Statistical Inference Assignment

Part 1

The exponential distribution describes the time between events in a process where such events occur independently and at a constant average rate (a “Poisson process”). Lambda, the single parameter of the distribution, defines the average rate of the events. The mean and variance of this distribution are both equal to 1/lambda.

In R, we can obtain n samples from an exponential distribution with the following code:

rexp(5,0.2)

## [1] 1.0156533 0.4406902 8.5772410 1.5403781 2.5494587

For a larger we need tidyverse too

library(tidyverse)
large <- rexp(100000,0.55)
summary(large)

##      Min.   1st Qu.    Median      Mean   3rd Qu.      Max. 
##  0.000008  0.523749  1.264569  1.824301  2.540215 18.985635

sd(large)

## [1] 1.819019

qplot(large,col="black", xlab = "large", ylab = "count")

In this exercise, we investigate the Central Limit Theorem by simulating 1000 independent exponential distributions (with identical lambda) and calculating their mean. These 1000 means form a distribution themselves. By the CLT, as the sample size of this new distribution increases, the following things should happen:

• The distribution becomes more Normal
• The mean of this new distribution approaches that of the individual distributions: (1/lambda)
• The standard deviation should approach 1/lambda/sqrt(n)

n <- 40
lambda <- 0.2
mean_expected <- 1/lambda
sd_expected <- 1/(lambda*sqrt(n))
c(mean_expected, sd_expected)

## [1] 5.0000000 0.7905694

means <- NULL
for(i in 1:1000) means = c(means,mean(rexp(40,0.2)))
qplot(means, col="black",xlab="means", ylab="count")

The mean seems to be around 5, lets check it

round(c(mean_expected, mean(means)),2)

## [1] 5.00 5.03

round(c(sd_expected, sd(means)),2)

## [1] 0.79 0.77

After looking at those close numbers we conclude that CLT works fine.