Question

Let X₁ , X₂ , … , X_n(n ≥ 2) be a random sample from Exp(\(\frac{1}{\theta}\)) distribution, where θ > 0 is unknown. If \(\overline{X}=\frac{1}{n}\sum^n_{i=1}X_i\), then which one of the following statements is NOT true ?

• A. \(\overline{X}\) is the uniformly minimum variance unbiased estimator of θ
• B. \(\overline{X}^2\) is the uniformly minimum variance unbiased estimator of θ²
• C. \(\frac{n}{n+1}\overline{X}^2\) is the uniformly minimum variance unbiased estimator of θ²
• D. \(Var(E(X_n|\overline{X}))\le Var(X_n)\)

Answer 1

The Correct Option is B

To solve the given problem, we need to analyze each option based on the statistical properties of estimators of a random sample from the exponential distribution with parameter \(\frac{1}{\theta}\).

We know that for a random sample \(X_1, X_2, \ldots, X_n\) from the exponential distribution with rate parameter \(\frac{1}{\theta}\):

• The sample mean \(\overline{X}\) is an unbiased estimator for \(\theta\) and is the Minimum Variance Unbiased Estimator (MVUE) due to the properties of exponential distribution.
• The variance of the exponential distribution is \(\theta^2\).
• The mean of the sample mean \(\overline{X}\) is \(\theta\), hence it cannot directly estimate \(\theta^2\).
• \(\overline{X}^2\) is not an unbiased estimator for \(\theta^2\). This is because if \(\overline{X}\) is an unbiased estimator for \(\theta\), then \(\overline{X}^2\) is not unbiased for \(\theta^2\).
• The transformation \(\frac{n}{n+1}\overline{X}^2\) is used to find an unbiased estimator for \(\theta^2\). This corrects the bias posed by directly using \(\overline{X}^2\).

Let's evaluate each statement:

• \(\overline{X}\) is the uniformly minimum variance unbiased estimator of \(\theta\): This is true as \(\overline{X}\) provides the MVUE for \(\theta\) in an exponential distribution.
• \(\overline{X}^2\) is the uniformly minimum variance unbiased estimator of \(\theta^2\): This is false. As explained, \(\overline{X}^2\) is biased for \(\theta^2\).
• \(\frac{n}{n+1}\overline{X}^2\) is the uniformly minimum variance unbiased estimator of \(\theta^2\): This is true as it corrects the bias of \(\overline{X}^2\).
• \(\text{Var}(E(X_n|\overline{X})) \le \text{Var}(X_n)\): This is true based on the conditional variance property.

Conclusion: The statement that is NOT true is: \(\overline{X}^2\) is the uniformly minimum variance unbiased estimator of \(\theta^2\).