Maximum Likelihood Estimator

**blueturkey** · May 8th 2010, 11:30 PM

Hi,
I have a solution to an MLE problem (picture attached). However, I am not sure how to get from line 1 to line 2 - with these particular questions in mind:

1) Why isnt the TT sign removed in the second line (and replaced with a sum from 0 to n)
2) Why would we get 1/theta^(2n) but not exp(-1/2n.theta)

Thanks for your time.

**Moo** · May 9th 2010, 12:04 AM

Hello,

Going back to basics... What you have to see is that :

$\prod_{i=1}^n \frac{x_i}{\theta^2} \cdot e^{-x_i/\theta}$

can be written :

$\left(\frac{x_1}{\theta^2} \cdot e^{-x_1/\theta}\right)\cdot\left(\frac{x_2}{\theta^2} \cdot e^{-x_2/\theta}\right) \cdots\left(\frac{x_n}{\theta^2} \cdot e^{-x_n/\theta}\right)$

these are products, so why would you want to transform it into a sum ??

grouping in some way, we get :

$\left(x_1\cdot x_2\cdots x_n\right)\cdot\bigg(\underbrace{\frac{1}{\theta^2 }\cdots \frac{1}{\theta^2}}_{n ~times}\bigg) \cdot \left(e^{-x_1/\theta}\cdots e^{-x_n/\theta}\right)$

The first parenthesis is exactly $\prod_{i=1}^n x_i$

The second parenthesis will be $\left(\frac{1}{\theta^2}\right)^n=\frac{1}{(\theta ^2)^n}=\frac{1}{\theta^{2n}}$

The third parenthesis is - remember that $e^a\cdot e^b=e^{a+b}$ - $\exp\left(-\frac{x_1}{\theta}-\frac{x_2}{\theta}-\dots-\frac{x_n}{\theta}\right)=\exp\left(-\frac 1\theta \cdot (x_1+\dots+x_n)\right)=\exp\left(-\frac1\theta \sum_{i=1}^n x_i\right)$

better now ?

**blueturkey** · May 9th 2010, 03:08 AM

Thanks! Yes, that was very clear. I had originally thought that the products could eventually be expressed as a sum but this clears it up

Thread: Maximum Likelihood Estimator

LinkBack

Thread Tools

Display

Maximum Likelihood Estimator

Similar Math Help Forum Discussions

Maximum likelihood estimator

Maximum Likelihood Estimator

maximum likelihood estimator

Maximum likelihood estimator

Maximum likelihood estimator

Search Tags