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Assignment 3
The second part of this course focuses on interval estimation. Actually understanding confidence intervals at this level helps to explain why your STAT 200 teacher had troubles giving a good definition of a confidence interval. It looked as though there should be a probability definition to it, but your prof kept using the term “confident.” So confusing.
This section shows where probability enters the calculations for these intervals. As such, this assignment focuses on calculating — and better understanding — those confidence intervals. In this assignment, you will work with a well known distribution and be introduced to a new one, the Rayleigh distribution.
Again, do not use Mathematica; I want you to immerse yourself in the mathematics to see if you can make connections amongst the parts.
Part A: The Exponential Distribution
The Exponential distribution is the default distribution for random variables that measure the (continuous) time until an event happens. It assumes a constant rate of occurrence, which it infrequently true outside physics. It also models the interarrival time for a Poisson process.
Let \(X_1, X_2, X_3, \ldots, X_n\) be a random sample from an Exponential distribution with
rate \(\lambda\); that is, let \(X_i \stackrel{iid}{\sim} EXP(\lambda)\) for \(i = \{1, 2, \ldots, n\}\). Recall that the probability density function for an Exponentially distributed random variable is defined as
\[
f(x;\ \pi) = \left\{\hspace{1em} \begin{array}{ll} \lambda e^{-\lambda x} & 0 \le x \\[1em] 0 & \text{otherwise} \end{array} \right.
\]
Problem 1: Sufficiency
Find a sufficient estimator for the parameter \(\lambda\).
Problem 2: Pivotal Quantity
Find a pivotal quantity that can be used to obtain a confidence interval for \(\lambda\) using all of the data.
Problem 3: Endpoints
Use that pivotal quantity to calculate the endpoints of an exact 95% confidence interval for \(\lambda\).
Problem 4: Check Your Work
Use simulation to check that you actually did obtain a 95% confidence interval for \(\lambda\). To help, here is the code to generate a random sample of size \(n = 10\) from a \(EXP(\lambda = 0.125)\) distribution:
x = rexp(10, rate=0.125)
I am sure you will check more than just this specific instance.
Part B: The Rayleigh Distribution
The Rayleigh distribution is used to model magnitudes of two-dimensional vectors— including complex numbers. Specifically, if \(P\) and \(Q\) are iid Normally distributed random variables with zero mean, then \(\sqrt{P^2 + Q^2}\) follows a Rayleigh distribution. It is very handy in distribution of wind speeds and in complex statistics [the latter being very near and dear to my heart].
Let \(X_1, X_2, \ldots, X_n\) be a random sample from a Rayleigh distribution. The probability density function of a Rayleigh-distributed random variable is
\[
f(x;\ \theta) = \left\{\hspace{1em} \begin{array}{ll} \frac{x}{\theta^2}\ e^{-x^2/(2\theta^2)} & 0 \le x \\[1em] 0 & \text{otherwise} \end{array} \right.
\]
Problem 5: Sufficiency
Show that \(U = \sum X_i^2\) is sufficient for \(\theta\).
Problem 6: An Easy Pivotal Quantity
Let us also know, without proof, that
\[
\frac{X_i^2}{\theta^2} \sim \chi^2_{\nu=2}
\]
if \(X_i\) comes from a Rayleigh distribution. That is, the random variable \( X_i^2 / \theta^2 \) follows a chi-square distribution with two degrees of freedom.
Find a pivotal quantity that can be used to obtain a confidence interval for \(\theta\) using all of the data.
Problem 7: Endpoints
Use that pivotal quantity to calculate the endpoints of an exact 95% confidence
interval for \(\theta\).
Problem 8: Check Your Work
Use simulation to check that you actually did obtain a 95% confidence interval for \(\theta\). To help, here is the code to generate a random sample of size \(n = 10\) from a Rayleigh\((\theta = 2)\) distribution:
library(VGAM)
x = rrayleigh(10, scale=2)
Again, I am sure you will check more than just this specific instance.
This page was last modified on 21 April 2025.