Why the variance of Maximum Likelihood Estimator(MLE) will be less than Cramer-Rao Lower Bound(CRLB)? Consider this example. Suppose we have three events to happen with probability $p_1=p_2=\frac{1}{2}\sin ^2\theta ,p_3=\cos ^2\theta $ respectively. And we suppose the true value $\theta _0=\frac{\pi}{2}$. Now if we do $n$ times experiments, we will only see event 1 and event 2 happen. Hence the MLE should be
$$\mathrm{aug} \underset{\theta}{\max}\left[ \frac{1}{2}\sin ^2\theta \right] ^{m_1}\left[ \frac{1}{2}\sin ^2\theta \right] ^{m_2}$$
where $m_1$ is how many times event 1 happened and $m_2$ is how many times event 2 happened.
Obviously, the solution to the above optimization problem(also it's our MLE) is $\pi/2$ which is a constant and has nothing to do with $m_1,m_2$.
But let's see what is our fisher information, whose inverse should give us a lower bound of the variance of any unbiased estimator:
$$\begin{align}
&\frac{\left[ \partial _{\theta}\frac{1}{2}\sin ^2\theta \right] ^2}{\frac{1}{2}\sin ^2\theta}+\frac{\left[ \partial _{\theta}\frac{1}{2}\sin ^2\theta \right] ^2}{\frac{1}{2}\sin ^2\theta}+\frac{\left[ \partial _{\theta}\cos ^2\theta \right] ^2}{\cos ^2\theta}
\\
&=2\frac{\left[ \partial _{\theta}\frac{1}{2}\sin ^2\theta \right] ^2}{\frac{1}{2}\sin ^2\theta}+\frac{\left[ \partial _{\theta}\cos ^2\theta \right] ^2}{\cos ^2\theta}
\\
&=4\cos ^2\theta +4\sin ^2\theta 
\\
&=4
\end{align}$$
Hence there's a conflict, where do I understand wrong?
Thanks in advance!
Edit
If I calculate MLE not based on the observed data, but instead, based on all the possible outcomes before we do experiments, it should be:
$$\mathrm{aug}\underset{\theta}{\max}\left[ \frac{1}{2}\sin ^2\theta \right] ^{m_1}\left[ \frac{1}{2}\sin ^2\theta \right] ^{m_2}\left[ \cos ^2\theta \right] ^{m_3}$$
taking ln we will have
$$\left( m_1+m_2 \right) \left( 2\ln\sin \theta -\ln 2 \right) +2m_3\ln\cos \theta $$
taking derivative w.r.t. $\theta$ and set derivative to be zero we will get
$$2\left( m_1+m_2 \right) \frac{\cos \theta}{\sin \theta}=2m_3\frac{\sin \theta}{\cos \theta}$$
Hence if true value $\theta_0=\pi/2$, $m_3$ will always be $0$, and we will always have a conclusion that $\hat\theta=\pi/2$ which will have no variance at all. Hence the variance of $\hat\theta$ is $0$ which against the rule of Cramer-Rao bound.
 A: The first issue you have here is that your likelihood function does not appear to match the description of your sampling mechanism.  You say that you only observe events 1 and 2 happen, but if the sample size $n$ is known then this still fixes the number of times that event 3 happens (since $m_1 + m_2 + m_3 = n$).  Taking an observation $\mathbf{m} \sim \text{Mu}(n, \mathbf{p})$ gives the likelihood function:
$$L_\mathbf{m}(\theta)
= \bigg( \frac{1}{2} \sin^2 (\theta) \bigg)^{m_1 + m_2} \bigg( \cos^2 (\theta) \bigg)^{n - m_1 - m_2},$$
which gives the log-likelihood:
$$\ell_\mathbf{m}(\theta)
= \text{const} + (m_1 + m_2) \log (\sin^2 (\theta)) + (n - m_1 - m_2) \log (\cos^2 (\theta)).$$
As you can see, the likelihood function has an extra term that you have not included in your question.  We can see that $M_* = M_1+M_2 \sim \text{Bin}(n, \sin^2 (\theta))$ is a sufficient statistic in this problem so the problem is essentially one of binomial inference with a transformed probability parameter.  With a bit of calculus it can be shown that the MLE solves:
$$\sin^2 (\hat{\theta}) = \frac{m_1 + m_2}{n}
\quad \quad \quad \implies \quad \quad \quad 
\hat{\theta} = \text{arcsin} \bigg( \sqrt{\frac{m_1 + m_2}{n}} \bigg).$$
This estimator for the parameter is generally biased (it is unbiased in the case where $\sin^2 (\theta) = \tfrac{1}{2}$) so the applicable version of the Cramér–Rao lower bound in this case is the generalisation for biased estimators:
$$\mathbb{V}(\hat{\theta}) \geqslant \frac{|\psi'(\theta)|}{I(\theta)}
\quad \quad \quad \quad \quad
\psi(\theta) \equiv \mathbb{E}(\hat{\theta}).$$
The expectation function is:
$$\begin{align}
\psi(\theta)
&= \sum_{m=0}^n \text{arcsin} \bigg( \sqrt{\frac{m}{n}} \bigg) \cdot \text{Bin} (m|n,\sin^2 (\theta)) \\[6pt]
&= \sum_{m=0}^n {n \choose m} \cdot \text{arcsin} \bigg( \sqrt{\frac{m}{n}} \bigg) \cdot \sin^{2m} (\theta) \cdot \cos^{2(n-m)} (\theta). \\[6pt]
