I have, say, $Y_i \stackrel{iid}{\sim} \mathcal{B}(1, p)$ and I want to calculate the lower bound of variance of an estimator of $\theta = p(1 - p)$ (which happens to be $Var[Y_i]$).
If $\theta = p$, this would be straightforward:
- Calculate log-likelihood $\mathcal{l}(p | \vec{Y}) = \log f_{\vec{Y}}(\vec{y})$;
- Calculate Fisher Information $I(p) = - E \left[ \frac{\partial^2}{\partial p^2} \mathcal{L}(p | \vec{Y}) \right]$
- Voila: $Var\left[\hat{p}\right] \geq 1 / I(p)$
For me $\theta = p(1 - p)$ is less clear, because I don't see how to write $f_{\vec{Y}}(\vec{y})$ as a function of $\theta$, which I think I would need to do in order to take the partial derivatives w.r.t $\theta$ later on.
I think I've read elsewhere that if $\theta = g(p)$ (as is the case here), then $Var[\hat{\theta}] = g'(p)^2 / I(p)$. But I can't find this now, and even if I could, I don't understand why it would be true.