How do I use the method of moments to find the estimate of #theta# in the pdf #f(x) = {[theta e^(–thetax)",", x >= 0],[0",","otherwise"]:}#?

1 Answer
Nov 13, 2017

Assuming we let #X_i" "(i=1,..., n)# be a sample from this distribution, then #hattheta_(MM)=barX.#

Explanation:

The pdf #f(x)=thetae^(-thetax)# for #x>=0# is that of an exponential random variable. Let #X_i# be iid exponential random variables #(i=1,...,n)#; then the pdf of each #X_i# is the #f(x)# given, with common mean #theta#.

The method of moments tells us to equate theoretical moments to their corresponding sample moments (about the origin), and then solve for the parameter.

For this question, we begin by setting the first theoretical moment #(E[X^1]=theta)# equal to the first sample moment #(M_1=1/n sum_(i=1)^nX_i):#

#E[X]=theta=1/nsum_(i=1)^nX_i=M_1#

or simply

#color(white)(E[X]=)theta=1/nsum_(i=1)^nX_i#

Since there is only one parameter to estimate, we do not need to use any 2nd moments, etc.

The equation above gets solved for #theta#; once that's done, the other side is the method of moments estimator for #theta#, which will be denoted as #hattheta_(MM).#

Looks like the equation is already sovled for #theta#, so our method of moments estimator for #theta# is:

#hattheta_(MM)=1/nsum_(i=1)^nX_i#
#color(white)(hattheta_(MM))=barX#