In what limit does the estimator sample variance converge to the Cramer-Rao bound?

In the context of a single phase estimation problem of a quantum photonics experiment (related post). For example consider a 3-photon quantum circuit (such as the Mach-Zehnder which depends on some phase shift operator which encodes a parameter $$\theta$$) with a photon counting measurement (two detectors) at the end of the circuit with measurement probabilities:

• P(0,2): 0 photons detected in Detector 1, 2 photons detected in Detector 2.
• P(1,1): 1 photon detected in each detector.
• P(2,0): 2 photons detected in Detector 1, and none in Detector 2.

Consider that at a given time we carry out $$\nu$$ total measurements. We will get some set of measurement outcomes {$$m_{02},~m_{11},~m_{20}$$}, where $$\nu = m_{02}+m_{11}+m_{20}$$. We can define the corresponding likelihood function $$L(\theta|\text{data})$$ by: $$L(\theta|\text{data}):= P(0,2)^{m_{02}}P(1,1)^{m_{11}}P(2,0)^{m_{20}}.$$

Using Baye's rule, given some assumed prior, $$P(\theta)$$ and the likelihood function $$L(\theta | \text{data})$$ above, we can update the posterior (our knowledge about the distribution of the phase $$\theta$$) iteratively, from the data of the measurement outcomes, by $$P(\theta| \text{data}) = \frac{L(\theta| \text{data})P(\theta)}{P(\text{data})} ,$$ where $$P(\text{data})$$ is the normalization constant. We can instead consider many (of the order of 100) unnormalized Log-Posterior updates of the form $$\log(P(\theta| \text{data})) = \log(L(\theta| \text{data}) + \log(P(\theta)).$$ Where after each update round $$\log(P(\theta| \text{data}))$$ replaces $$\log(P(\theta))$$ for the next update round. This is commonly done for more efficient numerical simulation. We can then consider the MAP estimator (to estimate the encoded phase $$\theta$$) which is defined as $$\hat{\theta}:= \text{max arg}_{\theta}\log(\theta|\text{data}).$$

If we consider the sample variance $$(\Delta \hat{\theta})^2$$ of the estimation of the true $$\theta$$ by carrying out this process $$\mu$$ independent times (with $$\nu$$ measurements for every 100 update rounds), yielding a set of $$\mu$$ different estimation values, am I correct that based on the Cramer-Rao bound we expect that, for $$\mu \to \infty$$, the sample variance of this set of estimations for an efficient estimator should converge to $$\frac{1}{\nu \cdot \text{FI}}$$, where FI is the Fisher information? An inefficient estimator should yield some sample variance $$\geq \frac{1}{\nu \cdot \text{FI}}$$. Are these expectations valid? Thanks for your time and assistance.

The answer is the same as for the other question of yours: no, the efficiency is asymptotic in $$\nu$$ not in $$\mu$$. The MLE is not efficient (in general) for any finite finite number of observations.
Say $$X_i\sim\operatorname{Bern}(p)$$ are IID random variables, and you collect a statistic of $$N$$ such samples. To be clear, I mean that $$X_i\in\{0,1\}$$ for each $$i$$, and $$p$$ is the probability of getting $$1$$ at each coin toss, so that $$\mathbb{E}[X_i^k]=p$$ for all integers $$k\ge1$$. Say you want to estimate the value of the parameter $$p^2$$. The corresponding MLE is $$\hat p_N^2$$ with $$\hat p_N$$ the MLE for $$p$$, which is the standard one given by the fraction of times you observed the outcome $$1$$: $$\hat p_N^2 = \left(\frac{\hat N_1}{N}\right)^2, \qquad \hat N_1 \equiv \sum_{i=1}^N X_i.$$ Let's compute the variance of $$\hat p_N^2$$. This requires computing the expectation values $$\mathbb{E}[(\sum_i X_i)^2]$$ and $$\mathbb{E}[(\sum_i X_i)^4]$$. Using standard multinomial expansions considerations you get $$\mathbb{E}\left[\left(\sum_{i=1}^N X_i\right)^4\right] = N p + 7N(N-1)p^2 \\+ 6 N(N-1)(N-2) p^3 + N(N-1)(N-2)(N-3)p^4, \\ \mathbb{E}\left[\left(\sum_{i=1}^N X_i\right)^2\right] = Np + N(N-1)p^2.$$ It follows that $$\mathbb{E}[\hat p_N] = p^2 + \frac{p(1-p)}{N},$$ $$\operatorname{Var}[\hat p_N^2] = \frac{\mathbb{E}[(\sum_i X_i)^4] - \mathbb{E}[(\sum_i X_i)^2]^2}{N^4} \\= \frac{4p^3(1-p)}{N} + \frac{2p^2(1-p)(3-5p)}{N^2} + O(1/N^3).$$ In the large $$N$$ limit, you thus get an unbiased estimator with variance $$\operatorname{Var}[\hat p_N^2] \sim \frac{4p^3(1-p)}{N},$$ which is exactly what you should expect, being the Fisher information associated with estimating the parameter $$p^2$$ equal to $$\frac{1}{4p^2 p(1-p)}$$ (see e.g. the wiki page and this post on math.SE).
In conclusion, this shows you clearly that while in the $$N\to\infty$$ limit the MLE indeed becomes efficient, it is never actually efficient for finite $$N$$. This is probably what you're observing: unless you compute the MLE for $$\nu\to\infty$$ (in your notation), you'll always have a variance larger than the minimum imposed by the CR bound. Sending $$\mu\to\infty$$ only serves to accurately estimate the variance at fixed $$N$$ or $$\nu$$.