Section 6.5 (p.346)

 

2.         The likelihood function for the given sample is p58 (1-p)12. Among all values of p in the interval 1/2 £ p £ 2/3, this function has a maximum when p = 2/3. Hence MLE is 2/3.

 

5.         Let q = s2. Then the likelihood function is

*  

*   If we let L(q) = log fn(x|q), then

*

*The maximum of L(q) will be attained at a value of q for which this derivative is equal to 0. Therefore,

*

7.         The likelihood function is

(a) For any given x. fn(x|q) will be a maximum when q is made as large as possible subject to the strict inequality q < min{x1, …, xn}. But there is no maximum q for the given constraint, so MLE doesn’t exist.

(b) Simple, just construct the constraint on q so that its maximum can be attained. That is

In this case the likelihood has constraint q £ min{x1, …, xn} and the MLE is min{x1, …, xn}.

 

8.         The likelihood function is

Let L(q) = log fn(x|q), then

Let the above equal to 0 and solve it, we get MLE

10.       The likelihood function is

fn is maximized if (q2 - q1) is minimized. Therefore the MLE of q1 is the largest possible value min{x1, …,xn} and the MLE of q2 is the smallest possible value max{x1, …,xn}.