1. ## maximum likelihood estimator!

tried to solve a problem on MLE of the this type below;

to find the MLE of $\theta$

$f(x;\theta) = \theta x^{\theta-1}$ $0

$L(x;\theta) = \prod_{i=1}^{n}\theta x^{\theta-1}$

if i were to take the In of the above, how will the next step look like? cause am confuse here

thanks

2. ## Re: maximum likelihood estimator!

Originally Posted by lawochekel
tried to solve a problem on MLE of the this type below;

to find the MLE of $\theta$

$f(x;\theta) = \theta x^{\theta-1}$ $0

$L(x_1,...,x_n;\theta) = \prod_{i=1}^{n}\theta x_i^{\theta-1}$

if i were to take the In of the above, how will the next step look like? cause am confuse here

thanks

$LL(x_1,...,x_n;\theta) = \sum_{i=1}^{n}\log(\theta)+ (\theta-1)\log(x_i) = n\log(\theta) + (\theta-1)\sum_{i=1}^n\log(x_i)$

.

3. ## Re: maximum likelihood estimator!

$f(x;\theta) = \frac{1}{2} \exp^{-{\mid x-\theta \mid}}$

pls the mod in the above equation seem to confuse me, if i take In of both sides, what will it be?

thanks

4. ## Re: maximum likelihood estimator!

Originally Posted by lawochekel
$f(x;\theta) = \frac{1}{2} \exp^{-{\mid x-\theta \mid}}$

pls the mod in the above equation seem to confuse me, if i take In of both sides, what will it be?

thanks
The log of a single term in the log-likelihood will be $\log(1/2) -|x-\theta|$

.

5. ## Re: maximum likelihood estimator!

pls to differentiate this function $f(x;\theta) = -\mid x-\theta \mid$ w.r.t $\theta$

how do i go about it to remove the modulus sign.

thanks