# Math Help - Maximum Likehood Estimator

1. ## Maximum Likehood Estimator

If f_Y(y;theta) = (theta*k^theta)*(1/y)^(theta+1), y>=k, theta>=1,

a) Find the maximum likelihood estimator for theta if information has been collected on a random sample of 25 people.

b) Find the method of moments estimator for theta if information has been collected on a random sample of 25 people.

Assume k is unknown.

Any help would be appreciated.

2. Originally Posted by eigenvector11
If f_Y(y;theta) = (theta*k^theta)*(1/y)^(theta+1), y>=k, theta>=1,

a) Find the maximum likelihood estimator for theta if information has been collected on a random sample of 25 people.

[snip]

Assume k is unknown.

Any help would be appreciated.
For an example of what to do, read this: http://www.mathhelpforum.com/math-he...estimator.html

3. Originally Posted by eigenvector11
If f_Y(y;theta) = (theta*k^theta)*(1/y)^(theta+1), y>=k, theta>=1,

[snip]

b) Find the method of moments estimator for theta if information has been collected on a random sample of 25 people.

Assume k is unknown.

Any help would be appreciated.
The only thought I have at the moment is that $E(Y) = \frac{k \theta}{\theta - 1}$ and so let $\frac{k \theta}{\theta - 1} = \overline{y}$ and solve for $\theta$.

4. For question a), I attached what I am getting so far. Am I on the right track here? I'm not sure where to go now.

Edit merge: Actually I already spotted an error, it should be 25lnk in the derivative..oops.

5. How did you come up with that expected value in question b)? And if you solve for theta in that equation, won't you get theta=(ybar-ybar)/k, which means that theta equals zero?

6. Originally Posted by eigenvector11
How did you come up with that expected value in question b)? And if you solve for theta in that equation, won't you get theta=(ybar-ybar)/k, which means that theta equals zero?
$\overline{y}$ is the sample mean. $E(Y)$ is NOT the same as $\overline{y}$. I had thought this would be clear. So you get $\theta$ in terms of the sample eman.

I found E(Y) using the usual formula: $E(Y) = \int_k^{+\infty} y \, f_Y (y, \theta) \, dy$. Again, this is something that I thought would be clear.