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# Thread: method of maximum likelihood

1. ## method of maximum likelihood

HI, i'm struggling with a question on maximum likelihood and lookin for help here it is:

Let x1, x2, . . . , xn be an iid sample from a probability density with
parameter θ.
(a) Show that the value that maximizes the log-likelihood also maxi-
mizes the likelihood. 
Hint: This does not require extensive calculations. Remember
that the log() function is increasing

(b) Suppose that ϕ = g(θ) where g is a one-to-one function that is
di erentiable. Show that that the maximum likelihood estimate
for ϕ is g(^θ).
Hint 1: Look at a speci c example rst. For example, if the data
are Poisson(θ) distributed and ϕ = log(θ). In this case, you can
check directly.
Hint 2: Don't forget about the chain rule for di ffrentiation 2. '''...now thats tough... Did you try to refer any Credibility Theory text book... It might be explained in detail in that book. Gud luck!!:laugh: 3. Don't they tell you the answer to the first pretty much in the hint?

Find your likelihood function, call it L(\theta). It will be a product of f(x_i|\theta). Now take log of it, call it l(\theta).

Now fill in the blanks, log of a product = _______ of logs.

Now how do you get MLE? You differentiate next, if you have the fact they're allowing you to use about logs... the rest should follow no? 4. Originally Posted by NoMoreExams Don't they tell you the answer to the first pretty much in the hint?

Find your likelihood function, call it L(\theta). It will be a product of f(x_i|\theta). Now take log of it, call it l(\theta).

Now fill in the blanks, log of a product = _______ of logs.

Now how do you get MLE? You differentiate next, if you have the fact they're allowing you to use about logs... the rest should follow no?
so the answer is basically just the log-likelihood is an increasing function and also a transformation of the original function then it obtains its maximum for the same value of theta??? 5. Originally Posted by ucd90 so the answer is basically just the log-likelihood is an increasing function and also a transformation of the original function then it obtains its maximum for the same value of theta???
Prove to me that whatever maximizes L(.), would also maximize l(.) = ln[L(.)] 6. Originally Posted by NoMoreExams Prove to me that whatever maximizes L(.), would also maximize l(.) = ln[L(.)]
you've got me on this one, the only way I can think of proving it is via a diagram, not quite sure of a more formal approach 7. Originally Posted by ucd90 you've got me on this one, the only way I can think of proving it is via a diagram, not quite sure of a more formal approach
Suppose that value is x, then L(x) > L(y) for all y.

Since ln(.) is an increasing function, ln[L(x)] > ln[l(y)] for all y as well.

That's the basic sketch of the proof, I don't think it's super formal... 8. Originally Posted by NoMoreExams Suppose that value is x, then L(x) > L(y) for all y.

Since ln(.) is an increasing function, ln[L(x)] > ln[l(y)] for all y as well.

That's the basic sketch of the proof, I don't think it's super formal...
yeah putting it like that it's clear now thanks, while i'm at it would you know how to calculate the bias of an MLE for a generalized scale parameter on a gamma distribution? My MLE turned out to be (shape parameter*n)/(sum of the xi) Thanks for any help 9. Originally Posted by ucd90 yeah putting it like that it's clear now thanks, while i'm at it would you know how to calculate the bias of an MLE for a generalized scale parameter on a gamma distribution? My MLE turned out to be (shape parameter*n)/(sum of the xi) Thanks for any help
Well how do you normally calculate bias? 10. Originally Posted by NoMoreExams Well how do you normally calculate bias?
expected value of your estimate - the parameter your estimating, i can't seem to work out a sensible looking answer, getting the expected value of my estimate seems to be the problem There are currently 1 users browsing this thread. (0 members and 1 guests)

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