Results 1 to 5 of 5

Math Help - Rayleigh distribution and unbiased estimator

  1. #1
    Junior Member
    Joined
    Apr 2010
    Posts
    36

    Rayleigh distribution and unbiased estimator

    X_1 ... X_n is random sample from Rayleigh distribution f(x;\Theta)

    1. Show that E(X^2) = 2\Theta and than construct unbiased estimator of parameter \Theta based on \sum_{k=1}^n X_k^2

    2. Estimate parameter \Theta from following n=10 observations:
    Code:
    16.88 10.23 4.59 6.66 13.68 14.23 19.87 9.40 6.51 10.95
    ---

    1. I have just plugged in 2 theta in Rayleigh's variance formula and it evaluates to true, but I'm not sure about correct way of constructing unbiased estimator \hat{\Theta}

    Var(X)=E(X^2) - E^2(X)

    Var(X)=2\Theta - E^2(X)

    \frac{4-\pi}{2}\Theta=2\Theta - \sqrt{\Theta\cdot\frac{\pi}{2}}^2

    4\Theta - \pi\Theta = 2(2\Theta - \frac{\pi\Theta}{2})

     \Theta = \Theta


    2. I need help with this one
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by losm1 View Post
    X_1 ... X_n is random sample from Rayleigh distribution f(x;\Theta)

    1. Show that E(X^2) = 2\Theta and than construct unbiased estimator of parameter \Theta based on \sum_{k=1}^n X_k^2

    2. Estimate parameter \Theta from following n=10 observations:
    Code:
    16.88 10.23 4.59 6.66 13.68 14.23 19.87 9.40 6.51 10.95
    ---

    1. I have just plugged in 2 theta in Rayleigh's variance formula and it evaluates to true, but I'm not sure about correct way of constructing unbiased estimator \hat{\Theta}

    Var(X)=E(X^2) - E^2(X)

    Var(X)=2\Theta - E^2(X)

    \frac{4-\pi}{2}\Theta=2\Theta - \sqrt{\Theta\cdot\frac{\pi}{2}}^2

    4\Theta - \pi\Theta = 2(2\Theta - \frac{\pi\Theta}{2})

     \Theta = \Theta


    2. I need help with this one
    Use the unbiased estimator found in part 1!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2010
    Posts
    36
    First, how do I know that it is unbiased?
    Second, basically I should just calculate E(X^2) with given dataset?

    Thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    May 2010
    Posts
    1,028
    Thanks
    28
    An estimate is unbiased if its expected value is equal to the true value of the parameter being estimated. So, you can confirm the estimate is unbiased by taking its expectation.

    So, assuming your estimate was
    \hat{\Theta} = \frac{\sum{x^2}}{2n}
    E(\hat{\Theta}) = E \left(\frac{\sum{x^2}}{2n} \right)
    E(\hat{\Theta}) = 0.5n^{-1}1 E \left(\sum{x^2} \right)
    E(\hat{\Theta}) = 0.5n^{-1} \left(\sum{E(x_i^2)} \right)
    E(\hat{\Theta}) = 0.5n^{-1} \left(\sum{E(X^2)} \right)
    E(\hat{\Theta}) = 0.5n^{-1} * n{E(X^2) \right)
    E(\hat{\Theta}) = 0.5 E(X^2) \right)
    E(\hat{\Theta}) = \Theta

    And the estimator is unbiased.

    I have been told, but i never managed to prove, that all method of moments estimates are unbiased and you dont need to check each time. But im not convinced...
    Last edited by SpringFan25; June 20th 2010 at 03:49 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2013
    From
    United States
    Posts
    11

    Re: Rayleigh distribution and unbiased estimator

    Guys, what am I missing here? I ran monte carlo simulations for Rayleigh samples and this estimator does not come close for small n. For example, with sigma = 1.0, using sample sizes of 2 the average value of the estimate is 0.50 (over 1MM iterations). Even with sample sizes of 10 the estimate is still only .90! Shouldn't an unbiased estimator work better than this for small n?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Gamma distribution, mgf and unbiased estimator
    Posted in the Advanced Statistics Forum
    Replies: 8
    Last Post: February 13th 2011, 05:05 AM
  2. unbiased estimator
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: November 23rd 2010, 07:36 AM
  3. Unbiased estimator in binomial distribution
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: April 30th 2010, 09:23 PM
  4. Estimator Unbiased?
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: February 18th 2010, 04:58 PM
  5. Unbiased estimator involving chi square distribution
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: January 6th 2010, 03:45 PM

Search Tags


/mathhelpforum @mathhelpforum