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Chapter 5: Problem 8

Determine a method to generate random observations for the Cauchy distribution with pdf $$ f(x)=\frac{1}{\pi\left(1+x^{2}\right)}, \quad-\infty

Short Answer

Expert verified
The given method to generate random observations for the Cauchy distribution is to use the relationship between a uniform random variable and a Cauchy random variable. A function for this in R might look like: rcauchy_sample <- function(n) { u <- runif(n); x <- tan(pi * (u - 0.5)); return(x) }; where 'n' is the sample size.

Step by step solution

01

Generation Method

The first step is to devise a method to generate random observations for the Cauchy distribution. The way to achieve this is to compute the random observations through the relationship between a uniform distribution and the Cauchy distribution; specifically, if \(U\) is a random variable uniformly distributed on \( (0,1) \), then \( X = tan( \pi (U - 1/2) ) \) is a random variable with the standard Cauchy distribution.
02

Writing the R function

Based on the generation method described in Step 1, an R function can be written to create random sample observations from a Cauchy distribution. Keep in mind that the inverse tangent function in R is atan, and the pi constant is simply pi. The function might be something like this: \n\n rcauchy_sample <- function(n) { \n u <- runif(n) \n x <- tan(pi * (u - 0.5)) \n return(x) \n }
03

Explanation

In the proposed function, 'n' indicates the size of the sample you desire. The function first generates 'n' random variables 'u' from a uniform distribution on the interval (0,1) using the runif function in R. The function then calculates 'x', a random variable with the standard Cauchy distribution, using the computed 'u'. Then it returns 'x'. So, by calling rcauchy_sample(n), you can generate 'n' observations from a standard Cauchy distribution.

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Most popular questions from this chapter

Let \(\bar{X}\) be the mean of a random sample from the exponential distribution, \(\operatorname{Exp}(\theta)\) (a) Show that \(\bar{X}\) is an unbiased point estimator of \(\theta\). (b) Using the mgf technique determine the distribution of \(\bar{X}\). (c) Use (b) to show that \(Y=2 n \bar{X} / \theta\) has a \(\chi^{2}\) distribution with \(2 n\) degrees of freedom. (d) Based on Part (c), find a \(95 \%\) confidence interval for \(\theta\) if \(n=10 .\) Hint: Find \(c\) and \(d\) such that \(P\left(c<\frac{2 n \bar{X}}{\theta}

Proceeding similar to Example 5.8.6, use the Accept-Reject Algorithin to generate an observation from a \(t\) distribution with \(r>1\) degrees of freedom ussing the Cauchy distribution.

Compute \(P\left(Y_{3}<\xi_{0.5}

Let a random sample of size 17 from the normal distribution \(N\left(\mu, \sigma^{2}\right)\) yield \(\bar{x}=4.7\) and \(s^{2}=5.76 .\) Determine a 90 percent confidence interval for \(\mu .\)

. Let \(Y_{1}
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