# Maximum Likehood Estimator

• Feb 6th 2009, 06:12 PM
eigenvector11
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.
• Feb 7th 2009, 02:24 AM
mr fantastic
Quote:

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
• Feb 7th 2009, 02:27 AM
mr fantastic
Quote:

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 $\displaystyle E(Y) = \frac{k \theta}{\theta - 1}$ and so let $\displaystyle \frac{k \theta}{\theta - 1} = \overline{y}$ and solve for $\displaystyle \theta$.
• Feb 7th 2009, 11:53 AM
eigenvector11
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.
• Feb 7th 2009, 12:40 PM
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?
• Feb 7th 2009, 12:57 PM
mr fantastic
Quote:

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?

$\displaystyle \overline{y}$ is the sample mean. $\displaystyle E(Y)$ is NOT the same as $\displaystyle \overline{y}$. I had thought this would be clear. So you get $\displaystyle \theta$ in terms of the sample eman.

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