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Given a probability density say $f(x)$ with parameters $\theta$ and a sample of size n say $x_1,\ldots,x_n$ we can compute the MLE estimate say $\theta_n$ by passing $f$ and $x_1,\dots,x_n$ to optim in R.

We will have by the asymptotic normality of the MLE that,

$\sqrt n (\theta_n - \theta) $ converges in distribution to $\mathcal N(0,V)$ where $V$ is the inverse of the Hessian.

Now suppose we want to make sure that the parameter $\theta$ needs to satisfy a constraint, say $0\leq \theta \leq1$. Then we define
g($\omega$) = $\exp(\omega)/(1+\exp(\omega))$. And we consider the function $f(g(\omega))$. Now we pass $f \circ g$ and $x_1,\ldots,x_n$ to optim. Can we claim that

$\sqrt n ( \omega_n - \omega) $ will converge in distribution to $\mathcal{N}(0,V1)$ where $g(\omega)= \theta $. It's not clear to me if I can apply the asymptotic normality of MLE to $f \circ g$. Can someone explain this step in detail. I do realize that the next step will be the application of the delta method. I need clarity on this step though.

Lucas Roberts
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user2338823
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  • By the invariant property of the MLE, if $\hat \theta$ is the MLE of $\theta$ then $g(\hat \theta)$ is the MLE of $g(\theta)$. – Chamberlain Mbah Jul 27 '17 at 04:18
  • That is not my query. My query is how is $\omega_n$ the MLE of $\omega$. – user2338823 Jul 27 '17 at 04:22
  • Your question is confusing because you say $g(\omega)=\theta$. – Chamberlain Mbah Jul 27 '17 at 04:29
  • We have parameterized $\theta$ to be $g(\omega)$ so that \theta is between 0 and 1. How do I show that $\omega_n$ is the MLE of $\omega$ in this parameterization – user2338823 Jul 27 '17 at 04:31
  • Here is my query. Let $\omega_n$ be the value which maximizes f(x_i | g($\omega$)). How do I show that $\omega_n$ is the MLE of $\omega$ – user2338823 Jul 27 '17 at 05:05
  • So what you mean is $\theta(\omega)=g(\omega)$? and you know the MLE of the function $\theta(\omega)$, what is the MLE of $\omega$?. If the inverse of $g$ exist which it does, then the MLE of $\omega$ is $g^{-1}(\theta_n)$ where $\theta_n$ is the MLE of $\theta(\omega)$. – Chamberlain Mbah Jul 27 '17 at 05:05
  • where do we have that g is invertible ? – user2338823 Jul 27 '17 at 06:56

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