Maximum Likelihood Estimator

Aug 2009
16
0
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.
 

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Moo

MHF Hall of Honor
Mar 2008
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P(I'm here)=1/3, P(I'm there)=t+1/3
Hello,

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

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

can be written :

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

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

grouping in some way, we get :

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

The first parenthesis is exactly \(\displaystyle \prod_{i=1}^n x_i\)

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

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

(Whew) better now ?
 
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Aug 2009
16
0
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 :)