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Thread: 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 $\displaystyle t = h(\theta)$ and $\displaystyle \hat{\theta}$ is the MLE of $\displaystyle \theta$, then the MLE of $\displaystyle t$ is $\displaystyle \hat{t} = h(\hat{\theta})$

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

    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 $\displaystyle t=h(\theta)=\theta^2$ then

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

    But that's only going to hold if we define $\displaystyle t$ = $\displaystyle \theta^{2}$. That $\displaystyle 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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