#StackBounty: #estimation #inference #fisher-information #efficiency Deriving C-R inequality from H-C-R bound

Bounty: 50

As mentioned in the title, I want to derive the Cramer-Rao Lower bound from the Hammersly-Chapman-Robbins lower bound for the variance of a statistic $T$.
The statement for the H-C-R lower bound is the following,

Let $mathbf{X} sim f_{theta}(.)$ where $theta in Theta subseteq mathbb{R}^k.$ Suppose $T(mathbf{X})$ is an unbiased estimator of $tau(theta)$ where $tau colon Theta to mathbb{R}$. Then we have,
begin{equation}
text{Var}{theta}(T) ge displaystyle sup{Delta in mathcal{H}{theta}}, displaystyle frac{[tau(theta + Delta)]^2}{mathbb{E}{theta}left(frac{f_{theta + Delta}}{f_{theta}} – 1right)^2}
end{equation
}

where $mathcal{H}_{theta} = {alpha in Theta colon text{ support of } f text{ at } theta + alpha subseteq text{ support of } f text{ at } theta}$

Now when $k = 1$ and the regularity conditions hold, taking $Delta to 0$ gives the following inequality,
begin{equation}
text{Var}{theta}(T) ge displaystyle frac{[tau'(theta)]^2}{mathbb{E}{theta} left( frac{partial }{partial theta} log f_{theta}(mathbf{X}) right)^2}
end{equation
}

which is exactly the C-R inequality for univariate case.

However, I want to derive the general form of C-R inequality from the H-C-R bound, i.e. when $k > 1$. But, I have not been able to do it. Though, I was able to figure out that we would have to use $mathbf{0} in mathbb{R}^k$ instead of $0$ and $|Delta|$ to obtain the derivatives, which was obvious anyways, I couldn’t get to any expression remotely similar to the C-R inequality. One of the difficulty arises while dealing with the squares. Since for the univariate case, we were able to take the limit inside and as a result got the square of the derivative. While, for the latter case, we cannot take the limit inside, because the derviate in this case would be a vector and we will have the expression containg the square of a vector which is absurd.

I want to know how to derive the C-R inequality in the latter case?


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