Maximum Likelihood Estimator + Prior

[Scootertaj]

1.Suppose that X~B(1,∏). We sample n times and find n₁ ones and n₂=n-n₁zeros
a) What is ML estimator of ∏?
b) What is the ML estimator of ∏ given 1/2≤∏≤1?
c) What is the probability ∏ is greater than 1/2?
d) Find the Bayesian estimator of ∏ under quadratic loss with this prior
2. The attempt at a solution
a) L=\pi^n_1 *(1-\pi)^{n_2}
Do \frac{d(logL)}{d\pi} = \frac{n_1}{\pi} - \frac{n_2}{1-\pi} → \pi_{ML} = \frac{n_1}{n}
b) not sure how to go about
c) not sure
d) I think I know how.

 

[Ray Vickson]

1.Suppose that X~B(1,∏). We sample n times and find n₁ ones and n₂=n-n₁zeros
a) What is ML estimator of ∏?
b) What is the ML estimator of ∏ given 1/2≤∏≤1?
c) What is the probability ∏ is greater than 1/2?
d) Find the Bayesian estimator of ∏ under quadratic loss with this prior



2. The attempt at a solution
a) L=\pi^n_1 *(1-\pi)^{n_2}
Do \frac{d(logL)}{d\pi} = \frac{n_1}{\pi} - \frac{n_2}{1-\pi} → \pi_{ML} = \frac{n_1}{n}
b) not sure how to go about
c) not sure
d) I think I know how.

In (b) you are asked to maximize L (or log L) subject to 1/2 ≤ π ≤ 1. Your solution to )a)_ may, or may not work in this case. When does it work? When does it fail? If it fails, what then must be the solution?

RGV

 

[Scootertaj]

What do you mean fail?
Intuitively, \pi_{ML}=\frac{n_1}{n} would "fail" in the case that it is \frac{n_1}{n} < 1/2
But, I'm not sure what our solution must be then if it fails.

 

[Ray Vickson]

What do you mean fail?
Intuitively, \pi_{ML}=\frac{n_1}{n} would "fail" in the case that it is \frac{n_1}{n} < 1/2
But, I'm not sure what our solution must be then if it fails.

"Fail" = does not succeed = is wrong = does not work. When that is the case, something must have happened; what was that? What does that tell you about the behaviour of L(π)? (Hint: draw a hypothetical graph.)

RGV

 

[Scootertaj]

Well, based off the graph of \pi^{n_1}(1-\pi)^{n_2} with several different n1 and n2 values plugged in that the best choice would be \pi=n1/n when 1/2≤n1/n≤1, else we choose \pi=1/2 since we usually look at the corner points (1/2 and 1)