement is correct.

All four statements are correct.