# Estimating h(theta)

• February 2nd 2014, 03:39 PM
alyosha2
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.
• February 2nd 2014, 04:41 PM
romsek
Re: Estimating h(theta)
Quote:

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 $t = h(\theta)$ 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

$\hat{t}=h(\hat{\theta})=\hat{\theta}^2$
• February 3rd 2014, 02:44 AM
alyosha2
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?