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A complete sufficient statistic is independent of every ancillary statistic. To prove this, let $S$ be complete and sufficient, let $A$ be ancillary, and take any bounded measurable function $h$. By sufficiency, $g(S)=E_{\theta }[h(A)\mid S]$ has a parameter-independent version; by ancillarity, $c=E_{\theta }h(A)$ is independent of $\theta$. Thus $E_{\theta }[g(S)-c]=0$ for every $\theta$. Completeness gives $g(S)=c$ almost surely. Choosing $h$ to be the indicator of each measurable event in the ancillary space proves independence. Completeness is essential: a sufficient statistic may itself contain a nonconstant ancillary, as in the minimum and maximum of a uniform location sample.

When an experiment's configuration is observed, inference conditions on the experiment actually performed rather than averaging over its possible configurations. Conditioning on an ancillary statistic is a central application; freedom from a nuisance parameter must still be checked in the resulting conditional distribution.

In a parametric statistical model, the likelihood function compares parameter values, while sampling distributions calibrate tests and confidence intervals. A sufficient statistic retains the parameter information. Conditioning on an observed ancillary statistic restricts that calibration to experiments with the same ancillary configuration, instead of averaging over configurations whose distribution is unrelated to the parameter. Since the ancillary's marginal law is parameter-free, conditional and unconditional likelihoods have the same parameter-dependent factor. With a nuisance parameter, a conditional law or pivotal quantity must remove it before inference for the parameter of interest can use a nuisance-free reference law.

In a location-scale family, $Y=\mu +\sigma Z$ for a fixed base probability distribution of $Z$, with $\mu \in \mathbb{R}$ and $\sigma >0$. Location translates observations, and scale multiplies their deviations. For an independent sample with a base density and at least two observations, the normalized residual vector $(Y_i-\bar{Y})/s$, where $s^2=n^{-1}{\sum }_i(Y_i-\bar{Y}{)}^2$, is an ancillary statistic. Indeed, $\bar{Y}=\mu +\sigma \bar{Z}$ and $s=\sigma s_Z$, making the residual vector a function of the base sample alone. This algebraic result needs no population moments.

A maximal ancillary statistic has no strictly finer ancillary partition of the sample space, but different maximal ancillary partitions can exist. Let $(U,V)$ take values in $\{0,1{\}}^2$, with probabilities $\theta /2$ on each equal pair and $(1-\theta )/2$ on each unequal pair, for $0<\theta <1$. Both $U$ and $V$ have the Bernoulli distribution with success probability $1/2$ for every $\theta$. Their joint law depends on $\theta$, so combining them is not ancillary. Each partition is maximal: splitting one of its two-point cells singles out a point whose probability depends on $\theta$. There is no common ancillary refinement, and conditioning on one ancillary choice need not specify a unique conditioning principle for all models.

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.