Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 1 / 19H (Statistics)

A sufficient statistic has a conditional distribution of the full data given its value that is independent of the unknown parameter. A minimal sufficient statistic is sufficient and is a function of every other sufficient statistic, up to null sets. For a dominated family with common positive support, the likelihood-ratio criterion for minimal sufficiency states that $T$ is minimal sufficient if $T(x)=T(y)$ holds exactly when $p_{\theta }(x)/p_{\theta }(y)$ is independent of $\theta$.

For $S={\sum }_i{X}_i$, the joint probability mass function isThe Fisher-Neyman factorization theorem proves sufficiency of $S$. For two samples, the likelihood ratio is $(1-1/\theta {)}^{S(x)-S(y)}$, independent of $\theta >1$ exactly when their sums agree. Thus $S$ is minimal sufficient. Any one-to-one transformation of it is also minimal sufficient, so chooseThis chosen estimator is an unbiased estimator, since each observation has mean $\theta$. If instead the statistic is reported as $S$, its expectation is $n\theta$, and as an estimator of $\theta$ it has bias $(n-1)\theta$ (zero only when $n=1$). Being a minimal sufficient statistic is unchanged by this rescaling; being an unbiased estimator is not.

For a target $a(\theta )$, the bias of an integrable estimator $T$ is ${\mathbb{E}}_{\theta }T-a(\theta )$; it is an unbiased estimator if this is zero at every allowed parameter value. Here $Y$ has a zero-truncated Poisson distribution, $\theta >0$, and $p=1-e^{-\theta }\in (0,1)$. Write an estimator based only on $Y$ as $t(Y)$, assuming a finite expectation for every $\theta >0$. Unbiasedness requiresAbsolute integrability at every positive parameter ensures that the power series on the left converges absolutely on every complex disc. Uniqueness of coefficients in a power series therefore gives $t(y)=0$ for odd $y$ and $t(y)=2$ for positive even $y$. Conversely these values give the displayed identity, sois the unique unbiased estimator of this form. The uniqueness claim concerns nonrandomized functions of $Y$; allowing external randomness would permit addition of independent mean-zero noise.

For this parity estimator for a zero-truncated Poisson count, usefulness depends on the loss, but under squared-error loss it performs poorly despite being an unbiased estimator. Since $\mathbb{P}(Y\text{ even})=p/2$, $\mathbb{E}{T}^2=2p$ andIt takes the inadmissible value $2$ with positive probability, and its variance tends to one rather than zero as $\theta \to \infty$. Clipping to the parameter interval gives $T_c=\min (T,1)=1_{\{Y\text{ even}\}}$. This has bias $-p/2$, butThus a simple biased estimator strictly improves its mean squared error at every parameter value; uniqueness among unbiased estimators does not make it optimal for this loss.