1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?