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Thread: MVUE help

  1. #1
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    MVUE help

    Question:

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

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

    My work:

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

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

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

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

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

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

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