Question

Let 𝑋₁,𝑋₂, … , 𝑋_𝑛 be a random sample from a population having the probability density function
\(f(x;μ) =\begin{cases} \frac{1}{2}e-(\frac{x-2μ}{2}), & \quad \text{if }0>2μ,\\ 0, & \quad Otherwise \end{cases}\)
where −∞ < 𝜇 < ∞. For estimating 𝜇, consider estimators
\(T_1=\frac{\overline{X}-2}{2}\) and  \(T_2=\frac{nX_{(1)}-2}{2n}\)
where 𝑋̅ =\(\frac{1 }{𝑛} ∑^n_{i=1} x_i\) and X_i and X_(i)=min{𝑋₁, 𝑋₂, … , 𝑋_𝑛}. Then, which one of the following statements is TRUE?

• A. 𝑇₁ is consistent but 𝑇₂ is NOT consistent
• B. 𝑇₂ is consistent but 𝑇₁ is NOT consistent
• C. Both 𝑇₁ and 𝑇₂ are consistent
• D. Neither 𝑇₁ nor 𝑇₂ is consistent

Answer 1

The Correct Option is C

The problem involves testing the consistency of two statistical estimators, \(T_1\) and \(T_2\), for estimating the parameter \(\mu\) in a given probability density function. 

Firstly, let's understand what consistency means in statistical estimation:

• An estimator is said to be consistent if, as the sample size \(n\) increases, the estimator converges in probability to the true parameter value. In more formal terms, an estimator \(\hat{\theta}_n\) for the parameter \(\theta\) is consistent if: \(\lim_{n \to \infty} P(|\hat{\theta}_n - \theta| > \epsilon) = 0\) for every \(\epsilon > 0\).

Now, let's analyze each estimator:

1. The estimator \(T_1 = \frac{\overline{X} - 2}{2}\), where \(\overline{X} = \frac{1}{n} \sum_{i=1}^{n} X_i\), is based on the sample mean. The sample mean \(\overline{X}\) is a consistent estimator for the population mean under the given density function. As \(n \rightarrow \infty\), \( \overline{X} \rightarrow E(X)\). Therefore, \(T_1\) being a linear transformation of \(\overline{X}\), is also consistent.
2. The estimator \(T_2 = \frac{nX_{(1)} - 2}{2n}\), where \(X_{(1)} = \min \{X_1, X_2, \ldots, X_n\}\) is based on the minimum order statistic. For large \(n\), \(X_{(1)}\) converges in distribution to the left endpoint of the support of the distribution, which is \(2\mu\) for the given density function. Hence, when appropriately scaled, \(T_2\) also converges to \(\mu\) as \(n\) increases, making \(T_2\) consistent.

Therefore, both \(T_1\) and \(T_2\) are consistent estimators of \(\mu\), making the correct answer: Both 𝑇₁ and 𝑇₂ are consistent.