Results 1 to 7 of 7

Math Help - Unbiased estimators?? Can't do it.

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    46

    Unbiased estimators?? Can't do it.

    Given X1,...,X7, a random sample from a population of mean mu and variance sigma^2. Prove that the following estimators are unbiased:

    estimator 1 = (X1+...X7)/7 = mean = unbiased by definition,

    my problem is with the second one:

    estimator 2 = (2X1 - X6 + X4) / 2


    Then they ask which estimator is the best, so that I kinda know, it will be the estimator with the lower variance but I can't solve for the variance because I have no numbers.

    And then they ask for efficiency...needless to tell that I don't know that 1 either.


    Much appreciate it guys!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
     E\Big(\frac{2X_{1}-X_{6}+X_{4}}{2}\Big) = E(X_{1}) - \frac{1}{2}E(X_{6}) + \frac{1}{2}E(X_{4})

     = \mu - \frac{1}{2} \mu + \frac{1}{2} \mu = \mu


     V\Big(\frac{2X_{1}-X_{6}+X_{4}}{2}\Big) = V(X_{1}) + \Big(-\frac{1}{2}\Big)^{2} V(X_{6})  +  \Big(\frac{1}{2}\Big)^{2} V(X_{4})

     = \sigma^{2} + \frac{1}{4}\sigma^{2} + \frac{1}{4}\sigma^{2} = \frac{3}{2} \sigma^{2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2009
    Posts
    46
    Thanks a lot. Show me the following if you don't mind :P

    I wana calculate relative efficiency of the 2, they say it's:

    MSE(estimator) = E( estimator - teta)^2

    and relative efficiency is: MSE(estimator1)/MSE(estimator2)

    I got trouble calculating the MSE of each estimator, those symbols mix me up badly lol :P

    tyy
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    Let  \hat{\theta_{1}} be the first unbiased estimator. and let  \hat{\theta_{2}} be the second unbiased estimator.


    then  \text{eff} \ (\hat{\theta_{1}}, \hat{\theta_{2}}) =\frac{V(\hat{\theta_{1}})}{V(\hat{\theta_{2}})}

     = \frac{(1/7) \sigma^{2}}{(3/2) \sigma^{2}} = \frac{2}{21} \approx 0.0952
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2009
    Posts
    46
    Quote Originally Posted by Random Variable View Post
    Let  \hat{\theta_{1}} be the first unbiased estimator. and let  \hat{\theta_{2}} be the second unbiased estimator.


    then  \text{eff} \ (\hat{\theta_{1}}, \hat{\theta_{2}}) =\frac{V(\hat{\theta_{1}})}{V(\hat{\theta_{2}})}

     = \frac{(1/7) \sigma^{2}}{(3/2) \sigma^{2}} = \frac{2}{21} \approx 0.0952
    Oh so I don't have to use MSE I get it. But how did you get 1/7, when I calculated the V(mean) I did:

    V( (X1+X2+X3+X4+X5+X6+X7)/7) = V(X1)/7 +V(X2)/7 + ...V(X7)/7 = sigma^2/7 * 7 = sigma^2

    (
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    Quote Originally Posted by Provoke View Post
    Oh so I don't have to use MSE I get it. But how did you get 1/7, when I calculated the V(mean) I did:

    V( (X1+X2+X3+X4+X5+X6+X7)/7) = V(X1)/7 +V(X2)/7 + ...V(X7)/7 = sigma^2/7 * 7 = sigma^2

    (
    In general  MSE(\hat{\theta}) = Var(\hat{\theta}) + \big( E(\hat{\theta})-\theta)\big)^{2}

    There's probably a proof in your textbook.

    In our case the estimators are unbiased, so  E(\hat{\theta})-\theta = \theta - \theta =0

    and  MSE(\hat{\theta}) = Var(\hat{\theta})
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
     V(aX) = a^{2}V(X)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. unbiased estimators
    Posted in the Advanced Statistics Forum
    Replies: 6
    Last Post: March 16th 2011, 08:10 AM
  2. unbiased estimators
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: February 2nd 2010, 11:54 PM
  3. Unbiased estimators
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: August 3rd 2009, 07:16 AM
  4. Unbiased estimators
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: April 30th 2009, 04:09 AM
  5. Unbiased Estimators
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: February 7th 2009, 08:45 PM

Search Tags


/mathhelpforum @mathhelpforum