DATA 605 - Discussion 8 - 8.2 Exercise 16
Jim Mundy
Let Z = X/Y where X and Y have normal densities with mean 0 and standard deviation 1. Then it can be shown that Z has a Cauchy density.
- Write a program to illustrate this result by plotting a bar graph of 1000 samples obtained by forming the ratio of two standard normal outcomes. Compare your bar graph with the graph of the Cauchy density. Depending upon which computer language you use, you may or may not need to tell the computer how to simulate a normal random variable. A method for doing this was described in Section 5.2.
Here is the histogram of a normal distribution x/y ratio.
hist(rnorm(1000)/rnorm(1000), breaks=10000, xlim=c(-10,10),
xlab="", ylab="", col = 'blue',
main="Histogram of Normal X/Y Distribution")
Here is the Cauchy Distribution histogram.
hist(rcauchy(1000), breaks=10000, xlim=c(-10,10),
xlab="", ylab="", col = 'blue',
main="Histogram of Cauchy X/Y Distribution")
The two histograms are very similar.
- We have seen that the Law of Large Numbers does not apply to the Cauchy density. Simulate a large number of experiments with Cauchy density and compute the average of your results. Do these averages seem to be approaching a limit? If so can you explain why this might be?
Now I run a simulation 5,000 times with each iteration producing 1,000 outcomes. I calculate the average for all iterations. Per the law of large numbers I would expect the average to converge to zero.
# Initialize vectors
trials <- 5000
dist_cauchy <- rep(0,trials)
avg_cauchy <- rep(0,trials)
dist_norm <- rep(0,trials)
avg_norm <- rep(0,trials)
# Simulated n trials and record average as number of trials increases
for (n in 1:trials) {
dist_cauchy[n]<-mean(rnorm(1000)/rnorm(1000))
dist_norm[n]<-mean(rnorm(1000))
avg_cauchy[n]<-mean(dist_cauchy[1:n])
avg_norm[n]<-mean(dist_norm[1:n])
}The normal distribution converges to zero, as expected.
The Cauchy distribution does not converge to zero and appears to be disjointed and jumps around.
