Question

Let \( X_1, X_2, \dots, X_n \) be a random sample from \( U(0, \theta) \), where \( \theta > 0 \) is the unknown parameter. Let \( X_{(n)} = \max{\{X_1, X_2, \dots, X_n\}}. \) Then which of the following is (are) consistent estimator(s) of \( \theta^3 \)? 
 

• A. \( 8 X_{n}^3 \)
• B. \( X_{(n)}^3 \)
• C. \( (\frac{2}{n} \sum_{i=5}^{n} X_i)^3 \)
• D. \( \frac{n X_{(n)}^3 + 1}{n+1} \)

Hint:

For uniform distributions, the maximum statistic is often used to estimate the parameter, and scaling it appropriately gives a consistent estimator.

Answer 1

The Correct Option is B, C, D

Given: $X_1, X_2, ..., X_n$ is a random sample from $U(0, \theta)$, where $\theta > 0$.

Let $X_{(n)} = \max{X_1, X_2, ..., X_n}$.

Key Facts about $X_{(n)}$:

For $U(0, \theta)$, the CDF of $X_{(n)}$ is: $$F_{X_{(n)}}(x) = \left(\frac{x}{\theta}\right)^n, \quad 0 \leq x \leq \theta$$

The PDF is: $$f_{X_{(n)}}(x) = \frac{n x^{n-1}}{\theta^n}, \quad 0 \leq x \leq \theta$$

The expected value is: $$E[X_{(n)}] = \int_0^\theta x \cdot \frac{n x^{n-1}}{\theta^n} dx = \frac{n}{\theta^n} \int_0^\theta x^n dx = \frac{n}{\theta^n} \cdot \frac{\theta^{n+1}}{n+1} = \frac{n\theta}{n+1}$$

Important: As $n \to \infty$, $X_{(n)} \xrightarrow{P} \theta$ (converges in probability to $\theta$).

Option (A): $8X_{(n)}^3$

Since $X_{(n)} \xrightarrow{P} \theta$, by the continuous mapping theorem: $$X_{(n)}^3 \xrightarrow{P} \theta^3$$

Therefore: $$8X_{(n)}^3 \xrightarrow{P} 8\theta^3 \neq \theta^3$$

Option (A) is NOT consistent 

Option (B): $X_{(n)}^3$

Since $X_{(n)} \xrightarrow{P} \theta$, by the continuous mapping theorem: $$X_{(n)}^3 \xrightarrow{P} \theta^3$$

Option (B) is consistent 

Option (C): $\left(\frac{2}{n}\sum_{i=5}^n X_i\right)^3$

First, note that $E[X_i] = \frac{\theta}{2}$ for $X_i \sim U(0, \theta)$.

By the Law of Large Numbers: $$\frac{1}{n}\sum_{i=5}^n X_i = \frac{1}{n}\sum_{i=1}^n X_i - \frac{1}{n}\sum_{i=1}^4 X_i$$

As $n \to \infty$: $$\frac{1}{n}\sum_{i=1}^n X_i \xrightarrow{P} \frac{\theta}{2}$$

and $\frac{1}{n}\sum_{i=1}^4 X_i \to 0$ (finite sum divided by $n$).

Therefore: $$\frac{1}{n}\sum_{i=5}^n X_i \xrightarrow{P} \frac{\theta}{2}$$

By continuous mapping: $$\left(\frac{2}{n}\sum_{i=5}^n X_i\right)^3 = \left(2 \cdot \frac{1}{n}\sum_{i=5}^n X_i\right)^3 \xrightarrow{P} \left(2 \cdot \frac{\theta}{2}\right)^3 = \theta^3$$

Option (C) is consistent 

Option (D): $\frac{n X_{(n)}^3 + 1}{n+1}$

We can rewrite this as: $$\frac{n X_{(n)}^3 + 1}{n+1} = \frac{n}{n+1} X_{(n)}^3 + \frac{1}{n+1}$$

As $n \to \infty$:

• $\frac{n}{n+1} \to 1$
• $\frac{1}{n+1} \to 0$
• $X_{(n)}^3 \xrightarrow{P} \theta^3$

Therefore: $$\frac{n X_{(n)}^3 + 1}{n+1} \xrightarrow{P} 1 \cdot \theta^3 + 0 = \theta^3$$

Option (D) is consistent 

Answer: (B), (C), and (D)