Why Selection Changes the Likelihood
A response model describes outcomes in a source population. A sampling design may then make some outcomes more likely than others to enter the observed sample. When inclusion depends on the response, the response distribution among sampled units is generally not the source distribution. The correct likelihood for selected responses must account for both the source model and the probability of being selected.
The key operation is simple but consequential: multiply each source response probability by its selection probability, then normalize across all possible responses. This tutorial develops that operation and its likelihood consequences. It distinguishes a known selection mechanism from an unknown one; knowing the source model alone does not make the selection mechanism known.
The normalizing constant \(Z_\theta(x)\) is the source-population probability of inclusion conditional on \(X=x\). Responses with selection probability zero cannot appear in the selected sample, even if they have positive probability in the source population. The adjusted distribution is therefore supported only where both the source model and the selection mechanism assign positive probability.
Deriving the Selected-Response Distribution
The formula follows directly from conditional probability. For any response value \(y\) with positive selection probability, $$ P(Y=y\mid X=x,S=1) =\frac{P(Y=y,S=1\mid X=x)}{P(S=1\mid X=x)}. $$ Under the sampling mechanism, the numerator is \(f_\theta(y\mid x)s(y,x)\), while the denominator is the sum of that quantity over all possible responses. The result is a probability distribution because its probabilities sum to one.
Proof. By the definition of conditional probability, $$ P(Y=y\mid X=x,S=1) =\frac{P(Y=y,S=1\mid X=x)}{P(S=1\mid X=x)}. $$ Conditioning first on \(Y=y\), the numerator is $$ P(Y=y\mid X=x)P(S=1\mid Y=y,X=x) =f_\theta(y\mid x)s(y,x). $$ Summing over the response support gives $$ P(S=1\mid X=x) =\sum_u P(Y=u\mid X=x)P(S=1\mid Y=u,X=x) =\sum_u f_\theta(u\mid x)s(u,x) =Z_\theta(x). $$ The assumed positivity makes the conditional probability well-defined. Substitution gives the stated formula. Moreover, $$ \sum_y f_{\theta,S}(y\mid x,S=1) =\frac{\sum_y f_\theta(y\mid x)s(y,x)}{Z_\theta(x)} =1, $$ so the formula is normalized. \(\square\)
If \(n\) selected observations have responses \(y_1,\ldots,y_n\) and covariates \(x_1,\ldots,x_n\), and the units are independent conditional on their covariates and selection, the conditional likelihood is $$ L_S(\theta)=\prod_{i=1}^{n} \frac{f_\theta(y_i\mid x_i)s(y_i,x_i)} {\sum_u f_\theta(u\mid x_i)s(u,x_i)}. $$ This is a likelihood for the selected responses conditional on the observed covariates. It is not automatically a likelihood for the source covariate distribution, nor does it include information from the number of units that were not selected.
If selection probabilities are known, they enter this expression as specified quantities. If they depend on unknown parameters \(\gamma\), the conditional likelihood instead contains \(s(y_i,x_i;\gamma)\) and a normalizing constant that also depends on \(\gamma\). Treating an unknown selection mechanism as if it were known changes the likelihood and can change what is identifiable.
Worked Examples
Worked Example: A Binary Response with Known Selection Probabilities
In a source population, let \(Y=1\) denote disease and suppose \(P(Y=1)=0.25\). Assume that a person with disease is selected with probability \(s(1)=0.8\), while a person without disease is selected with probability \(s(0)=0.2\). Find the disease probability among selected people.
The unnormalized selected probability for a person with disease is $$ P(Y=1)s(1)=0.25(0.8)=0.20. $$ For a person without disease it is $$ P(Y=0)s(0)=0.75(0.2)=0.15. $$ Thus the overall selection probability is $$ Z=0.20+0.15=0.35. $$ The disease probability among selected people is consequently $$ P(Y=1\mid S=1)=\frac{0.20}{0.35} =\frac{4}{7}\approx 0.5714. $$ Selection has raised the disease proportion from \(0.25\) in the source population to about \(0.5714\) in the selected sample. The increase follows from the larger inclusion probability for people with disease; it does not mean that the source disease probability was \(0.5714\).
