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Math Help - Maximum Likelihood Estimator

  1. #1
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    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.
    Attached Thumbnails Attached Thumbnails Maximum Likelihood Estimator-picture-3.png  
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  2. #2
    Moo
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    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 ?
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  3. #3
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    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
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