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Consider an approximate conditional likelihood in which a paired total $t_i$ leads to a structural zero with probability $\theta$, and otherwise a binomial distribution with success probability $p$. For an observed zero, the posterior cure weight is $q_i=\theta /[\theta +(1-\theta )(1-p{)}^{t_i}]$; for a positive count it is zero. Differentiating the log-likelihood gives the displayed equations at an interior maximum-likelihood estimator, with the weights evaluated at that estimator. Recomputing the weights and then updating the parameters is an expectation-maximization algorithm. Boundary fits must also be considered. These equations describe a conditional mixture: if cure is instead specified before two Poisson distributions are conditioned on their total, its conditional probability generally depends on the total and on the baseline nuisance rate. A zero observed in a short interval is not by itself proof of permanent biological cure.

For conditionally independent paired counts with means ${\lambda }_i$ and ${\lambda }_i{e}^{\beta }$, condition on every pair total $N_i=Y_{i0}+Y_{i1}$. The joint conditional likelihood removes all pair-specific rates, leaving the displayed independent binomial distributions. A zero-total pair contributes no information. With unrestricted fixed baseline rates, this conditional likelihood has the same parameter-dependent part as the ordinary profile likelihood; a specified Gamma–Poisson hierarchical model can retain additional information by imposing structure on those rates.

A conditional likelihood is the conditional density of the observed data given a chosen statistic, viewed as a function of the parameter. Conditioning can eliminate a nuisance parameter if the statistic is sufficient for it within the relevant family. The conditioning statistic need not be ancillary; conditioning can also discard some information about the parameter of interest.

For the canonical family, let $T_1={\sum }_i{\tau }_1(Y_i)$ and $s=n^{-1}{\sum }_i{\tau }_2(Y_i)$. Normalization of the density gives the cumulant-generating function of one nuisance statistic:For the sample mean $S$, it is $K_S(t)=n[d(\psi ,\lambda +t/n)-d(\psi ,\lambda )]$. Differentiation gives $d_{\lambda }=\mathbb{E}{\tau }_2(Y_i)$ and $d_{\lambda \lambda }=\mathrm{Var}({\tau }_2(Y_i))$.

Assume an interior saddle with positive curvature and the usual smooth nonlattice density conditions. Define ${\hat{\lambda }}_{\psi }$ byThis is the constrained maximum-likelihood estimator of $\lambda$ with $\psi$ held fixed. The exponential tilt parameter for one summand is $\hat{t}={\hat{\lambda }}_{\psi }-\lambda$. Tilting puts the sample mean at $s$, where its tilted density has leading normal value $\sqrt{n/[2\pi d_{\lambda \lambda }(\psi ,{\hat{\lambda }}_{\psi })]}$. Undoing the tilt gives the saddlepoint density approximationThis is the density of the mean, not the sum; their factors differ by $n$.

The sample log-likelihood is $\ell (\psi ,\lambda )=\psi T_1+n\lambda s-nd(\psi ,\lambda )-{\sum }_{i}Q(Y_i)$. Conditional on $S=s$, the factor involving $\lambda$ is constant on the conditioning surface, so the conditional distribution is free of $\lambda$. Subtracting the saddlepoint log-density from the joint log-likelihood now cancels all the original nuisance terms:The final term, and any conditioning-coordinate factor, are data-only constants. Thus the saddlepoint conditional likelihood adjustment isThe absolute-value notation agrees with the question; positive variance makes it unnecessary in this regular setting. The sign of this correction is positive, because a nuisance-statistic density with inverse-square-root curvature has been divided out. A lattice-valued statistic requires the corresponding mass approximation, and a boundary or zero-variance saddle is outside the stated regular density calculation.

Let $A=R_{11}+2R_{12}+R_{22}$. Among the ten complete trios, six have two heterozygous parents: three have child $11$, two have child $12$ and one has child $22$. Their conditional likelihood contribution isTwo more trios have a heterozygous parent and a $22$ parent, with child $12$ in each; they contribute $[R_{12}/(R_{12}+R_{22}){]}^2$. The remaining two parent pairs can produce only child $12$, so their conditional probabilities are one. Multiplying givesThe factor $4$ does not depend on the genotype relative risks, so the conventional proportional likelihood function isThis is the expression in the original PDF, up to the parameter-independent constant. Its last denominator is $R_{12}+R_{22}$; the converted TeX's $R_{11}+R_{22}$ is a transcription error. The factor $4$ in the exact probability comes from the two heterozygous children of two heterozygous parents, each of which has two labelled transmission routes.

