Ch4 p.16 Distribution of μ

We could use Cramér-Rao lower bound, short CRLB, (CT3 ch10 p26) to prove the variance of MLE \( \tilde{\mu} \) instead of two pages of proof.

\[ \tilde{\mu} \backsim N \left( \mu, CRLB \right) \]

Keep in mind that the likelihood function is \( L\left( \mu; d, v \right) = e^{-\mu v} \mu^d \).

\begin{align}
CRLB &= \frac{1}{-\mathbb{E}\left[\frac{\partial^2}{\partial\mu^2} \log L\left( \mu; d, v \right) \right]} \\
&= \frac{1}{\mathbb{E}\left[-\frac{D}{\mu^2}\right]}\\
&=\frac{\mu^2}{\mathbb{E}\left[D\right]}\\
&=\frac{\mu}{\mathbb{E}\left[V\right]} \space \left(since \space \mathbb{E} \left[ D \right] =\mu \mathbb{E} \left[V \right] \right)\\
\end{align}

What do you think?

(Apologise, paragraph 5 just does that...)

 

First check whether "u(tilde)" is an unbiased estimator of "u" or in other words,
E[ u(tilde) ] = "u"
Only then you can use your suggested method.

 

I thought that every estimator is unbiased when is found using MLE? CRLB is the lower bound for variance of this unbiased estimator. However, if estimator is biased then CRLB is not sure that it is the lower bound of variance i.e. we can have variance lower than the one found using CRLB.

 

I'm not familiar with the context of this since I destroyed my CT4 notes many moons ago, but the short answer is no, the MLE is not in general an unbiased estimator. It is asymptotically unbiased though. Therefore CRLB is frequently invoked for MLEs where the sample size is large.

 

Yep. You are right! Tks a lot