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Math Help - MVUE help

  1. #1
    Junior Member
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    MVUE help

    Question:

    Let \bar{X_1} and \bar{X_2} be the means of two independent random samples of size n_1 > 1 and n_2 > 1 from an infinite population that has mean \mu and finite variances \sigma_{1}^{2} and \sigma_{2}^{2} where \sigma_{1}^{2}>0,  \sigma_{2}^{2}>0, \sigma_{1}^{2} \neq \sigma_{2}^{2}.

    If we want \hat\theta=w \bar{X_1}+(1-w) \bar{X_2} to be the minimum variance unbiased estimator of \mu, what must w be?

    My work:

    V \hat\theta = w^2 \frac{\sigma_{2}^{1}}{n_1}+(1-w)^2 \frac{\sigma_{2}^{2}}{n_2}

    Let \frac{dV \hat\theta}{dw} = 0:

    2w\frac{\sigma_{2}^{1}}{n_1}-2(1-w)\frac{\sigma_{2}^{2}}{n_2}=0

    2w \left(\frac{\sigma_{2}^{1}}{n_1}+\frac{\sigma_{2}^  {2}}{n_2} \right)=2 \frac{\sigma_{2}^{2}}{n_2}

    w=\frac{ \frac{ \sigma_{2}^{2} }{n_2} } {\left( \frac{\sigma_{2}^{1}}{n_1}+\frac{\sigma_{2}^{2}}{n  _2} \right)}

    Look good?
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  2. #2
    MHF Contributor matheagle's Avatar
    Joined
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    you messed up on the sub/super scripts of \sigma_1^2

    Otherwise, it looks ok.
    But you should obtain the second derivative to prove that you have a min.
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