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Homework Help: Maximum likelyhood extimater(MLE)

  1. Feb 16, 2005 #1
    so by definition, the likelyhood function(w, theta) is the product of the pdf fw(w, theta) evalutated at n data points.

    but I dont know how they do those calculations.....

    so for example:

    fy(y, theta) = [tex]1/\theta^2 ye^{-y/\theta}[/tex]

    L(theta) = [tex] \theta^{-2n}\prod y_{i}e^{-1/\theta}\sum y_{i} [/tex]

    so first of all, I'm looking at this but I dont know how they went from this to that....I look at another problme

    how did they go from [tex] e^{-(y-\theta)} [\tex]to [tex] \pro e^{-(y_{i}- \theta)} [/tex] [/theta]

    I dotn see a pattern...I compared it w/ the definitoin, but I just dont get it....
    I mean, when they did the L(theta) it seems that they added some "n' and i's somewhere....and I dotn know where they added these things.
     
    Last edited: Feb 16, 2005
  2. jcsd
  3. Feb 16, 2005 #2
    It looks like there are some typos in your expression.

    You are trying to estimate [tex]\theta[/tex] using n data points, labelled as [tex]y_i[/tex]. The single likelihood for [tex]\theta[/tex] given one data point[tex]y_i[/tex] is:

    [tex]1/\theta^2 y_{i}e^{-y_{i}/\theta}[/tex]

    In order to get the likelihood of [tex]\theta[/tex]for all n data points, then you need to multiply the single likelihoods together. And that's just a simple matter of multiplying terms and summing up what's in the exponential term.
     
  4. Feb 16, 2005 #3
    The indices are a shorthand for indeterminately long products. For example, given your definition of L, we evaluate the pdf at n values of y, {y1, y2, ..., yn} and take the product. So if
    [tex]f(y) = \frac{1}{\theta^2} ye^{-\frac{y}{\theta}}[/tex]
    we get
    [tex]f(y_1)f(y_2)...f(y_n) &= \prod_{i=1}^n f(y_i)[/tex]

    [tex] = \prod_{i=1}^n \frac{1}{\theta^2} y_i e^{-\frac{y_i}{\theta}}[/tex]

    [tex] = \frac{1}{\theta^{2n}} \prod_{i=1}^n y_i e^{-\frac{y_i}{\theta}}[/tex]

    [tex] = \frac{1}{\theta^{2n}} \left(y_1 e^{-\frac{y_1}{\theta}}...y_n e^{-\frac{y_n}{\theta}}\right)[/tex]

    [tex] = \frac{1}{\theta^{2n}} \left(y_1...y_n e^{-\frac{y_1+y_2+...+y_n}{\theta}}\right) [/tex]

    [tex] = \frac{1}{\theta^{2n}} \left(\prod_{i=1}^n y_i\right) \left(e^{-\frac{1}{\theta}\sum_{i=1}^n y_i}}\right) [/tex]
     
  5. Feb 19, 2005 #4
    ok, thanx, now that makes a little bit more sense, but i'll think about it some more...still a bit confusing.

    what about to get ln L(theta)? the book does some wierd stuff an I dont know what it did.

    ln L(theta) = [tex]-2n ln \theta + ln \prod yi - 1/\theta \sum yi [/tex]


    how did that happen?

    and also, i'm not understanding where they put the n's. maybe i'm having trouble w/ the definition. like, on the first problem, why did it become [tex]\theta^{-2n} [/tex]?

    and for ths problem,

    we have fy (y, theta) = [tex] e^{-(y-\theta)}, [/tex] so L(theta) = [tex] e^{\sum y + n\theta} [/tex]
     
    Last edited: Feb 19, 2005
  6. Feb 19, 2005 #5
    Because [tex]\theta^{-2}[/tex] multiplied by itself n times is [tex]\theta^{-2n}[/tex]. They just took it out of the n-product by associativity.

    What happens when you multiply ea with eb ? That's all that's going on here. :)
     
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