= Solution
In a <parametric statistical model> $\{P_\theta:\theta\in\Theta\}$, the <likelihood function> $L(\theta;y)$ measures the relative support given by the observed sample for different parameter values. <Maximum likelihood estimation> supplies a point estimate; the sampling distributions of suitable <statistics> calibrate significance tests and <confidence intervals>. A <sufficient statistic> retains all parameter-dependent information in the <likelihood function>, so one first seeks a reduction to such a <statistic> rather than discarding information arbitrarily.
The characteristic conditioning principle in <Fisherian conditional inference> is to calibrate inference conditional on the observed value of an <ancillary statistic>. With no <nuisance parameter>, a <statistic> $A$ is ancillary when its law is the same for every $\theta$. Its observed value can nevertheless describe the experimental configuration or precision. Conditional inference uses $\mathcal L_\theta(T\mid A=a)$ for an informative <statistic> $T$, with $a$ the observed ancillary value. Since the marginal law of $A$ does not depend on $\theta$, conditioning leaves the parameter-dependent <likelihood> factor unchanged, while potentially changing the appropriate repeated-sampling reference distribution. Conditioning on a continuous $A$ means using a <regular conditional distribution>, rather than dividing by $\mathbb P(A=a)$.
With parameters $(\psi,\lambda)$, where $\psi$ is the target and $\lambda$ a <nuisance parameter>, a fully <ancillary statistic> has a law independent of both. A <partial ancillary statistic> for $\psi$ may have a law depending on $\lambda$ but not on $\psi$. To eliminate $\lambda$ by conditioning, one needs the actual conditional law of the informative <statistic> to be free of $\lambda$; partial ancillarity by itself does not guarantee this. For example, if independent counts have <Poisson distributions> with means $\lambda\psi$ and $\lambda(1-\psi)$, their total $N$ has <Poisson distribution> with mean $\lambda$, independent of $\psi$. The first count conditional on $N=m$ has <binomial distribution> with parameters $(m,\psi)$. Conditioning removes the nuisance intensity $\lambda$ exactly. More generally, nuisance-free <pivotal quantities> or <conditional likelihoods> can be used where available; there is no universal conditioning device that removes every <nuisance parameter>.
For a concrete example of <non-uniqueness of maximal ancillary statistics>, consider one observed pair $(U,V)\in\{0,1\}^2$, with
$$
P_\theta(0,0)=P_\theta(1,1)=\frac\theta2,\qquad
P_\theta(0,1)=P_\theta(1,0)=\frac{1-\theta}{2},\qquad0<\theta<1.
$$
Both $U$ and $V$ have <Bernoulli distribution> with success probability $1/2$, independent of $\theta$, so each is an <ancillary statistic>. Their generated partitions are different and incomparable. Moreover, each is maximal: refining either two-point cell singles out a point whose probability is $\theta/2$ or $(1-\theta)/2$, which is parameter-dependent. Their joint <statistic> is not ancillary, since $P_\theta(U=V)=\theta$. Thus there is no common finer ancillary partition that reconciles the two choices. The conditional laws differ as labelled experiments: $P_\theta(V=U\mid U)=\theta$ and $P_\theta(U=V\mid V)=\theta$, but they condition on different observed information. This illustrates why “condition on a maximal ancillary” need not specify a unique procedure.
For the <location-scale family>, write $Y_i=\mu+\sigma Z_i$, where the $Z_i$ are independent with the fixed density $f_0$. Let $s_Z^2=n^{-1}\sum_i(Z_i-\bar Z)^2$. The transformation gives
$$
\widehat\mu=\bar Y=\mu+\sigma\bar Z,\qquad
\widehat\sigma=\sigma s_Z,
$$
and therefore
$$
\boxed{A_i=\frac{Y_i-\widehat\mu}{\widehat\sigma}
=\frac{Z_i-\bar Z}{s_Z}.}
$$
For $n\geq2$, $s_Z>0$ <almost surely>, because an independent sample from a density has probability zero of all observations coinciding. The right side depends only on a sample from the fixed base law, so the entire vector $A$ has a law independent of $(\mu,\sigma)$: $\boxed{A\text{ is ancillary}.}$ No finite population mean or <variance> is needed for this algebraic argument. With one observation the displayed standardization is undefined, so at least two observations are implicitly required.
The pair $(\widehat\mu,\widehat\sigma)$ together with $A$ reconstructs the full sample through $Y_i=\widehat\mu+\widehat\sigma A_i$. Thus <conditional inference in a location-scale family> uses
$$
\boxed{\mathcal L_{\mu,\sigma}(\widehat\mu,\widehat\sigma\mid A=a),}
$$
where $a$ is the observed residual configuration. Equivalently one can use the parameter-free conditional law of $((\widehat\mu-\mu)/\sigma,\widehat\sigma/\sigma)$ given $A=a$. To see this explicitly, the centered residual subspace has dimension $n-1$, so its radial coordinate $t=\widehat\sigma$ contributes a <Jacobian determinant> factor proportional to $t^{n-2}$. The <conditional density> of $\widehat\mu=m$, $\widehat\sigma=t>0$ is therefore proportional to
$$
t^{n-2}\sigma^{-n}\prod_{i=1}^nf_0\!\left(\frac{m+ta_i-\mu}{\sigma}\right).
$$
Setting $u=(m-\mu)/\sigma$ and $v=t/\sigma$ gives a <conditional density> proportional to $v^{n-2}\prod_i f_0(u+va_i)$, with no unknown parameter. The normalizing factor can depend on $a$, but not on $\mu$ or $\sigma$. These statements concern almost every attainable ancillary value.
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