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Math Help - Estimating h(theta)

  1. #1
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    Estimating h(theta)

    I'm having a really hard time understanding this.

    The MLE of a function of a parameter is the function of the MLE of the parameter.

    That is if t = h(\theta) and \hat{\theta} is the MLE of \theta, then the MLE of t is \hat{t} = h(\hat{\theta})

    Ok, so the MLE of a function of a parameter is the same function with the MLE of the parameter fed into it. That's my understanding.

    But the next claim is that for any square of a parameter the MLE of the square of the parameter will be the square of the MLE of the parameter. I just can't see how this follows.
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  2. #2
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    Re: Estimating h(theta)

    Quote Originally Posted by alyosha2 View Post
    I'm having a really hard time understanding this.

    The MLE of a function of a parameter is the function of the MLE of the parameter.

    That is if t = h(\theta) and \hat{\theta} is the MLE of \theta, then the MLE of t is \hat{t} = h(\hat{\theta})

    Ok, so the MLE of a function of a parameter is the same function with the MLE of the parameter fed into it. That's my understanding.

    But the next claim is that for any square of a parameter the MLE of the square of the parameter will be the square of the MLE of the parameter. I just can't see how this follows.
    let t=h(\theta)=\theta^2 then

    \hat{t}=h(\hat{\theta})=\hat{\theta}^2
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  3. #3
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    Re: Estimating h(theta)

    But that's only going to hold if we define t = \theta^{2}. That h({\theta}) squares it's input. What if the input is squared before hand? How do we know the MLE will be a square of the non squared input? How does it hold in the general case?
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