Question

R1. Now that you've seen both MME and MLE, we might begin comparing the two worlds. In class, we studied X Uni (0.0) and showed the MLE is aMLE = max Xi One can show the MME for this setup is θΜΜΕ 2 X. As seen in HW2, these estimators are RVs, and each will have its own (sampling) distribution. The sampling distribution gives a good sense of what types of values you'll get from θ when you draw a random sample. Use the internet to learn how to create overlapping histograms in R. Then, on the same set of axes, make a red histogram that shows 10000 values of θΜΜΕ based on 10000 random samples of size 40, Next, overlay a green histogram that shows 10000 values of eMLE based on 10000 new random samples of size 40. Draw from the Unif (0,6) distribution Include your code and a sketch of the overlapping histograms.

Answer 1

R-code( Debugged with comments included)

#Random numbers
tmme=c()
tmle=c()

#Generating the Method of Moments estimator and Maximum Likelihood Estimator

for (i in 1:10000)
{
s<-runif(40,min = 0,max=6)
tmme[i]=2*mean(s)
tmle[i]=max(s)
}

# Histogram Colored (red and green)

hist(tmme, col="red",main="Overlapping Histogram", xlab="Estimates of theta")
hist(tmle, col="green",xlim=c(0,10),add=T)
legend("topright",c("MLE","MME"),col=c("green","red"),lty=c(1,1))
box()