Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 29L (Principles of Statistics)

The delta method givesIf ${\phi }^{\prime }(\psi )=0$, this denotes the degenerate distribution at zero.

The log likelihood isIts strictly concave maximizer is

Since $\mathbb{E}{X}_i=1/\theta$ and $\mathrm{Var}(X_i)=1/{\theta }^2$, the central limit theorem givesApplying the delta method to $g(x)=1/x$ yieldsThus, with $z_{1-\alpha /2}={\Phi }^{-1}(1-\alpha /2)$, an asymptotic interval centered at the MLE isoptionally intersected with $(0,\infty )$. Slutsky's theorem gives coverage tending to $1-\alpha$.

For $y=0,1,\dots$,so $Y_i$ is geometric with success probability $1-e^{-\theta }$. The log likelihood isIts score equation givesand henceWhen $\overline{Y}=0$, the extended MLE is $+\infty$; this event has probability tending to zero.

Put $r=e^{\theta }$. The geometric moments areFor $g(y)=\log ((1+y)/y)$, one has $g(m)=\theta$ andThe central limit theorem for $\overline{Y}$ followed by the delta method therefore gives