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

Question

Suppose that the dataset \(x_{1}, x_{2}, \ldots, x_{n}\) is a realization of a random sample from an \(\operatorname{Exp}(\lambda)\) distribution with distribution function \(F_{\lambda}\), and that \(\bar{x}_{n}=5\). a. Check that the median of the \(\operatorname{Exp}(\lambda)\) distribution is \(m_{\lambda}=(\ln 2) / \lambda(\) see also Exercise 5.11). b. Suppose we estimate \(\lambda\) by \(1 / \bar{x}_{n} .\) Describe the parametric bootstrap \(\operatorname{sim}-\) ulation for \(\operatorname{Med}\left(X_{1}, X_{2}, \ldots, X_{n}\right)-m_{\lambda}\).

Step-by-Step Solution

Answer

a. The median is \(m_{\lambda} = \frac{\ln 2}{\lambda}\). b. Use \( \hat{\lambda} = 0.2 \) to simulate medians and find differences.

Step 1: Verify Median of Exponential Distribution

The median of the exponential distribution \( \operatorname{Exp}(\lambda) \) is defined as the value \( m_{\lambda} \) such that \( F_{\lambda}(m_{\lambda}) = 0.5 \). The cumulative distribution function is \( F_{\lambda}(x) = 1 - e^{-\lambda x} \). We solve the equation:\[ 1 - e^{-\lambda m_{\lambda}} = 0.5 \]Thus,\[ e^{-\lambda m_{\lambda}} = 0.5 \]Taking the natural logarithm of both sides:\[ -\lambda m_{\lambda} = \ln(0.5) \]Hence, the median is given by:\[ m_{\lambda} = -\frac{\ln(0.5)}{\lambda} = \frac{\ln 2}{\lambda} \], verifying the provided median.

Step 2: Estimate \\lambda Using Sample Mean

Given that \( \bar{x}_{n} = 5 \), we are estimating \( \lambda \) using the formula \( \hat{\lambda} = \frac{1}{\bar{x}_{n}} \). Therefore:\[ \hat{\lambda} = \frac{1}{5} = 0.2 \].

Step 3: Perform Parametric Bootstrap Simulation

To perform a parametric bootstrap simulation, follow these steps:1. Using the estimated \( \hat{\lambda} = 0.2 \), generate a large number of bootstrap samples (e.g., 1000 samples) of size \( n \) from an \( \operatorname{Exp}(0.2) \) distribution.2. For each bootstrap sample, compute the median \( \operatorname{Med}(X_1, X_2, \ldots, X_n) \).3. Calculate the difference \( \operatorname{Med}(X_1, X_2, \ldots, X_n) - m_{\lambda} \) for each sample, where \( m_{\lambda} = \frac{\ln 2}{0.2} \).4. Analyze the distribution of these differences, which provides a bootstrap estimate of the sampling distribution of \( \operatorname{Med}(X_1, X_2, \ldots, X_n) - m_{\lambda} \).

Key Concepts

Parametric Bootstrap

The Parametric Bootstrap is a statistical technique used to estimate the distribution of a statistic by resampling from an assumed population model. It provides a way to assess the variability of sample estimates in situations where analytical solutions are complex or unknown. In the given problem, we assume that our data follows an Exponential Distribution with parameter \( \lambda \).
To perform a parametric bootstrap:

• First, estimate the population parameter(s). Here, \( \lambda \) is estimated as \( \hat{\lambda} = 0.2 \) using the sample mean.
• Next, generate numerous bootstrap samples from the assumed distribution, in this case, \( \text{Exp}(0.2) \).
• Calculate the statistic of interest, such as the median, for each sample.
• Finally, analyze these statistics to understand their variability and construct confidence intervals or hypothesis tests.

This technique helps students comprehend how random variability affects statistical estimates, enhancing intuition around inferential statistics.

Sample Mean

The sample mean is a basic concept in statistics and represents the average value of a sample. It is denoted as \( \bar{x} \) and calculated by summing all observations and dividing by the number of observations. In the context of an Exponential Distribution, the sample mean helps estimate the rate parameter \( \lambda \). If you have a sample \( x_1, x_2, \ldots, x_n \) from an \( \text{Exp}(\lambda) \) distribution, the sample mean is particularly useful.
For an exponential distribution, the expected population mean is \( \frac{1}{\lambda} \). Therefore, we can estimate the rate parameter \( \lambda \) as the reciprocal of the sample mean:

• Calculate the sample mean \( \bar{x}_n \).
• Estimate \( \lambda \) using \( \hat{\lambda} = \frac{1}{\bar{x}_n} \).

In our example, given \( \bar{x}_n = 5 \), we estimate \( \lambda \) to be \( 0.2 \). This connects the data to the theoretical distribution, forming the basis for further analysis like the bootstrap procedure.

Cumulative Distribution Function

The Cumulative Distribution Function (CDF) is a critical concept in understanding probability distributions. For a random variable \( X \), the CDF at a value \( x \) is given by \( F(x) = P(X \leq x) \), representing the probability that the variable is less than or equal to \( x \).
In the context of the Exponential Distribution, the CDF is explicitly expressed as \( F_{\lambda}(x) = 1 - e^{-\lambda x} \). This function provides insight into the properties of the exponential distribution, such as the probability of observing a value less than a particular threshold.
The CDF is used to derive other statistical measures like the median. The median is the value for which the CDF is 0.5, meaning half the observations are below this value. By solving the CDF equation \( F_{\lambda}(m_{\lambda}) = 0.5 \), we find that \( m_{\lambda} = \frac{\ln 2}{\lambda} \).
This application of the CDF shows its utility in determining central tendencies and contributes to understanding the statistical behavior of a dataset based on its distribution.