Results 1 to 5 of 5

Thread: MLE estimation

  1. #1
    Super Member Anonymous1's Avatar
    Joined
    Nov 2009
    From
    Big Red, NY
    Posts
    517
    Thanks
    1

    MLE estimation

    Suppose $\displaystyle X$ is exponentially distributed with density $\displaystyle f_{X}(x) = e^{-x}1_{x>0}.$ Consider $\displaystyle Z = \frac{h}{X},$ where $\displaystyle h > 0.$ Our goal is to estimate $\displaystyle h.$

    (a) Find the density function of $\displaystyle Z.$
    (b) Let $\displaystyle Z1,...,Zn$ be an i.i.d. sample from $\displaystyle f_{Z}(z),$ the density of $\displaystyle Z.$ Find the MLE of $\displaystyle h,$ denoted by $\displaystyle \hat{h}.$
    (c) Find the asymptotic variance of $\displaystyle \hat{h}.$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Use the Jacobian transformation to get the pdf of Z.

    You should get $\displaystyle f_Z(z)=e^{-h/z}\cdot \frac{z^2}{h} \cdot 1_{z>0}$

    The likelihood function is then the pdf of the n-tuple, which gives :

    $\displaystyle L(\bold{z};h)=\exp\left(-h\sum_{i=1}^n \frac{1}{z_i}\right) \cdot \frac{\left(\prod_{i=1}^n z_i\right)^2}{h^n}$ (where all the zi are >0, and where $\displaystyle \bold{z}=(z_1,\dots,z_n)$)

    So the log likelihood function is $\displaystyle \log L=-h\sum_{i=1}^n \frac{1}{z_i}-n\log h+\eta(\bold{z})$

    The function $\displaystyle \eta$ is not important, since we will differentiate the log likelihood function with respect to h.

    Now find its maximum and you'll get $\displaystyle \hat h$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Anonymous1's Avatar
    Joined
    Nov 2009
    From
    Big Red, NY
    Posts
    517
    Thanks
    1
    Quote Originally Posted by Moo View Post
    $\displaystyle L(\bold{z};h)=\exp\left(-h\sum_{i=1}^n \frac{1}{z_i}\right) \cdot \frac{\left(\prod_{i=1}^n z_i\right)^2}{h^n}$
    Thank you for your wonderful answer. Quick question though, why is this not the summation of the $\displaystyle z_i?$

    Also, notable shortcut: instead of taking the Jacobian. Consider the cdf, $\displaystyle P(Z\leq z),$

    $\displaystyle => Z$ ~ $\displaystyle exp(h/z)$

    Or, wait does this not work?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Anonymous1 View Post
    Thank you for your wonderful answer. Quick question though, why is this not the summation of the $\displaystyle z_i?$
    Because in the pdf, it's 1/z, so you're summing 1/z_i. You just have to know that e^x e^y = e^(x+y)

    Also, notable shortcut: instead of taking the Jacobian. Consider the cdf, $\displaystyle P(Z\leq z),$
    You can, but you have to find the cdf of X first.

    $\displaystyle => Z$ ~ $\displaystyle exp(h/z)$
    What is z ??? It's nothing near from a parameter...


    You should read again some basics...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Anonymous1's Avatar
    Joined
    Nov 2009
    From
    Big Red, NY
    Posts
    517
    Thanks
    1
    Quote Originally Posted by Moo View Post
    You should read again some basics...
    I'll admit I go to fast to keep up with myself sometimes, but I do know these things...., buried somewheres or another.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Estimation
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: Feb 3rd 2011, 07:07 PM
  2. Estimation
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: Sep 21st 2010, 05:55 AM
  3. Estimation
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: Apr 12th 2010, 01:42 AM
  4. estimation
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: May 28th 2009, 11:50 PM
  5. Estimation
    Posted in the Statistics Forum
    Replies: 0
    Last Post: Apr 22nd 2009, 03:19 AM

Search Tags


/mathhelpforum @mathhelpforum