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Math Help - Cramer Rao

  1. #1
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    Cramer Rao

    hello there,

    I need some help with this question.

    X1,...Xn~f(x,theta)

    f(x,theta)=exp{-(x-theta)}exp{-exp{-(x-theta)}}

    ( look picture attached )

    find the Cramer Rao lower bound for the unbiased estimator of theta.

    cheers !
    Attached Thumbnails Attached Thumbnails Cramer Rao-crlb.jpg  
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  2. #2
    Moo
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    Hello,

    Assuming the Xi are independent... This is how I've been taught. Dunno how you've learnt it...

    We have the joint pdf of the Xi equal to :

    g(x,\theta)=\prod_{i=1}^n \exp\left(-(x_i-\theta)-\exp\left(-(x_i-\theta)\right)\right)=\prod_{i=1}^n \exp\left(\theta-x_i-\exp\left(\theta-x_i\right)\right)

    =\exp\left(\sum_{i=1}^n [\theta-x_i-e^{\theta-x_i}]\right)

    =\exp\left(n\theta-\sum_{i=1}^n x_i+e^{\theta-x_i}\right)

    \Rightarrow \log(g(x,\theta))=n\theta-\sum_{i=1}^n x_i-\sum_{i=1}^n e^{\theta-x_i}


    \Rightarrow \frac{\partial \log(x,\theta)}{\partial \theta}=n-\sum_{i=1}^n e^{\theta-x_i}=n-e^{\theta} \sum_{i=1}^n e^{-x_i}

    An estimator \hat{\theta} will be solution of \frac{\partial \log(x,\hat{\theta})}{\partial \theta}=0


    Thus e^{\hat{\theta}}\sum_{i=1}^n e^{-x_i}=n \Rightarrow \hat{\theta}=\log\left(\frac{n}{\sum_{i=1}^n e^{-x_i}}\right)


    Wow...now, that's ugly !

    I've been taught to find Fisher's information, I(\theta) and h such that h(\theta)=\mathbb{E}(\hat{\theta})
    (if it's an unbiased estimator, then h=Id)

    And then Cramer Rao's lower bound is \frac{[g'(\theta)]^2}{I(\theta)}
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  3. #3
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    it is ugly !


    anyone knows how to finish it ?
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