Why Model the Distribution of Test Statistics?
A genomic analysis may produce thousands of test statistics, some from null hypotheses that are true and others from hypotheses with genuine effects. Looking at the full collection as if every statistic came from one common distribution obscures this distinction. A two-group mixture model represents the overall distribution as a weighted combination of a null component and an alternative component. Its central purpose here is to describe that marginal distribution; it does not, by itself, label any particular test as null or alternative.
The model is useful only when its pieces are stated clearly. The null proportion determines the mixture weights, while the component densities describe how test statistics behave within each group. The null component is tied to the null test-statistic distribution. The alternative component summarizes test-statistic behavior among non-null hypotheses, which may include many different effect sizes or directions.
Assumptions and Meaning of the Components
Each component density must be nonnegative and integrate to one over the support of the test statistic. The weight \(\pi_0\) is a proportion, not a density, and the two weights sum to one. In a pooled model, the tests are treated as draws from a population with a common null proportion and common component distributions. This is a modeling description of the collection of tests; it need not mean that genomic features were literally sampled at random from a biological population.
A common choice for \(f_0\) is the theoretical null distribution of the test statistic, provided the test-statistic construction and its assumptions justify that distribution. For example, a standard normal null density is appropriate for a statistic whose null distribution is \(N(0,1)\). The alternative density \(f_1\) is conditional on the null being false. It need not be a single distribution for one particular effect. It can summarize a range of effects, including effects in both directions, and may therefore be more dispersed or asymmetric than \(f_0\).
The mixture describes the marginal distribution of one statistic selected from the modeled collection. It does not imply that the statistics from different tests are independent. Genomic tests can be dependent, for example because of linkage disequilibrium, while their one-statistic marginal distribution still has a mixture representation. Independence would be an additional assumption needed for some joint-probability calculations, not a consequence of the mixture formula.
Deriving the Marginal Density
The derivation is an application of the law of total probability for densities. For a small interval around \(t\), the probability of observing a statistic in that interval is the sum of the probabilities contributed by the null and alternative groups. Dividing by the interval width and taking the limit gives the density formula.
The same reasoning applies to any measurable region \(A\) of test-statistic values. Integrating the density over that region gives the marginal probability:
This also verifies that \(g\) is a proper density. Since each component density integrates to one,
At a particular value \(t\), \(g(t)\) is a density, not the probability that \(T=t\). For a continuous statistic, the probability of exactly one value is zero; probabilities are obtained by integrating over an interval. Density values have units reciprocal to the units of \(T\), while probabilities are dimensionless.
Worked Example: Marginal Density at One Test-Statistic Value
Suppose, hypothetically, that \(80\%\) of the tests are null, with \(f_0\) the standard normal density, and that the alternative density is \(N(1.5,1)\). Find the marginal density at \(t=2.5\). Here the second parameter in \(N(\mu,\sigma^2)\) is the variance.
The component densities are \(f_0(2.5)=\phi(2.5)\approx0.0175283\) and \(f_1(2.5)=\phi(2.5-1.5)=\phi(1)\approx0.241971\), where \(\phi\) is the standard normal density. The mixture weight for the alternative is \(1-0.80=0.20\).
Substituting the unrounded component values into the mixture formula gives
Thus the marginal density at 2.5 is approximately \(0.06242\) per test-statistic unit. It is not a \(6.242\%\) probability of observing exactly 2.5; a probability requires an interval and integration.
Worked Example: Probability of a Test Statistic in an Interval
Continue with a hypothetical model having \(\pi_0=0.70\), \(f_0=N(0,1)\), and \(f_1=N(1.5,1)\). Find the marginal probability that \(1\leq T\leq2\).
Under the null, the interval probability is \(P(1\leq T\leq2\mid C=0)=\Phi(2)-\Phi(1)\approx0.9772499-0.8413447=0.1359052\). Under the alternative, standardizing by subtracting 1.5 gives \(P(1\leq T\leq2\mid C=1)=\Phi(0.5)-\Phi(-0.5)\approx0.6914625-0.3085375=0.3829250\).
