Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 4 / 28L (Principles of Statistics)

$I(\theta )=E_{\theta }[({\partial }_{\theta }\log f(X;\theta ){)}^2]$. The biased Cramer-Rao bound bound is ${\mathrm{Var}}_{\theta }\delta \ge (1+b^{\prime }(\theta ){)}^2/I(\theta )$.

$R(\delta ,\theta )=E_{\theta }[(\delta -\theta {)}^2]={\mathrm{Var}}_{\theta }\delta +b(\theta {)}^2$.

Combining the risk assumption with Cramer-Rao bound gives $(1+b^{\prime }{)}^2\le \gamma -I{b}^2$. Taking the branch compatible with super-efficiency yields $b^{\prime }\le -[1-\sqrt{\gamma -I{b}^2}]\le -(1-\sqrt{\gamma })$, equivalent to the displayed sharper pointwise inequality.

Integrating gives $b({\theta }_2)-b({\theta }_1)\le -(1-\sqrt{\gamma })L$. Risk at the endpoints implies $|b({\theta }_i)|\le \sqrt{\gamma }$, so the left side is at least $-2\sqrt{\gamma }$. Hence $L\le 2\sqrt{\gamma }/(1-\sqrt{\gamma })$.

For ${\delta }_c=cX$, $R=c^2+(1-c{)}^2{\theta }^2$. Thus $I_{\gamma }=[-\sqrt{\gamma -c^2}/(1-c),\sqrt{\gamma -c^2}/(1-c)]$, with the stated length. Substitution (or elementary algebra) verifies it is at most the bound in part (d).