Worked Example: Selection Across Three Response Categories
Suppose a source response \(Y\) has categories A, B, and C, with probabilities \(0.50\), \(0.30\), and \(0.20\). The known selection probabilities are \(0.20\), \(0.50\), and \(1.00\), respectively. Find the response distribution among selected units.
Multiply the source probability of each category by its selection probability: $$ \begin{aligned} \text{A: }&0.50(0.20)=0.10,\\ \text{B: }&0.30(0.50)=0.15,\\ \text{C: }&0.20(1.00)=0.20. \end{aligned} $$ The normalizing constant is \(Z=0.10+0.15+0.20=0.45\). Therefore, $$ \begin{aligned} P(Y=\mathrm{A}\mid S=1)&=\frac{0.10}{0.45}=\frac{2}{9},\\ P(Y=\mathrm{B}\mid S=1)&=\frac{0.15}{0.45}=\frac{1}{3},\\ P(Y=\mathrm{C}\mid S=1)&=\frac{0.20}{0.45}=\frac{4}{9}. \end{aligned} $$ As a check, \(2/9+1/3+4/9=2/9+3/9+4/9=1\). Category C rises from a source probability of \(0.20\) to a selected probability of \(4/9\), because it has the largest selection probability.
Worked Example: A Poisson Response Selected Only When Positive
Let \(Y\mid X=x\) have a Poisson distribution with mean \(\lambda(x)>0\), so that $$ f(y\mid x)=e^{-\lambda(x)}\frac{\lambda(x)^y}{y!}, \qquad y=0,1,2,\ldots. $$ Suppose units are included if and only if \(Y>0\). Derive the response mass function among included units.
Here \(s(0,x)=0\) and \(s(y,x)=1\) for \(y\geq 1\). The selection probability is $$ Z(x)=P(Y>0\mid X=x) =1-P(Y=0\mid X=x) =1-e^{-\lambda(x)}. $$ For \(y\geq 1\), divide the source probability by this selection probability: $$ P(Y=y\mid X=x,S=1) =\frac{e^{-\lambda(x)}\lambda(x)^y/y!} {1-e^{-\lambda(x)}}. $$ At \(y=0\) the selected probability is zero. The denominator is positive because \(\lambda(x)>0\), and summing the displayed mass function over \(y\geq1\) gives $$ \frac{\sum_{y=1}^{\infty}e^{-\lambda(x)}\lambda(x)^y/y!} {1-e^{-\lambda(x)}} =\frac{1-e^{-\lambda(x)}}{1-e^{-\lambda(x)}}=1. $$ Thus a Poisson source response observed only when positive has a zero-truncated Poisson distribution. The likelihood for selected responses uses this truncated mass, not the unmodified Poisson mass.
What Changes When Selection Probabilities Are Unknown?
An unknown selection probability cannot simply be inserted into the adjustment as though its value were available. It must be modeled, and the observed data must contain enough information to distinguish the source response parameters from the selection parameters. Selected responses alone often identify only a combination of the two.
Worked Example: Unknown Binary Selection and Nonidentification
Let the source disease probability be \(p=P(Y=1)\), and write \(a=P(S=1\mid Y=1)\) and \(b=P(S=1\mid Y=0)\). The selected disease probability is $$ q=P(Y=1\mid S=1) =\frac{ap}{ap+b(1-p)}. $$ Suppose the selected sample has disease proportion \(q=0.60\), but neither \(p\) nor \(a,b\) is known. The selected-data likelihood, apart from its binomial coefficient, depends on \(p,a,b\) through \(q\): $$ L(p,a,b)\ \propto\ q^k(1-q)^{n-k}, $$ where \(k\) is the number of selected people with disease among \(n\) selected people.
