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 θMLE max Xi. One can show the MME for this set up is all -2 . X. As seen in HW2, these estinators 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 θΜΙΕ 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 to get the histograms is below (all statements starting with # are comments)

#set the random seed
set.seed(123)

#set the sample size
n<-40
#set the number of samples
r<-10000
#draw n*r samples from uniform (0,6)
x<-runif(n*r,0,6)
#convert into a matrix of size rxn
x<-matrix(x,nrow=r,ncol=n)
#get r MME estimates of theta, which is 2*mean(x)
thetaMME<-2*apply(x,1,mean)

#draw a new n*r samples from uniform (0,6)
x<-runif(n*r,0,6)
#convert into a matrix of size rxn
x<-matrix(x,nrow=r,ncol=n)
#get r MLE estimates of theta, which is max(x)
thetaMLE<-apply(x,1,max)

#histograms
hist(thetaMME,col="red",ylim=c(0,5000),xlab=expression(hat(theta)),main="Histograms of MME and MLE estimates")
hist(thetaMLE,col="green",add=TRUE)
box()
legend("topleft",c("MME Estimates","MLE Estimates"),col=c("red","green"),lty=c(1,1),lwd=5)

#get this plot

Histograms of MME and MLE estimates -MME Estimates 1-MLE Estimates 4

We know that the true value of $\theta$ is 6.

From the graph we can say the following

$\hat{\theta}_{MME}$

• Strength. The mean of the histogram is around 6. Hence $\hat{\theta}_{MME}$ is an unbiased estimator of $\theta$
• Weakness: The spread of the histogram, (that is variance of the estimator) seems to be high. This means that a randomly selected sample of size 40 may give us an estimate which is far away from the true value of $\theta$.

$\hat{\theta}_{MLE}$

• Strength: The spread of the estimates is lower than the estimates obtained using MME. That means given a sample of size 40, we are more likely to get a value close to the true value of $\theta$
• Weakness: The estimator is biased. The mean of the histogram is under estimated and is less than 6.
  • Of course we know that the unbiased estimator is $\hat{\theta}_{MLE}=\frac{n&plus;1}{n}\times \max X_i$ where n is the sample size