In performing the operations suggested in the last post and simplifying to find L, I got that L=Exp(theta-(xbar*n)) by using properties of products and exponents. (i.e. the product of Exp(x_i) from i to n is the same as saying Exp(sum of x_i from i to n) In order to maximize this function theta...
Fixing the MOM i now have theta = xbar - 1, thank you for catching my error.
I am a little confused on the MLE still. Do you mean that x should be evaluated as (x1, ..., xn)/n (xbar)? This gives theta = xbar. Is it a problem that the MLE and MOM give different estimates?
So for the MLE, The conditions are that x >= theta. Visualizing what this graph would look like, if x = theta, then you have the function equaling a value of 1, if If theta = x-1 you would get e^-1, which would be less that if x = theta, and this continues. So would it be correct to say that...
Homework Statement
f(x;theta)=Exp(-x+theta)
Find parameter estimates for variable 'theta' using maximum likelihood Estimator and Method of Moments.
Homework Equations
Log(x; theta) = Log(Exp(-x + theta)) -- For MLE
Integral from theta to infinity of (x*Exp(-x + theta)) = xbar -- For Method of...