Cramer-Rao inequality in model-based pricing

Question

Given a constant vector $\mathbf{h}^* \in R^d$, define ${{\mathbf{h}}^{\delta}}={{\mathbf{h}}^{*}}+\mathbf{w}$, where $\mathbf{w}\sim N(\mathbf{0},(\delta /d)\cdot {{\mathbf{I}}_{d}}$.

Define a function $g:{\mathcal{H}^{k}}\to \mathcal{H}$ such that $\mathbf{\tilde{h}}=g(\mathbf{h}^{{\delta}_1},\mathbf{h}^{{\delta}_2},\cdots, \mathbf{h}^{{\delta}_k})$,

s.t. $E(\mathbf{\tilde{h}})={{\mathbf{h}}^{*}}$

Question: How to prove the following inequality based on Cramer-Rao bound?

$$E\left( \left\| \mathbf{\tilde{h}}-{{\mathbf{h}}^{*}} \right\|_{2}^{2} \right)\ge \frac{1}{\frac{1}{{\delta}_1}+\frac{1}{{\delta}_2}+\cdots+\frac{1}{{\delta}_k}}$$.

I cannot figure out how to derive the sum term in the right side of the inequality.

Answer 1

Sufficient statistics for this problem: $$T(\{h^{\delta_i} : i \in [k]\}) = \sum_i \frac{h^{\delta_i}}{\delta_i}$$

Assuming this is complete Sufficient statistics, By Rao Blackwood theorem, any $g(T)$ such that $E(g(T)) = h^*$ is a minimum variance unbiased estimator. i.e., $h_{Minimum \ Variance \ Unbiased \ Estimator} = g(T)$

Hence , $h_{Minimum \ Variance \ Unbiased \ Estimator} = \frac{ \sum_i \frac{h^{\delta_i}}{\delta_i}}{ \sum_i \frac{1}{\delta_i}}$.

Now variance of this estimator is the cramer Rao lower bound since this is a minimum variance unbiased estimator.

This variance is exactly as you stated in the question. So in this particular case, the cramer Rao lower bound is achieved.

So now use the equation: $$ I(h^*)= -E\left(\frac{d^2 \ln P(\{h^{\delta_i}\})}{d{h^*}^2}\right)$$

Since the ditribution is iid, we can derive component wise: $$\ln P\left(\{h^{\delta_1},h^{\delta_2},...,h^{\delta_k}\}\right) = \text{constant} - \sum_i \frac{|h^{\delta_i}-h^*|^2}{\delta_i}$$

$$\frac{d \ln P\left(\{h^{\delta_i}\}\right)}{dh^*} = \sum_i \frac{h^{\delta_i}}{\delta_i}-\sum_i \frac{1}{\delta_i} h^*$$

$$-\frac{d^2 \ln P\left(\{h^{\delta_i}\}\right)}{d{h^*}^2} = \sum_i \frac{1}{\delta_i}$$

$$I(h^*) = -E\left(\frac{d^2 \ln P\left(\{h^{\delta_i}\}\right)}{d{h^*}^2}\right) = \sum_i \frac{1}{\delta_i} $$.

So now you have the Cramer Rao bound: $Var(h^*) \geq \frac{1}{I(h^*)}$.