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Math Help - Unbiased estimators

  1. #1
    Senior Member chella182's Avatar
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    Unbiased estimators

    I think that's what this question's gearing towards anyway... I'm not 100% sure, hence how I haven't a clue what to do, even after skimming through my course notes. The question goes like this...

    Suppose that X_{1}, X_{2},..., X_{n} are a random sample from a population with mean \mu and variance \sigma^{2}.

    i) Verify that X_{1}+X_{2}-2X_{3}+X_{4} is unbiased and has variance 7\sigma^{2};

    ii) Verify that \frac{2X_{1}+X_{2}+X_{3}+...+X_{n}}{n+1} is unbiased and has variance \frac{(n+3)\sigma^{2}}{(n+1)^{2}}
    Last edited by chella182; April 29th 2009 at 05:14 PM. Reason: I want to subscribe to the thread so I know when someone's replied
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  2. #2
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by chella182 View Post
    I think that's what this question's gearing towards anyway... I'm not 100% sure, hence how I haven't a clue what to do, even after skimming through my course notes. The question goes like this...

    Suppose that X_{1}, X_{2},..., X_{n} are a random sample from a population with mean \mu and variance \sigma^{2}.

    i) Verify that X_{1}+X_{2}-2X_{3}+X_{4} is unbiased and has variance 7\sigma^{2};


    ii) Verify that \frac{2X_{1}+X_{2}+X_{3}+...+X_{n}}{n+1} is unbiased and has variance \frac{(n+3)\sigma^{2}}{(n+1)^{2}}

    The expected value of each x_i is mu hence
    E\biggl(X_{1}+X_{2}-2X_{3}+X_{4}\biggr)=EX_1+EX_2-2EX_3+EX_4=\mu+\mu-2\mu+\mu=\mu
    The variance of each is sigma^2 and by independence..

    V\biggl(X_{1}+X_{2}-2X_{3}+X_{4}\biggr)=V(X_1)+V(X_2)+4V(X_3)+V(X_4)=7  \sigma^2.

    I'll leave 2 for you.
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  3. #3
    Senior Member chella182's Avatar
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    Ah right, cheers sounds simple, I feel daft for posting it now

    & I've done part (ii) successfully too. Thankyouu!
    Last edited by chella182; April 30th 2009 at 05:13 AM.
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