This problem uses computer simulations to demonstrate that a

This problem uses computer simulations to demonstrate that a) samples from this distribution often have extreme outliers (a consequence of the heavy tails of the distribution), and b) the sample mean is prone to the same type of outliers. Below is a graph of the pdf The K commands x-raunchy(500); summary(x) generate a random sample of size 500 from the Cauchy distribution and display the sample\'s five number summary; Report the five number summary and the interquartile range, and comment on whether or not the smallest and largest numbers generated from this sample of 500 are outliers. Repeat this 10 times. The R commands m matrix(raunchy(50000). Nrow = 500); xb = apply(m, 1. mean), summary (xb) generate the matrix m that has 500 rows, each of which is a sample of size n-100 from the Cauchy distribution, compute the 500 sample means and store them in xb, and display the five number summary xb. Repeat these commands 10 times, and report the 10 sets of five number summaries. Compare with the 10 sets of five number summaries from part (a), and comment on whether or not the distribution of the averages seems to be more prone to extreme outliers as that of the individual observations. Why does this happen? and does the LLN and CLT apply for samples from a Cauchy distribution?

Solution

a) the R-code and answer

> x=rcauchy(500)
> summary(x)
Min. 1st Qu. Median Mean 3rd Qu. Max.
-144.600 -0.983 0.110 13.270 0.983 6260.000
> IQR(x)
[1] 1.965498

this sample shows that a huge number of outlier as it takes sample value as 6260

b)

sample  1
Min. 1st Qu. Median Mean 3rd Qu. Max.
-4435.000 -0.852 0.030 -6.193 1.162 971.100
sample  2
Min. 1st Qu. Median Mean 3rd Qu. Max.
-1304.0000 -1.2700 -0.0504 -2.4620 0.8316 248.8000
sample 3
Min. 1st Qu. Median Mean 3rd Qu. Max.
-12130.000 -1.025 0.122 -25.030 0.991 79.390
sample 4
Min. 1st Qu. Median Mean 3rd Qu. Max.
-122.70000 -0.88870 -0.09102 -0.05400 0.94570 115.20000
sample 5
Min. 1st Qu. Median Mean 3rd Qu. Max.
-205.7000 -0.8019 0.0568 1.2690 1.0570 485.8000

in part (a) the sample value is 6260 but in this example we get maximum 971 so we can conclude that there is high chance of outlier in cauchy distribution. and also it it centrally( i.e. by mean and median) robast

c) if we notice cauchy distribution very carefully then we will observe that cauchy distribution is looks like Normal , but have heavy tail in this distribution compared to normal. so it takes so many outliers and it is centrally robustd

so here the property of central tendency fails

E(X) does`t exists and hence any higher order moment will exists so variance does`t exists. CLT is also not applicable here as E(X) and Var(X) is not exists