Different source and selection values can give the same \(q\). For example, \(p=0.50\), \(a=0.60\), and \(b=0.40\) give $$ q=\frac{0.60(0.50)}{0.60(0.50)+0.40(0.50)} =\frac{0.30}{0.50}=0.60. $$ A different set, \(p=0.25\), \(a=0.90\), and \(b=0.20\), gives $$ q=\frac{0.90(0.25)}{0.90(0.25)+0.20(0.75)} =\frac{0.225}{0.225+0.15}=0.60. $$ Both parameter sets imply the same selected response distribution. The selected responses alone therefore do not distinguish these source prevalences and selection probabilities.
Now suppose an external source-population study establishes \(p=0.25\), while the selected proportion remains \(q=0.60\). The relative selection probability can then be recovered: $$ 0.60=\frac{0.25a}{0.25a+0.75b} \quad\Longrightarrow\quad 0.60(0.25a+0.75b)=0.25a. $$ Thus \(0.15a+0.45b=0.25a\), so \(0.45b=0.10a\), and $$ \frac{a}{b}=\frac{0.45}{0.10}=4.5. $$ This identifies the ratio \(a/b\), not the two absolute probabilities separately. For instance, \(a=0.90,b=0.20\) has that ratio and is a valid pair, but multiplying both probabilities by the same positive constant, when they remain at most one, leaves the ratio and \(q\) unchanged. Additional information is needed to identify more than the ratio.
Likelihood Interpretation and Practical Limits
For known selection probabilities, the selected-data likelihood is built from the adjusted response distribution. In a model with unknown selection parameters, \(\theta\) and \(\gamma\), each observation contributes $$ \frac{f_\theta(y_i\mid x_i)s(y_i,x_i;\gamma)} {\sum_u f_\theta(u\mid x_i)s(u,x_i;\gamma)}. $$ The denominator matters: it depends on the source model and, in general, on the selection parameters. Dropping it does not give the conditional distribution of responses among selected units.
This conditional likelihood answers a particular question: how are responses distributed among included units with the observed covariates? A different data collection setup may also record the number of selected and unselected units in a known source frame, or provide external information about the source response distribution or selection rates. Those data can contribute additional likelihood terms. Which terms are justified depends on how the data were collected; the selected-response likelihood alone does not contain information that was never observed.
There is also a useful special case. If, at a fixed \(x\), the selection probability is constant across response values, then it cancels from the numerator and denominator: $$ \frac{f_\theta(y\mid x)c}{\sum_u f_\theta(u\mid x)c} =f_\theta(y\mid x), \qquad c>0. $$ Selection that varies across covariate values but not across responses at each fixed covariate value therefore does not alter this conditional response distribution. By contrast, response-dependent selection generally does.
Common Mistakes
- Using the source response probabilities unchanged. If selection depends on the response, the sampled response distribution is tilted by the selection probabilities.
- Multiplying without normalizing. The products \(f_\theta(y\mid x)s(y,x)\) are unnormalized weights; their sum is the probability of selection and must be used as the denominator.
- Assuming unknown selection is harmless. If source and selection parameters both vary freely, the selected distribution may not identify them separately. Extra data or justified constraints are required.
- Confusing the selected proportion with the source proportion. In the binary example, \(0.5714\) is the selected disease probability, while the source probability is \(0.25\).
- Ignoring the data collection scheme. A likelihood conditional on selected responses is not automatically a likelihood for unobserved units or for the covariate distribution.
Check Your Understanding
Use the normalization step explicitly in each calculation.
- A binary response has source probability \(P(Y=1)=0.10\). Selection probabilities are \(0.50\) for \(Y=1\) and \(0.10\) for \(Y=0\). What is \(P(Y=1\mid S=1)\)?
- In a three-category response model, why do the products of source probabilities and selection probabilities need to be divided by their sum?
- For the positive-only Poisson example, what is the selection probability, and why does the selected response distribution assign zero probability to \(Y=0\)?
- If the source disease probability and both binary selection probabilities are unknown, what feature of the selected-data likelihood explains the identification problem?
- At a fixed covariate value, under what condition does selection leave the conditional response distribution unchanged?