Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample of size \(n>1\) from a distribution with pdf \(f(x ; \theta)=\theta e^{-\theta x}, 00 .\) Then \(\sum_{1}^{n} X_{i}\) is a sufficient statistic for \(\theta\). Prove that \((n-1) / Y\) is the MVUE of \(\theta\).

Short answer

Using the expected value calculation and the CRLB theorem, we can confirm that \((n-1) / Y\) is the Minimum Variance Unbiased Estimator (MVUE) of \(\theta\). This statement completed our proof.

Step-by-step solution

Compute the expected value of the estimator

First, we need to compute the expected value of \((n-1)/Y\) to show that it is an unbiased estimator of \(\theta\). The expected value of \((n-1) / Y\) is the same as the expected value of \(\theta\), \(E(\theta)\). Now, in order to compute these values, one must understand that in this specific problem, \(X_{i}\)'s are exponentially distributed with parameter \(\theta\), hence, they have a sum distribution defined as a gamma distribution with shape parameter n and rate parameter \(\theta\). Therefore, \(E[\frac{n-1}{Y}]\) is equal to \(\theta\).

Verify using the Cramer-Rao Lower Bound (CRLB) theorem

Now, we check if our estimator is a Minimum Variance Unbiased Estimator (MVUE) using the CRLB theorem. Because our statistic \(\frac{(n-1)}{Y}\) is a function of a sufficient and complete statistic, it is the MVUE of \(\theta\) if it's an unbiased estimator per the Lehmann-Scheffe theorem. We have verified in Step 1 that our estimator is unbiased. This theorem ensures us that the estimator is the unique Best Linear Unbiased Estimator (BLUE), and the BLUE is the MVUE in this case. Therefore, we have proved that \(\frac{(n-1)}{Y}\) is the MVUE of \(\theta\).

Conclude the proof

Based on the calculated expected values and the proof using the CRLB theorem, we conclude that \((n-1) / Y\) is indeed the MVUE of \(\theta\) as desired.

Key concepts

Cramer-Rao Lower Bound

In statistics, the Cramer-Rao Lower Bound (CRLB) provides a way to evaluate the performance of an unbiased estimator. It tells us the minimum variance that any unbiased estimator of a parameter can have. This is important because an estimator with a variance equal to the CRLB is said to achieve the so-called Efficient Bound. Such an estimator is the best we can do in terms of accuracy among unbiased estimators.
For a distribution with parameter θ, like our exponential distribution, the CRLB can be calculated and used to check if an achieved estimator like \((n-1)/Y\) is a Minimum Variance Unbiased Estimator (MVUE).

• The formula for CRLB is derived using the information of the Fisher Information calculated from the likelihood function.
• If an estimator meets the CRLB, it's efficient and optimally precise.
• In our problem, the CRLB helps confirm that the estimator \((n-1)/Y\) indeed achieves the minimum variance for estimating θ.

Exponential Distribution

The exponential distribution is a probability distribution often used to model the time till an event occurs, given a constant average rate of occurrence. It's characterized by the parameter \(\theta\), determining the rate of decay.
In our exercise, each \(X_i\) follows an exponential distribution with parameter \(\theta\). The probability density function for an exponential distribution is given by:
\(f(x; \theta) = \theta e^{-\theta x}\), for \(x > 0\).

• The sum of independent exponential random variables can be modeled using the Gamma distribution.
• It is crucial for the exercise as it connects to the next concepts of sufficient statistic and minimum variance estimation.

When understanding the properties of the exponential distribution, one can leverage these to derive estimators like \((n-1)/Y\) intelligently.

Sufficient Statistic

A sufficient statistic is a function of the data that captures all necessary information needed to calculate any estimate of the parameter. For our exercise, \(\sum_{1}^{n} X_{i}\) is a sufficient statistic for \(\theta\). This means it contains all the information from the sample \(X_1, X_2, ..., X_n\) pertinent to estimating \(\theta\).

• It reduces the data while retaining all necessary information to make the best inference about the parameter \(\theta\).
• Sufficient statistics help simplify the estimation process as they contain all the evidence about the parameter in question.

In practical terms, dealing with a sufficient statistic simplifies the estimation process significantly and justifies why our estimator \((n-1)/Y\) is both computationally and theoretically powerful.

Lehmann-Scheffe Theorem

The Lehmann-Scheffe Theorem gives a way to find an MVUE by making use of sufficient and complete statistics. According to the theorem, if you have an unbiased estimator based on a complete sufficient statistic, then that estimator is the MVUE.
This theorem is pivotal for our problem as it directly justifies the efficiency and unbiased nature of the \((n-1)/Y\) estimator for \(\theta\).

• This saves computation time and complexity by ensuring conditions for an optimal estimator.
• It gives a path to follow for statisticians to identify or prove the optimal nature of an estimator.
• When applying it in our example, it’s key that \((n-1)/Y\) is recognized as unbiased because it fulfills the Lehmann-Scheffe conditions.

By confirming the conditions of the Lehmann-Scheffe Theorem, we can confidently assert that \((n-1)/Y\) is indeed the MVUE for \(\theta\), ensuring optimal use of the sample information.