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MLE (Statistics)

  1. Jan 22, 2008 #1
    This is my question: Find the Maximum Likelihood Estimator for
    f(y / x) = (x + 1)y^x, 0 < y < 1 and x > -1 OR 0, elsewhere.

    I think this is how you get started, but I get confused. I'm not sure how to continue.
    The likelihood function defined as the joint density of Y1, Y2, ...., Yn evaluated at y1, y2, ..., yn is given by
    L = product from i = 1 to n of (x + 1)(yi^x) = (x + 1)^n * product from i = 1 to n of (yi^x).
    I'm sorry for the notation. Any help is obviously appreciated.
     
  2. jcsd
  3. Jan 22, 2008 #2

    EnumaElish

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    Your question can be restated as:

    Find the MLE of x given:

    f(y | x) = (x + 1)y^x, 0 < y < 1 and x > -1
    f(y | x) = 0 elsewhere.

    I am assuming that (i) f is the pdf of "y given x" and (ii) the y's are independent.

    Then, the simplest procedure would be to take the log of L(x) = [itex]\prod_{i = 1}^n (x + 1)y_i^x[/itex] and maximize it with respect to x.
     
    Last edited: Jan 22, 2008
  4. Jan 22, 2008 #3
    So I get:

    ln[((x+1)^n)*(y1^x)*(y2^x)*....*(yn^x)], but I don't see how you maximize this.

    I would imagine you take the derivative and set it equal to zero, but I cannot solve for x. What is the maximum value for x?

    Thanks for the help from before.
     
  5. Jan 27, 2008 #4
    OK, I just want to check this.

    I get:
    ln[((x+1)^n)*(y1^x)*(y2^x)*....*(yn^x)] =
    nln(x+1) + x[ln(y1) + ln(y2) + ... + ln(yn)

    I take the derivative with respect to x and set it equal to zero:
    n/(x+1) + ln(y1) + ln(y2) + .... + y(n)
    and with algebra this implies

    x (the estimator) = [-n - ln(y1) - ln(y2) - ... - ln(yn)] / [ln(y1) + ... + ln(yn)

    Does this make sense? Thanks again for your help; it is appreciated.
     
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