The profile likelihood for the parameter of interest replaces the nuisance parameter by its constrained best fit: $L_p(\psi )={\sup }_{\lambda }L(\psi ,\lambda )$, with ${\ell }_p(\psi )=\ell (\psi ,{\hat{\lambda }}_{\psi })$. It is not in general a normalized sampling density. A conditional likelihood instead uses the sampling law conditional on a statistic carrying the nuisance dependence. In the canonical exponential family here, conditioning on $S$ removes $\lambda$ because the factor involving $\lambda$ is constant over the conditioning fibre.

Let ${\bar{T}}_1=n^{-1}{\sum }_i{\tau }_1(y_i)$ and let the observed $S$ equal $s$. For the cumulant-generating function of ${\tau }_2(Y)$ at fixed $\psi$,The saddlepoint solves $K^{\prime }(\hat{t})=s$, equivalently $d_{\lambda }(\psi ,{\hat{\lambda }}_{\psi })=s$ with ${\hat{\lambda }}_{\psi }=\lambda +\hat{t}$. This is also the constrained likelihood equation, since ${\ell }_{\lambda }=n(s-d_{\lambda })$. Assume an interior saddle, $d_{\lambda \lambda }>0$, and the usual smooth nonlattice conditions under which $S$ has a density. Exponential tilting to ${\hat{\lambda }}_{\psi }$ makes $s$ the tilted mean. A local normal approximation gives tilted density $\{n/(2\pi d_{\lambda \lambda }(\psi ,{\hat{\lambda }}_{\psi })){\}}^{1/2}$ at that mean; undoing the tilt gives the saddlepoint density approximationThe factor $n^{1/2}$ is for the sample mean, rather than the sum. A transform-based derivation gives the same expression by expanding $K$ to second order at its stationary point.

The joint log-likelihood is $n\psi {\bar{T}}_1+n\lambda s-nd(\psi ,\lambda )-{\sum }_{i}Q(y_i)$. Subtracting $\log {\hat{f}}_S(s;\psi ,\lambda )$ cancels every appearance of the original $\lambda$, leavingAny change from the ambient density to a conditional density on $S=s$ contributes only a data-dependent Jacobian determinant. Thus, up to a constant independent of $\psi$, the saddlepoint conditional likelihood adjustment isEquivalently $B=\frac{1}{2}\log j_{\lambda \lambda }(\psi ,{\hat{\lambda }}_{\psi })$ up to the constant $-\frac{1}{2}\log n$, where $j_{\lambda \lambda }=n{d}_{\lambda \lambda }$ is the nuisance observed information. The adjustment has a positive sign. The printed exponential-family form alone does not guarantee a density for $S$: for example a continuous $Y$ can have ${\tau }_2(Y)=1_{\{Y>0\}}$. In a lattice case one must approximate the probability mass of the sum instead; the displayed density calculation uses the regular continuous interpretation.

Let $T_i=X_i+Y_i$. Since the two counts are independent Poisson distributions with means ${\lambda }_i$ and $\beta {\lambda }_i$, their total is a Poisson distribution with mean $(1+\beta ){\lambda }_i$. For a fixed observed total $t_i=x_i+y_i$, divide the joint probability mass function by that of the total:Define $p=\beta /(1+\beta )$, so $1-p=1/(1+\beta )$. ThenFor independent patients, the paired Poisson conditional likelihood isEvery nuisance rate ${\lambda }_i$ cancels, while each patient's own baseline count level is retained through the conditioning total. A patient with total zero would contribute the constant one.

For these data, ${\sum }_i{x}_i=161$, ${\sum }_i{y}_i=23$, and ${\sum }_i{t}_i=184$. Maximizing this conditional likelihood givesThese are the fit under the model without a cure component, not the estimates of the mixture model in the later parts.

Work with the stipulated approximate conditional likelihood. Let $Z_0=\{i:y_i=0\}$ and $Z_{+}=\{i:y_i>0\}$, and putTerms independent of the parameters, including the positive-count binomial coefficients, can be omitted. The log-likelihood isFor an interior maximum-likelihood estimator, set its two score functions to zero. Differentiating with respect to $\theta$ givesDifferentiating with respect to $p$ givesHere there are seven zero-count patients and five positive-count patients, with ${\sum }_{Z_{+}}{y}_i=23$ and ${\sum }_{Z_{+}}(t_i-y_i)=36$. Thus the positive-count contribution to the second equation is $23/p-36/(1-p)$.

