Can the Cramér-Rao Lower Bound be constant?
Prologue

The CRLB of an unbiased estimator $\hat \theta$ for the paramater $\theta$ is 

$$
\text{var}(\hat \theta) \ge - \frac1 {\text E_x\left\{ \frac{\partial^2 (\ln f(x;\theta))}{\partial \theta^2} \right\}}
$$

with $f(x;\theta)$ being the likelihood function of $\theta$ under the observation $x$. It is the inverse of the so-called Fisher information.

Answer


  Is it possible that the CRLB remains constant for all SNR?


Yes. For that, the expectation (over all $x$, hence the index)

$$\text E_x\left\{ \frac{\partial^2 (\ln f(x;\theta))}{\partial \theta^2} \right\} $$

just needs to be constant.

The "easiest" explanation would be if noise just didn't have anything to do with the likelihood – ie. noise is just not actually observed by the estimator.

But that's kind of least exciting result – "well, if I add a lot of noise to something and then observe something completely unrelated, the noise has no effect" isn't that surprising.

A lot more complicated, is the option that the Fisher information is actually constant, although $f(x;\theta)$ actually depends on the noise.

The expectation operator $E_x$ is definable as

$$
\text E_x \{g(x)\} = \frac1{\mu (\xi)} \int\limits_{x\in\xi} g(x)\,d\mu\text,
$$

i.e. as the integral over all possible $x$ divided by the measure $\mu$ of the set $\xi$ from which $x$ can be chosen.

Assuming smoothness of $x$ and compactness of $\xi\subset \mathbb R^1$ (just to make all the derivates easier to handle and to avoid cascading integrals), this specializes to:

$$\begin{align}
\text E_x \{g(x)\} &amp;= \frac1{x_{max}-x_{min}} \int\limits_{x = x_{min}}^{x_{max}} g(x)\,dx\\
\text E_x\left\{ \frac{\partial^2}{\partial \theta^2}{ (\ln f(x;\theta))} \right\}  \cdot\left({x_{max}-x_{min}}\right) &amp;= \int\limits_{x = x_{min}}^{x_{max}} \frac{\partial^2}{\partial \theta^2}{ (\ln f(x;\theta))} \,dx\\
&amp;= \left[ \frac{\partial}{\partial \theta}{ (\ln f(x;\theta))}  \right]_{x = x_{min}}^{x_{max}}\\
&amp;= \left[ \frac1{f(x;\theta)}  \frac{\partial f(x;\theta)}{\partial \theta}  \right]_{x = x_{min}}^{x_{max}}\\
&amp;= \frac1{f(x_{max};\theta)}  \frac{\partial f(x;\theta)}{\partial \theta}_{x=x_{max}}
- \frac1{f(x_{min};\theta)}  \frac{\partial f(x;\theta)}{\partial \theta}_{x=x_{min}}\\
&amp;= c,&amp;c\text{ const.}\\
\end{align}$$