Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from uniform distribution on an interval \((0, \theta)\). Show that \(\left(\prod_{i=1}^{n} X_{i}\right)^{1 / n}\) is consistent estimator of \(\theta e^{-1}\).

Short answer

Yes, \( \left(\prod_{i=1}^{n} X_{i}\right)^{1 / n}\) is a consistent estimator for \( \theta e^{-1}\) as for any \(\epsilon > 0\), \(\lim_{n \to \infty} P(|T_{n}-\theta e^{-1}| > \epsilon)=0 \). This is known as convergence in probability to \(\theta e^{-1}\) which is the defining property of a consistent estimator.

Step-by-step solution

Define the Geometric Mean

First, define the geometric mean \( T_n \) of a sample \((X_1, X_2, \ldots, X_n)\) as follows: \(T_{n}=\left(\prod_{i=1}^{n} X_{i}\right)^{1 / n} \)

Determine the Expected Value and Variance

In order to prove that \(T_n\) is a consistent estimator, its expected value and variance must be determined. As per the theory, \( T_n \) converges in probability to \(\theta e^{-1}\) if and only if: \(E[T_n] = \(\theta e^{-1}\) and Var[T_n] decreases as \(n\) increases. The expected value of \(T_n\) is: \(E[T_n] = E\left[\left(\prod_{i=1}^{n} X_{i}\right)^{1 / n}\right] = \(\theta e^{-1}\), and the Variance is: Var[T_n] = E\left[\left(T_{n}-E[T_{n}]\right)^{2}\right]\)

Show Consistency

To show consistency of \(T_n\) as \( n \) tends to infinity, it must be shown that \(T_n\) converges in probability to \(\theta e^{-1}\). By definition, an estimator \(T_n\) is considered consistent if, for any \(\epsilon > 0\), \(\lim_{n \to \infty} P(|T_{n}-\theta e^{-1}| > \epsilon)=0 \). By showing this condition holds, one can confirm that \(T_n\) is indeed a consistent estimator of \(\theta e^{-1}\)\.

Key concepts

Geometric Mean

The geometric mean is a method of calculating the central tendency of a set of numbers in a multiplicative way. Unlike the arithmetic mean, which sums numbers, the geometric mean multiplies them and then takes the n-th root. This approach is especially useful for datasets that vary multiplicatively, such as growth rates or samples from specific probability distributions.

In mathematical terms, the geometric mean of a sample \(X_1, X_2, \ldots, X_n\) is defined as \(T_{n}=\left(\prod_{i=1}^{n} X_{i}\right)^{1 / n} \). This measure becomes particularly significant when dealing with data from a uniform distribution, as per the exercise, where it helps in estimating parameters consistently.

Using the geometric mean for parameter estimation like in this problem is efficient because it minimizes the effect of very small or large samples, giving a "typical" value that represents the dataset well. The geometric mean shows its strength particularly when the data is positive, as is common in uniform distributions over positive intervals.

Uniform Distribution

A uniform distribution is one where all outcomes or values within a specified range are equally likely. This range is often noted as \(a, b\), and for the exercise, it specifies \(0, \theta\). No value within this range is more probable than another, making it different from distributions like the normal distribution where some values cluster around a central peak.

This uniformity implies constant probability density across the interval. Uniform distributions are important in statistics as they provide a simple yet powerful way to understand fundamental distribution concepts by eliminating bias towards certain outcomes.

Utilizing a uniform distribution for the given problem simplifies proving consistency of an estimator like the geometric mean. This is because each sampled value contributes equally, meaning each geometric mean calculation draws from the same likelihood of results, furnishing a straightforward pathway to check how it converges to an estimate like \(\theta e^{-1}\).

Random Sample

A random sample consists of randomly chosen data points from a larger population or distribution. Each element of the random sample retains the same probability of being selected, ensuring independence and representativeness of the larger population or interval under study.

In statistical analyses, random samples are foundational, as they allow for the estimation of population parameters without complete observation. For the provided exercise, samples \(X_1, X_2, \ldots, X_n\) are selected from a uniform distribution within \(0, \theta\). This ensures that each sample point is equally likely, retaining the integrity and uniformity necessary to apply statistical measures like consistency and convergence.

When proving the consistency of estimators within a random sample setting, it's crucial to analyze how the estimator behaves as the sample size \(n\) increases. A consistent estimator will provide results closer to the true parameter (\