Proof that $ I(θ) = E[(s(θ;X))^2] $
Recall that:
$ \begin{align} \bar Var(Y) &= E[(Y-E(Y))^2]\\ &= \int_a^b g(x) dx \\ &= \frac{\mu_0}{2 \pi a \cdot b} \end{align} $
Recall that:
$ \begin{align} \bar Var(Y) &= E[(Y-E(Y))^2]\\ &= \int_a^b g(x) dx \\ &= \frac{\mu_0}{2 \pi a \cdot b} \end{align} $