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Let's say that $X_1, \dots, X_n$ has the joint distribution $f_\varphi(\mathbf{x})$ that belongs to the one-parameter exponential family

$$f_\varphi(\mathbf{x}) = \exp{\left\{ c(\varphi) T(\mathbf{x}) + d(\varphi) + s(\mathbf{x}) \right\}},$$

where $\mathbf{x} \in \text{supp}(f_\varphi)$, $\text{supp}(f_\varphi)$ does not depend on $\varphi$, and $c^\prime(\varphi)$ is continuous and does not vanish.

I am told that the results

$$E_\varphi [T(\mathbf{X})] = - \dfrac{d^\prime (\varphi)}{c^\prime(\varphi)}$$

and

$$\text{Var}_\varphi [T(\mathbf{X})] = \dfrac{c^ {\prime \prime}(\varphi) d^\prime(\varphi) - c^\prime(\varphi)d^{\prime \prime}(\varphi)}{c^\prime(\varphi)^3}$$

can be used to show that $T(\mathbf{X})$ is an unbiased estimator of $E_\varphi[T(\mathbf{X})]$ that achieves the Cramer-Rao lower bound. However, it is not at all clear to me how this is done. How do the results $E_\varphi [T(\mathbf{X})] = - \dfrac{d^\prime (\varphi)}{c^\prime(\varphi)}$ and $\text{Var}_\varphi [T(\mathbf{X})] = \dfrac{c^ {\prime \prime}(\varphi) d^\prime(\varphi) - c^\prime(\varphi)d^{\prime \prime}(\varphi)}{c^\prime(\varphi)^3}$ show that $T(\mathbf{X})$ is an unbiased estimator of $E_\varphi[T(\mathbf{X})]$ that achieves the Cramer-Rao lower bound?

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The unbiasedness part holds trivially from the fact that you have stipulated that the unknown quantity you are estimating is the expected value of the estimator. Denoting this by the parameter $\theta \equiv \mathbb{E}_\varphi(T(\mathbf{X}))$ and noting that your estimator is $\hat{\theta} \equiv T(\mathbf{X})$, you then clearly have:

$$\text{Bias}(\hat{\theta}, \theta) = \mathbb{E}_\varphi(T(\mathbf{X})) - \theta = \mathbb{E}_\varphi(T(\mathbf{X}))-\mathbb{E}_\varphi(T(\mathbf{X})) = 0.$$

So, it remains only to show that this estimator is efficient (i.e., it achieves the Cramér-Rao lower bound). To do this, we first observe that the Fisher information function for the distribution is:

$$\begin{align} \mathcal{I}(\varphi) &= -\mathbb{E}_\varphi \Bigg( \frac{d^2}{d\varphi^2} \log f_\varphi(\mathbf{x}) \Bigg) \\[6pt] &= -\mathbb{E}_\varphi \Bigg( \frac{d^2}{d\varphi^2} [c(\varphi) T(\mathbf{x}) + d(\varphi) + s(\mathbf{x})] \Bigg) \\[6pt] &= -\mathbb{E}_\varphi \Bigg( c''(\varphi) T(\mathbf{x}) + d''(\varphi) \Bigg) \\[12pt] &= - c''(\varphi) \mathbb{E}_\varphi(T(\mathbf{x})) + d''(\varphi) \\[12pt] &= \frac{c''(\varphi) d'(\varphi) - c'(\varphi) d''(\varphi)}{c'(\varphi)}. \\[6pt] \end{align}$$

(The final step here comes from substituting $\mathbb{E}_\varphi(T(\mathbf{x})) = - d'(\varphi)/c'(\varphi)$.) The corresponding information function for $\theta$ (still framed in terms of the parameter $\varphi$) is:

$$\begin{align} \mathcal{I}_\theta(\varphi) &= \mathcal{I}(\varphi) \Bigg/ \Bigg( \frac{d \theta}{d \varphi}(\varphi) \Bigg)^2 \\[6pt] &= \frac{c''(\varphi) d'(\varphi) - c'(\varphi) d''(\varphi)}{c'(\varphi)} \Bigg/ \Bigg( \frac{c''(\varphi) d'(\varphi) - c'(\varphi) d''(\varphi)}{c'(\varphi)^2} \Bigg)^2 \\[6pt] &= \frac{c'(\varphi)^3}{c''(\varphi) d'(\varphi) - c'(\varphi) d''(\varphi)}. \\[6pt] \end{align}$$

Consequently, the efficiency of the estimator $\hat{\theta}$ is:

$$\begin{align} \text{Eff}(\hat{\theta}, \theta) &\equiv \frac{1}{\mathcal{I}_\theta(\varphi)} \Bigg/ \mathbb{V}(\hat{\theta}) \\[6pt] &= \frac{c''(\varphi) d'(\varphi) - c'(\varphi) d''(\varphi)}{c'(\varphi)^3} \Bigg/ \frac{c''(\varphi) d'(\varphi) - c'(\varphi) d''(\varphi)}{c'(\varphi)^3} \\[6pt] &= 1. \\[6pt] \end{align}$$

This establishes that the estimator $\hat{\theta}$ is an efficient estimator for $\theta$ (i.e., it achieves the Cramér-Rao lower bound), which completes the demonstration.

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