1. ## 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.

2. ## Re: Estimating h(theta)

Originally Posted by alyosha2
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

$\displaystyle \hat{t}=h(\hat{\theta})=\hat{\theta}^2$

3. ## 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?