# When is Fisher Information the reciprocal of the variance?

I notice that for many common distributions (e.g., Bernoulli, Exponential), the Fisher Information is the same as the reciprocal of the population variance.

When is this true? Is there a theorem that states than if conditions X, Y, and Z are true, then we have that $$I(\theta ;X) = \frac{1}{\text{Var}(X)}$$

• I'm not seeing how Slutsky's theorem applies here...?
– max
Feb 21, 2021 at 21:40

Theorem: Consider a family of distributions $$\{ F_\theta | \theta \in \Theta \}$$. If the estimator $$\hat{\theta}(x) = x$$ (i.e., the identity estimator) is efficient, then we have:
$$\mathcal{I}(\theta) = \frac{1}{\mathbb{V}(X)}.$$
Proof: The variance of the identity estimator is $$\mathbb{V}(\hat{\theta}) = \mathbb{V}(X)$$. If the estimator is efficient then (by definition) it achieves the Cramér–Rao bound, so we then have:
$$\mathbb{V}(\hat{\theta}) = \frac{1}{\mathcal{I}(\theta)}.$$
Putting these two results together gives the required result. $$\blacksquare$$