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I learnt that the variance of the Maximum Likelihood estimator could be calculated as :

\begin{aligned} \operatorname{var}(\theta) &=[I(\theta)]^{-1} \\ &=(-E[H(\theta)])^{-1} \\ &=\left(-E\left[\frac{\partial^{2} \ln \mathcal{L}(\theta)}{\partial \theta \partial \theta^{\prime}}\right]\right)^{-1} \end{aligned}

I want to determine how bias a biased coin is. I've flipped it 3 times and I get 3 Heads.

The likelihood of getting 3 straight heads is $\mathcal{L} = \theta^3$ where $0 \leq \theta \leq 1$

$ \ln \mathcal{L} = 3 \ln \theta $

This means that my $p_{MLE} =1.0$ where $p$ is the probability of getting a heads in a single coin flip and represents the $\theta$ that I'm trying to estimate.

Now obviously because I've only flipped it 3 times, the variance $\text{Var}(p_{MLE})$ will be relatively high. However, what is the actual numerical value of this variance?

I want to show that if I keep flipping the coin and I keep getting heads that this variance goes to zero.

From a similar old post it was found that this variance is equal to:

$$\text{Var}(p_{MLE})=\frac{p(1-p)}{n}$$

However I can't see how I can get a numerical answer for the variance of $p_{MLE}$ when we get three straight heads. I know that n=3 but for $p$ what value am I supposed to substitute it with?

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$p_{MLE}$ is a random variable. More specifically, here $$p_{MLE} = p_{MLE} (X_1,X_2,X_3) = \frac{1}{3} (X_1 + X_2 + X_3)$$ where $X_i$'s are bernoulli random variables. In the sense in which you're defining your problem, there is no such thing as the variance of $p_{MLE}$ when you get three straight heads, because when you give the value of $p_{MLE}$ for 3 straight heads, it becomes a value rather than a random variable, so its variance is not defined.

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  • $\begingroup$ In my case $p_{MLE}$ is the value of p that maximises the likelihood. Now it is possible to get the variance of MLE estimate. See page 7/8 of sherrytowers.com/mle_introduction.pdf $\endgroup$ – piccolo Aug 14 at 6:41
  • $\begingroup$ I want to get some numerical value of the confidence of $p_{MLE}$ which is related to the curvature of the likelihood about $p= p_{MLE}$ $\endgroup$ – piccolo Aug 14 at 6:44

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