<s>[INST] 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.
 [/INST] 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.
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