# Thread: Maximum Likelihood Estimator

1. ## Maximum Likelihood Estimator

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.

2. 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 ?

3. 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