Problem 6 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

\square\( Consider the following dataset of lifetimes of ball bearings in hours. \begin{tabular}{rrrrrrrrrr} \hline \hline 6278 & 3113 & 5236 & 11584 & 12628 & 7725 & 8604 & 14266 & 6125 & 9350 \\ 3212 & 9003 & 3523 & 12888 & 9460 & 13431 & 17809 & 2812 & 11825 & 2398 \\ \hline \end{tabular} Source: J.E. Angus. Goodness-of-fit tests for exponentiality based on a lossof-memory type functional equation. Joumal of Statistical Planning and Inference, 6:241-251, 1982; example 5 on page \)249 .\( One is interested in estimating the minimum lifetime of this type of ball bearing. The dataset is modeled as a realization of a random sample \)X_{1}, \ldots, X_{n}\(. Each random variable \)X_{i}\( is represented as $$ X_{i}=\delta+Y_{i} $$ where \)Y_{i}\( has an \)\operatorname{Exp}(\lambda)\( distribution and \)\delta>0\( is an unknown parameter that is supposed to model the minimum lifetime. The objective is to construct an unbiased estimator for \)\delta\(. It is known that $$ \mathrm{E}\left[M_{n}\right]=\delta+\frac{1}{n \lambda} \text { and } \mathrm{E}\left[\bar{X}_{n}\right]=\delta+\frac{1}{\lambda}, $$ where \)M_{n}=\( minimum of \)X_{1}, X_{2}, \ldots, X_{n}\( and \)\bar{X}_{n}=\left(X_{1}+X_{2}+\cdots+X_{n}\right) / n\(. a. Check that $$ T=\frac{n}{n-1}\left(\bar{X}_{n}-M_{n}\right) $$ is an unbiased estimator for \)1 / \lambda\(. b. Construct an unbiased estimator for \)\delta\(. c. Use the dataset to compute an estimate for the minimum lifetime \)\delta\(. You may use that the average lifetime of the data is \)8563.5$.

Step-by-Step Solution

Answer

The minimum lifetime estimate \(\delta\) is 2398 hours.

Step 1: Understand the problem

We need to find an unbiased estimator for the minimum lifetime, \(\delta\), based on the given random variable model \(X_i = \delta + Y_i\) where \(Y_i\) follows an exponential distribution. Our task is divided into three parts: verifying a given estimator for \(1/\lambda\), constructing an unbiased estimator for \(\delta\), and then computing this estimator using provided data.

Step 2: Verify the given estimator for \(1/\lambda\)

To check if \(T = \frac{n}{n-1}(\bar{X}_n - M_n)\) is an unbiased estimator for \(1/\lambda\), we calculate \(\mathrm{E}[T]\) using the given expectations:\[ \mathrm{E}[T] = \frac{n}{n-1}(\mathrm{E}[\bar{X}_n] - \mathrm{E}[M_n]) = \frac{n}{n-1}\left(\delta + \frac{1}{\lambda} - (\delta + \frac{1}{n\lambda})\right) = \frac{n}{n-1}\left(\frac{1}{\lambda} - \frac{1}{n\lambda}\right) = \frac{1}{\lambda}. \]Hence, \(T\) is indeed an unbiased estimator for \(1/\lambda\).

Step 3: Construct an unbiased estimator for \(\delta\)

To find an unbiased estimator for \(\delta\), we utilize the difference in expected values. We know \(\mathrm{E}[\bar{X}_n] = \delta + \frac{1}{\lambda}\). Substituting the unbiased estimator for \(1/\lambda\) we get: \[ \hat{\delta} = \bar{X}_n - \frac{n}{n-1}(\bar{X}_n - M_n)M_n = M_n. \]Thus, an unbiased estimator for \(\delta\) is \(M_n\).

Step 4: Compute \(\delta\) using the dataset

From the dataset, the minimum value \(M_n\) is 2398 hours. Hence, the unbiased estimator for the minimum lifetime, \(\delta\), is \(2398\).

Key Concepts

Unbiased Estimator

An estimator is a rule or method used to estimate an unknown parameter. In statistics, an unbiased estimator is a special type of estimator that has an expected value equal to the true value of the parameter it estimates. This means it gives us estimates on average that are correct.

Unbiased estimators are important in statistics because they ensure that we're neither systematically overestimating nor underestimating the parameter. For example, if we have a random variable modeled by \(X_i = \delta + Y_i\) and \(Y_i\) follows an exponential distribution, our goal is to find an estimator \(\hat{\delta}\) for \(\delta\), the minimum lifetime, such that the expected value of \(\hat{\delta}\) is exactly \(\delta\).

In the original solution, the estimator \(M_n\) (the minimum value from the dataset) is shown to be unbiased for \(\delta\). This means that over repeated samples, \(M_n\) provides estimates for \(\delta\) that average out to the true value.

Exponential Distribution

The exponential distribution is a probability distribution that describes the time between events in a Poisson process. It is often used for modeling the waiting times between independent events that happen at a constant average rate.

A key property of the exponential distribution is its memoryless property. This means that the probability of an event occurring in the future is independent of how much time has already elapsed. In the context of the problem, \(Y_i\) follows an \(\text{Exp}(\lambda)\) distribution, capturing the randomness in the lifetime of the ball bearings beyond their minimum lifetime \(\delta\).

The exponential distribution has a parameter \(\lambda\), which is its rate parameter. This rate determines how rapidly the events occur. The mean of an exponential distribution is \(1/\lambda\), making calculations convenient since they can relate directly to estimates of the parameter, as seen in the solution where \(T\) is an unbiased estimator for \(1/\lambda\).

Goodness-of-Fit Tests

Goodness-of-fit tests are statistical tests that determine how well a sample of data matches the expectations of a specified distribution. These tests are crucial in validating whether the chosen model is appropriate for the data.

In our case, goodness-of-fit tests could be employed to verify that the exponential distribution is a good model for the data of lifetimes of ball bearings. When the model fits well, the resulting estimations, like those for \(\delta\) or \(\lambda\), are more reliable.

Common goodness-of-fit tests include the Chi-square test, Kolmogorov-Smirnov test, and Anderson-Darling test. Using such tests not only assists in confirming the model's appropriateness but also strengthens the robustness of our predictions, supporting decisions based on the tailored statistical model.