# Thread: How do I find the moment estimator and maximum likelihood from a denstiy function?

1. ## How do I find the moment estimator and maximum likelihood from a denstiy function?

Hi Guys,

Any help will be great.

Let
X1;X2; .......;Xn be a random sample from a distribution which has probability density function:

f
(x; µ) = 2(µ^2)/[x^3]

with
µ <= x < infinity;

Find both the moment estimator and the maximum likelihood estimator of the parameter
µ.

Thanks

2. The likelihood function is $\displaystyle {2^nu^{2n}\over \Pi_{i=1}^nx_i^3}I(X_{(1)}\ge u)$

Where that $\displaystyle X_{(1)}$ is the first order stat.
YOU do not differentiate this. You want the max wrt u.
The max occurs when u is as large as possible since u is in the numerator.
But the largest this can be is $\displaystyle X_{(1)}$, otherwise you have ZERO for that indicator.
HENCE the MLE of u is $\displaystyle X_{(1)}$.

AS for the MOM, I get 2u as the E(X), so get $\displaystyle \bar X=\hat E(X)$

or $\displaystyle \bar X=2\hat u$ or $\displaystyle \hat u=\bar X/2$.

3. Thanks, do you mind explaining how you got the likelihood function?

Also for the MOM how did you arrive at 2u for E(X)?

Cheers.

4. Originally Posted by matheagle
The likelihood function is $\displaystyle {2^nu^{2n}\over \Pi_{i=1}^nx_i^3}I(X_{(1)}\ge u)$

Where that $\displaystyle X_{(1)}$ is the first order stat.
YOU do not differentiate this. You want the max wrt u.
The max occurs when u is as large as possible since u is in the numerator.
But the largest this can be is $\displaystyle X_{(1)}$, otherwise you have ZERO for that indicator.
HENCE the MLE of u is $\displaystyle X_{(1)}$.

AS for the MOM, I get 2u as the E(X), so get $\displaystyle \bar X=\hat E(X)$

or $\displaystyle \bar X=2\hat u$ or $\displaystyle \hat u=\bar X/2$.
heyz, i was wondering how did u manage to get the second part of the likelihood function. i could somehow work out n i=1 2^n theta^ 2n over xi^3

5. Originally Posted by downdown
heyz, i was wondering how did u manage to get the second part of the likelihood function. i could somehow work out n i=1 2^n theta^ 2n over xi^3

You need all the x's to exceed u.
That's the same as the smallest of the data to exceed u.
Hence the smallest order stat is an important statistic for this parameter.