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!

A Question about Expected value

  1. Feb 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Suppose [itex]X_1, ... , X_n[/itex] are iid with pdf [itex]f(x,\theta)=2x/(\theta^2)[/itex], [itex] 0<x\leq\theta[/itex], zero elsewhere. Note this is a nonregular case. Find:

    a)The mle [itex]\hat{\theta}[/itex] for [itex]\theta[/itex].
    b)The constant c so that [itex]E(c\hat{\theta})=\theta[/itex].
    c) The mle for the median of the distribution.


    2. Relevant equations



    3. The attempt at a solution

    a) I got [itex]\hat{\theta} = max\{X_1, ... ,X_2\}=Y_n[/itex]
    b) I was stuck here...

    I need to find the pdf for [itex]Y_n[/itex] using the order statistics formula, right?

    So this is what I got...


    [itex]f_n(y_n)=\frac{n!}{(n-1)!(n-n)!}[F(y_n)]^{n-1}[1-F(y_n)]^{n-n}f(y_n)[/itex]

    = [itex]n(F(y_n))^{n-1}f(y_n)= n(\frac{-2x}{\theta})^{n-1}(\frac{2x}{\theta^2}) [/itex]

    = [itex]n(\frac{-2x}{\theta})^{n}(\frac{\theta}{-2x})(\frac{2x}{\theta^2}) [/itex]

    = [itex]n(\frac{-1}{\theta})(\frac{-2x}{\theta})^n [/itex]

    So now I want to find the expected value, but I'm not sure what the boundaries need to be for the integral...so can anybody help me?

    Thanks in advance
     
  2. jcsd
  3. Feb 10, 2013 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    The formula for ##f_n(y)## must not have x in it! Do you mean
    [tex] f_n(y) = n \left( \frac{-1}{\theta} \right) \left(\frac{-2x}{\theta}\right)^n?[/tex]
    I hope not, because for even n this whole expression is < 0, so you have a negative probability density.

    Why not just use [tex]P\{ \max(X_1,X_2, \ldots, X_n) \leq y \} = P\{ X_1 \leq y, X_2 \leq y, \ldots, X_n \leq y \}\\
    = F_X(y)^n [/tex] for iid ##X_i##?
     
  4. Feb 10, 2013 #3
    Thank you. So...

    [itex]F_X(y)=\frac{y^2}{\theta^2}[/itex], right? So...

    [itex]P\{ \max(X_1,X_2, \ldots, X_n) \leq y \} = P\{ X_1 \leq y, X_2 \leq y, \ldots, X_n \leq y \} = P(X_1 \leq y)P(X_2 \leq y)...P(X_n \leq y) = (\frac{y^2}{\theta^2})^n[/itex], right?

    But I'm still a bit confused about the boundaries for the intergral for the expected value...
     
  5. Feb 10, 2013 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Of course you need ##0 \leq y \leq \theta## in the above. For ##y > \theta## we have ##F_Y(y) = 1##.
     
  6. Feb 11, 2013 #5
    Thanks a lot.


    When I use the technique that you suggested, I get...

    [itex]P\{ \max(X_1,X_2, \ldots, X_n) \leq y \} = P\{ X_1 \leq y, X_2 \leq y, \ldots, X_n \leq y \} = P(X_1 \leq y)P(X_2 \leq y)...P(X_n \leq y) = (\frac{y^2}{\theta^2})^n[/itex]

    When I use the order statistics (I did it again, because the one I did before was wrong), I get...

    [itex]f_n(y_n)=\frac{n!}{(n-1)!(n-n)!}[F(y_n)]^{n-1}[1-F(y_n)]^{n-n}f(y_n)[/itex]

    = [itex]n(F(y_n))^{n-1}f(y_n)= n(\frac{{y_n}^2}{\theta^2})^n(\frac{\theta^2}{y_n})(\frac{2y_n}{\theta^2}) =(\frac{2n}{y_n})(\frac{{y_n}^2}{\theta^2})^n [/itex]

    This question is also solved in this link (question 2):

    http://www.math.harvard.edu/~phorn/362/362assn6-solns.pdf [Broken]

    All three of these answers are different. I keep repeating this to see if I did something wrong, but I just keep getting the same answers. It's really frustrating...
     
    Last edited by a moderator: May 6, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook