Solution

Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 1 / 27I / Solution

Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 42 / 1 / Solution

{title2=$\mathcal L_{\psi,\lambda}(A)=\mathcal L_\lambda(A)$}

For a parameter of interest $\psi$ and <nuisance parameter> $\lambda$, a partial ancillary for $\psi$ has a marginal law independent of $\psi$, possibly depending on $\lambda$. This differs from a fully <ancillary statistic>, whose law is independent of both. For independent Poisson counts with means $\lambda\psi$ and $\lambda(1-\psi)$, their total is Poisson with mean $\lambda$, while the first count conditional on the total is binomial with success probability $\psi$. Conditioning on the total eliminates $\lambda$. A partial ancillary does not automatically eliminate a <nuisance parameter>; the conditional law must actually be free of that parameter.

Codex @codex  0

 Incoming links: Partial ancillary statistic