An interpretable way to solve the equations uses a posterior cure weight for each observed patient:For a zero count, $1-q_i=(1-\theta )(1-p{)}^{t_i}/A_i$ is the posterior probability of belonging to the non-cured component under this approximate conditional model. The two score functions can then be rewritten asTherefore the conditional cure-weight equations for paired Poisson counts areThey can be solved by an expectation-maximization algorithm: calculate the weights at the current parameters, update $\theta$ to their mean and $p$ to the weighted event fraction, and iterate. A solution of the equations must be checked as a maximum; the equations alone do not guarantee a global maximum. In particular the boundary $\theta =0$ gives the no-cure estimate $p=1/8$ and should be compared. Boundaries $p=0$, $p=1$ or $\theta =1$ have zero likelihood for these positive counts with positive pre-treatment counts. Numerical maximization gives an interior fit about $(0.57549,0.38614)$, with larger log-likelihood than the no-cure boundary, consistent with the PDF's rounded estimates.

The word “approximate” is important. If cure is specified before conditioning in the original Poisson distributions, the conditional cure probability given $T_i=t_i$ would actually bewhich still depends on ${\lambda }_i$. This follows because a cured total has mean ${\lambda }_i$ and a non-cured total has mean $(1+\beta ){\lambda }_i$. The likelihood supplied in the question instead treats $\theta$ as the mixture probability after conditioning; the score and weight equations above maximize exactly that stipulated approximation, without claiming nuisance-rate elimination for the full cure model.

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 $(U,V)\in \{0,1{\}}^2$, withBoth $U$ 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 givesand thereforeFor $n\ge 2$, $s_Z>0$ 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 $(\hat{\mu },\hat{\sigma })$ together with $A$ reconstructs the full sample through $Y_i=\hat{\mu }+\hat{\sigma }{A}_i$. Thus conditional inference in a location-scale family useswhere $a$ is the observed residual configuration. Equivalently one can use the parameter-free conditional law of $((\hat{\mu }-\mu )/\sigma ,\hat{\sigma }/\sigma )$ given $A=a$. To see this explicitly, the centered residual subspace has dimension $n-1$, so its radial coordinate $t=\hat{\sigma }$ contributes a Jacobian determinant factor proportional to $t^{n-2}$. The conditional density of $\hat{\mu }=m$, $\hat{\sigma }=t>0$ is therefore proportional toSetting $u=(m-\mu )/\sigma$ and $v=t/\sigma$ gives a conditional density proportional to $v^{n-2}{\prod }_i{f}_0(u+v{a}_i)$, 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.

For pair $i$, use conditionally independent counts $Y_{i0}\sim \mathrm{Pois}({\lambda }_i)$ and $Y_{i1}\sim \mathrm{Pois}({\lambda }_i{e}^{\beta })$, with a common effect $\beta$ and a separate positive baseline rate for each physician. Equivalently the linear predictor is ${\alpha }_i+\beta x$, with ${\alpha }_i=\log {\lambda }_i$. This Poisson regression controls each physician's stable baseline propensity.

Condition on all twelve pair totals $N_i=Y_{i0}+Y_{i1}$. The paired Poisson conditional likelihood has independent factorsThis eliminates every ${\lambda }_i$ exactly. It is more useful as a nuisance-elimination device than the unpaired calculation, which removes only one common baseline: many pair-specific rates are otherwise estimated from just two measurements each.

A possible concern is loss of information through conditioning on totals that are not ancillary for $\beta$, especially with sparse pairs; a zero-total pair contributes no conditional information. There is an important qualification in this exact fixed-baseline Poisson model. Maximizing over ${\lambda }_i$ gives ${\hat{\lambda }}_i=N_i/(1+e^{\beta })$, and substitution leaves precisely the same $\beta$-dependent factor as the conditional likelihood. Hence there is no extra loss relative to ordinary profiling in this particular model; one should not automatically assert an incidental-parameter bias in its estimate of $\beta$.

Additional baseline observations, a calibrated distribution of baseline rates, or informative hierarchical constraints can retain information in the pair totals. A Gamma–Poisson hierarchical model is one possible structured model, but merely giving it a completely free common mean does not automatically create extra information about $\beta$: that mean can absorb the factor $1+e^{\beta }$ in the totals. Such remedies trade exact nuisance elimination for additional data or assumptions. Finally, paired sampling alone does not establish conditional Poisson independence or equidispersion: remaining within-physician dependence or overdispersion should be checked, and an appropriate joint count model or physician-level robust variance used if needed. The common multiplicative-effect assumption is also part of the analysis.

For independent aggregate Poisson distributions with means $n\lambda$ and $n\lambda e^{\beta }$, conditioning on the total eliminates $\lambda$. Independent Poisson conditioning gives the displayed binomial distribution. Its conditional likelihood is proportional to $e^{\beta Y_1}(1+e^{\beta }{)}^{-m}$, also the profile likelihood after maximizing over the unrestricted positive baseline rate. Unequal exposures replace the binomial success probability by $n_1{e}^{\beta }/(n_0+n_1{e}^{\beta })$.