&= \frac{\pi}{2} \cdot \sin^{2n} (\theta) - \frac{\pi}{2} \cdot \cos^{2n} (\theta) + \sum_{m=1}^{n-1} {n \choose m} \cdot \text{arcsin} \bigg( \sqrt{\frac{m}{n}} \bigg) \cdot \sin^{2m} (\theta) \cdot \cos^{2(n-m)} (\theta), \\[6pt]
\end{align}$$
and its derivative (which appears in the bound) is:
$$\begin{align}
\psi'(\theta) 
&= \sum_{m=0}^n {n \choose m} \cdot \text{arcsin} \bigg( \sqrt{\frac{m}{n}} \bigg) \frac{d}{d\theta} \bigg[ \sin^{2m} (\theta) \cdot \cos^{2(n-m)} (\theta) \bigg] \\[6pt]
&= n\pi \cdot \sin(\theta) \cos(\theta) [\sin^{2(n-1)} (\theta) +  \cos^{2(n-1)} (\theta)] \\[6pt]
&\quad + 
\sum_{m=1}^{n-1} {n \choose m} \cdot \text{arcsin} \bigg( \sqrt{\frac{m}{n}} \bigg) \sin^{2m-1} (\theta) \cos^{2(n-m)-1} (\theta) \\[6pt]
&\quad \quad \quad \quad \quad \times \bigg[ 2m \cos^2 (\theta) - 2(n-m) \sin^2 (\theta) \bigg]. \\[6pt]
&= n\pi \cdot \sin(\theta) \cos(\theta) [\sin^{2(n-1)} (\theta) +  \cos^{2(n-1)} (\theta)] \\[6pt]
&\quad + 
2 \sum_{m=1}^{n-1} {n \choose m} \cdot \text{arcsin} \bigg( \sqrt{\frac{m}{n}} \bigg) \sin^{2m-1} (\theta) \cos^{2(n-m)-1} (\theta) (m - n \sin^2 (\theta)). \\[6pt]
\end{align}$$
As you can see, this is a more complicated form for the Cramér–Rao lower bound.  Nevertheless, it should hold in this problem.
A: Variance of your MLE estimator should be computed while still considering the results to be random, i.e.:
$$
\hat{\theta} = \mathrm{arg} \underset{\theta}{\max}\left[ \frac{1}{2}\sin ^2\theta \right] ^{m_1}\left[ \frac{1}{2}\sin ^2\theta \right] ^{m_2} \left[ \cos ^2\theta \right] ^{m_3}
$$
$$
Var(\hat{\theta}) = Var \left( \mathrm{arg} \underset{\theta}{\max}\left[ \frac{1}{2}\sin ^2\theta \right] ^{m_1}\left[ \frac{1}{2}\sin ^2\theta \right] ^{m_2} \left[ \cos ^2\theta \right] ^{m_3} \right) = \dots
$$
where $m_3$ is the number of observation of the event 3. You can't just rule out $\left[ \cos ^2\theta \right] ^{m_3}$ just because you didn't observe the event 3 after the experiment. If you don't know the value of $\theta$, you must consider the possibility of event 3 happening.
Even if after doing $n$ experiments you would observe some events 3, i.e. $m_3 > 0$, the evaluation of the $\mathrm{arg}\max$ above would be a constant (if $m_1$, $m_2$, $m_3$ are some constant numbers). That would not mean the variance of the estimator is $0$ - we just evaluated the estimate accoring to results of experimenting, there is no randomness in it.
What is the variance of the estimator $\hat{\theta}$ of the random variable $\theta$? To compute it, you'd need some assumption about $\theta$ (e.g. which distribution does it come from?). However, what you know by the Cramer-Rao theorem, is that the variance will be higher or equal than $1/4$.
A: To make the equations easier I am gonna use only two states instead of three. $p_1 = f(\theta)$ and $p_2 = 1-f(\theta)$
You apply the formula  $$\mathcal{I}(\theta) = \sum_x{{ \overbrace{\left[ \frac{\frac{\partial}{\partial \theta} p\left( x;\theta \right) }{p\left( x;\theta \right) }\right] ^2}^{{logL^\prime}^2}}{p\left( x;\theta \right)}} \qquad \tag{1} $$
This derivative of the log likelihood is:
$$logL^\prime = \frac{ m_1 f^\prime(\theta) - m_2 f^\prime(\theta)}{m_1 f(\theta) + m_2 (1-f(\theta))}$$
The probability is
$$p(x;\theta) = f(\theta)^{m_1} (1-f(\theta))^{m_2}$$
And if we restrict to a single observation $m_1 = 1, m_2 = 0$ or $m_1 = 0, m_2 = 1$ then the sum in formula (1) becomes
$$\mathcal{I}(\theta) = \frac{f^\prime (\theta)^2}{f(\theta)} + \frac{f^\prime (\theta)^2}{1-f(\theta)} = \frac{f^\prime (\theta)^2}{f(\theta) (1-f(\theta))}  $$
then
$$\mathcal{I}(\theta)^{-1} = \frac{f(\theta) (1-f(\theta))}{f^\prime (\theta)^2}$$
and with $f(\theta) = \sin^2\theta$ this becomes the same as your result
$$\mathcal{I}(\theta)^{-1} =  \frac{ \sin^2\theta \cos^2\theta}{(2 \cos\theta \sin\theta )^2} = \frac{1}{4}$$
But with $f(\theta) = \theta$ then you get
$$\mathcal{I}(\theta)^{-1} =  \theta(1-\theta)$$
like in this question https://math.stackexchange.com/questions/396982/fisher-information-of-a-binomial-distribution
Maybe there is some regularity condition that is not satisfied? Doesn't the Cramer Rao bound only apply to continuous distributions and unbiased variables?
