Title: “Statistical Inference - Simulation Course Project Part 1”

Author: “Dharmpal Sharma”

Date: “Sunday, September 21, 2014”

output: html_document keep_md: yes

Illustrate via simulation and associated explanatory text the properties of the exponential distribution.

The mean of exponential distribution is 1/lambda and the standard deviation is also also 1/lambda.

Set lambda = 0.2 for all of the simulations.

Initialization:

nosim   <- 1000      # number of simulations
n      <- 40        # sample size of 40 as requested
lambda <- 0.2
mu     <- 1/lambda  # mu theoretical population mean = 5
s      <- mu        # s theoretical population standard deviation = 5
SE     <- s/sqrt(n) # SE is the theoretical standard error 
set.seed(7890)

The simulated samples are stored in a matrix, where each row contains one sample of 40 random exponential variables.
The mean and standard-deviation were calculated for each sample, and stored as vectors - meanx, and sdx :

x<-matrix(rexp(n*nosim,lambda),nosim) 
meanx<-apply(x, 1, mean)              #means of all samples        
sdx<-apply(x, 1, sd)                  #std-deviations of all samples
  1. Showing where the distribution is centered at, and comparing it to the theoretical center of the distribution:

The average mean of all the samples simulated is 4.997 compared with the theoretical mean of the exponential distribution, which is \( \mu=1/\lambda=5 \)

  1. Show how variable it is and compare it to the theoretical variance of the distribution.

The variance of all the samples simulated means is 0.6176 compared with the theoretical variance of the exponential distribution, for sample-size of n, which is \( s^2=\sigma^2/n=(1/\lambda)^2/n=5^2/40=0.625 \)

  1. Show that the distribution is approximately normal.

With the histogram of the means simulated, compared to the density function curve of normal distribution of \( \mu=1/\lambda \) and \( \sigma=1/\lambda/\sqrt{n} \) and the Q-Q plot of the sample means compared to random normal sample with the same expected parameters, we confirm the normalized distribution. 

par (mfrow = c(1,2), mar=c(5,2,4,2), mgp=c(2,.7,0), bty="l", tck=-.02)
hist(meanx, breaks=40, col="lightblue", freq=F, main="Histogram meanx vs. normal curve")  
xaxis<-seq(0,max(meanx),length=100) #prepare the x-axis 
lines(xaxis,
      dnorm(xaxis,1/lambda,sd=1/lambda/sqrt(n)),
      type="l", col="red", lwd=2)  #add ref normal curve

qqplot(rnorm(1000,1/lambda,sd=1/lambda/sqrt(n)),meanx, pch=20, cex = .5, col="black", main="Q-Q plot: meanx vs. normal dist.", xlab="normal")
abline(0,1, col = "red",lwd=1,lty=2)

plot of chunk normal_proof_graphs

  1. Evaluation of the coverage of the confidence interval for \( 1/\lambda \) using \( \overline{X}\pm 1.96*s/\sqrt{n} \):

First at all, figure the interval's lower and upper limit for each sample: 

ul<-meanx+sdx*1.96/sqrt(n) #upper limit vector
ll<-meanx-sdx*1.96/sqrt(n) #lower limit vector

And evaluate the good intervals, which contain the mean=5, with my settings, the coverage is 92.5%.