Maximum Likelihood Estimator + Prior

  • #1
Scootertaj
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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) [tex]L=\pi^n_1 *(1-\pi)^{n_2}[/tex]
Do [tex]\frac{d(logL)}{d\pi} = \frac{n_1}{\pi} - \frac{n_2}{1-\pi}[/tex] → [tex]\pi_{ML} = \frac{n_1}{n}[/tex]
b) not sure how to go about
c) not sure
d) I think I know how.
 
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  • #2
Scootertaj said:
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) [tex]L=\pi^n_1 *(1-\pi)^{n_2}[/tex]
Do [tex]\frac{d(logL)}{d\pi} = \frac{n_1}{\pi} - \frac{n_2}{1-\pi}[/tex] → [tex]\pi_{ML} = \frac{n_1}{n}[/tex]
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
 
  • #3
What do you mean fail?
Intuitively, [tex]\pi_{ML}=\frac{n_1}{n}[/tex] would "fail" in the case that it is [tex]\frac{n_1}{n} < 1/2[/tex]
But, I'm not sure what our solution must be then if it fails.
 
  • #4
Scootertaj said:
What do you mean fail?
Intuitively, [tex]\pi_{ML}=\frac{n_1}{n}[/tex] would "fail" in the case that it is [tex]\frac{n_1}{n} < 1/2[/tex]
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
 
  • #5
Well, based off the graph of [tex]\pi^{n_1}(1-\pi)^{n_2}[/tex] with several different n1 and n2 values plugged in that the best choice would be [tex]\pi=n1/n[/tex] when [tex]1/2≤n1/n≤1[/tex], else we choose [tex]\pi=1/2[/tex] since we usually look at the corner points (1/2 and 1)
 

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