Weight each component probability by its group proportion:
The model assigns approximately \(0.2100\), or \(21.00\%\), of the statistics to the interval from 1 to 2. The calculation checks out because it combines the null and alternative probabilities using weights that sum to one.
Alternative Components Can Combine Different Effects
A pooled alternative density need not be symmetric or unimodal. For example, if non-null hypotheses include positive and negative effects, the alternative distribution can combine statistics shifted in both directions. This lets the model describe heterogeneity without treating every non-null test as having the same effect.
Worked Example: A Two-Direction Alternative Component
Suppose a hypothetical collection has \(\pi_0=0.90\) and standard normal null density \(f_0\). Among the non-null tests, assume \(60\%\) have test statistics distributed as \(N(1.5,1)\) and \(40\%\) as \(N(-1.5,1)\). Find the marginal density at \(t=1.5\).
The pooled alternative density is the weighted average of its two subgroups: \(f_1(t)=0.60\phi(t-1.5)+0.40\phi(t+1.5)\). At \(t=1.5\),
The null density at that value is \(f_0(1.5)=\phi(1.5)\approx0.1295176\). Therefore,
The marginal density is approximately \(0.1407\) per test-statistic unit. Even though the alternative group contains two directions, its combined density enters the overall model as one component with total weight \(0.10\).
Checking the Model and Comparing Descriptions
A mixture model is a compact description, not a guarantee that its assumptions fit. Useful checks begin with the component definitions and then compare the implied marginal distribution with the observed collection of statistics. For example, if a theoretical null is standard normal, a central concentration that is markedly wider or shifted than expected may signal that the null model or the pooled assumptions need review. Such a discrepancy is a warning, not proof of a particular alternative explanation.
A pooled two-group model is simpler than assigning a separate null proportion and component densities to every test, but it can be misleading when tests differ systematically. Different test-statistic scales, covariate groups, or analysis strata may call for separate or stratified descriptions. The pooled model can also be compared with a direct description of the overall distribution alone: the latter may fit the observed histogram, but does not distinguish a null component from an alternative one.
That distinction may not be recoverable from the marginal density alone. Many combinations of \(\pi_0\), \(f_0\), and \(f_1\) can produce the same \(g\). Consequently, specifying a mixture formula does not establish that its components are uniquely identified or accurately estimated. Component assumptions and constraints matter, and estimates can be sensitive to them.
Other failure modes include using a theoretical null density when the statistics were computed under conditions that do not support that null distribution, treating a pooled alternative as though all effects were identical, or inferring independence from the mixture representation. A model that fits one portion of the distribution but misses another may be too restrictive or may combine unlike test groups. Conversely, a noisy histogram from a finite number of tests does not by itself show that the model is wrong; the discrepancy should be assessed with the sampling variability and dependence structure in mind.
The two-group description is most informative when the tested hypotheses have a defensible common null distribution, the null proportion is meaningful for the collection, and the alternative component can reasonably summarize the non-null statistics. It is a starting model for organizing the distribution, not a substitute for checking test construction, heterogeneity, or dependence.
Check Your Understanding
Try each question, then compare your reasoning with the solutions.
- If \(\pi_0=0.65\), write the marginal density in terms of \(f_0\) and \(f_1\). What is the alternative proportion?
- Suppose \(P(T\in A\mid C=0)=0.10\), \(P(T\in A\mid C=1)=0.40\), and \(\pi_0=0.80\). Find \(P(T\in A)\).
- Does the two-group mixture formula imply that statistics from different genomic tests are independent? Explain.
- Why is \(g(t)\) not the probability that a continuous test statistic equals \(t\)?
Solutions. (1) \(g(t)=0.65f_0(t)+0.35f_1(t)\); the alternative proportion is \(0.35\). (2) By weighting the conditional probabilities, \(P(T\in A)=0.80(0.10)+0.20(0.40)=0.08+0.08=0.16\). (3) No. The formula describes the marginal distribution of one statistic. Dependence across tests is a separate feature of their joint distribution. (4) \(g(t)\) is a density; for a continuous statistic, probability is obtained by integrating over a region, and the probability at one exact value is zero.