Polygenicity as a Source of Test-Statistic Inflation

Why Test Statistics Can Rise Without Confounding

A mixed-model GWAS is designed to account for covariance among individuals, including covariance associated with relatedness. When that adjustment is adequate, statistics for variants with no association should be calibrated under their null distributions. But a genome-wide scan contains more than null variants: many variants may have genuine, individually small effects on the phenotype. Their test statistics are then noncentral, and their presence can shift the overall distribution upward even if population confounding has been controlled.

This distinction matters because an elevated genome-wide distribution is not, by itself, evidence that the null model failed. In this tutorial, the central calculation is deliberately simple: it shows how genuine effects contribute to test statistics under an idealized, correctly calibrated model. Real scans also involve linkage disequilibrium (LD), estimated variance components, and varying information across markers, so the calculation is an explanation of the mechanism rather than a universal formula for every scan.

From a Genuine Effect to a Noncentral Statistic

Definition: A test statistic is central under a null hypothesis when it has its null distribution. For a one-degree-of-freedom association test, this is commonly a \(\chi^2_1\) distribution. Under a genuine effect, the statistic is often approximately noncentral: its distribution is shifted, with noncentrality parameter \(\delta\geq 0\). For a signed statistic \(Z_j\) that is approximately \(N(\mu_j,1)\), the squared statistic \(Z_j^2\) has a noncentral chi-square distribution with one degree of freedom and noncentrality \(\delta_j=\mu_j^2\).

The sign of an effect determines the direction of the signed statistic, but not the noncentrality of its square. If \(Z_j\sim N(\mu_j,1)\), then

E(Z_j^2)=\operatorname{Var}(Z_j)+[E(Z_j)]^2=1+\mu_j^2=1+\delta_j.

Thus, under this approximation, a genuine effect increases the expected squared statistic above its central-null expectation of 1. This does not mean every affected marker will have a large observed statistic: sampling variation remains, and small effects can produce test statistics close to the null range.

For a simple illustrative calculation, let \(X_j\) be a standardized genotype and \(Y\) a standardized quantitative phenotype, with marginal correlation \(a_j\). In a large-sample single-marker regression or correlation test, the signed statistic has the approximate mean

\mu_j \approx \frac{\sqrt{N}\,a_j}{\sqrt{1-a_j^2}}, \qquad \delta_j \approx \frac{N a_j^2}{1-a_j^2}.

Here \(N\) is the sample size, and \(a_j\) is signed: it can be positive or negative. Consequently, \(\mu_j\) is positive when \(a_j>0\) and negative when \(a_j<0\). Equivalently, when \(a_j\ne 0\), \(\mu_j\approx\operatorname{sign}(a_j)\sqrt{\delta_j}\). The squared statistic has noncentrality \(\delta_j\) in either direction. These expressions rely on the simple standardized single-marker approximation; in a mixed-model scan, the corresponding noncentrality depends on the test’s information and covariance model.

Worked Example: One Small Effect in a Large Sample

Suppose a hypothetical scan has \(N=10{,}000\) individuals. At one variant, the standardized marginal effect is \(a_j=-0.02\). Assume the single-marker approximation applies and the null test is correctly calibrated.

The signed-statistic mean is

\mu_j \approx \frac{\sqrt{10{,}000}(-0.02)}{\sqrt{1-(-0.02)^2}} = \frac{100(-0.02)}{\sqrt{0.9996}} \approx -2.000.

The noncentrality is the square of this mean:

\delta_j\approx\frac{10{,}000(0.02)^2}{1-0.02^2} =\frac{4}{0.9996}\approx 4.002.

Therefore the expected squared statistic is approximately \(1+4.002=5.002\). The negative effect gives a negative expected signed statistic, about \(-2.000\), while the squared statistic’s noncentrality is positive. The calculation describes a genuine association; it does not require population confounding.

Why Many Small Effects Matter Across a Genome

A polygenic architecture has effects spread across many variants. Each effect may be too small to create an extreme statistic on its own, yet the collection of affected tests can shift the genome-wide distribution. To see the arithmetic cleanly, consider an idealized scan in which markers are independent, all tests have the same sample size, and the phenotype is standardized. Suppose \(m\) of \(M\) tested markers have marginal effects, and the other \(M-m\) markers are null. For each affected marker, let its statistic have noncentrality \(\delta_j\); for each null marker, its statistic has expectation 1. The average expected squared statistic is then

\frac{1}{M}\sum_{j=1}^{M} E(Z_j^2) = 1+\frac{1}{M}\sum_{j:\,\text{effect}} \delta_j.

This average increases with the number of affected tests and with their noncentralities. It is not a formula for the median-based genomic inflation factor \(\lambda\): that factor is based on the observed median, and its value depends on the full distribution of statistics. The average calculation instead isolates the basic reason that many non-null tests can raise the overall scale.

In an especially simple independent-marker model, write \(Y=\sum_k\beta_k X_k+\varepsilon\), where the \(X_k\) are independent standardized genotypes, the phenotype has variance 1, and \(\varepsilon\) is independent noise. The genetic contribution to variance is \(\sum_k\beta_k^2\). If effects are distributed over many loci, that total can be appreciable even when each \(\beta_k\) is small. Increasing sample size also increases the noncentrality for a fixed nonzero effect, approximately in proportion to \(N\). Therefore larger studies can show more visible departures from a central-null distribution because they have more power to detect genuine small effects, not only because they are more vulnerable to confounding.

Worked Example: Many Small Effects Shift the Average

Consider a hypothetical scan of \(M=1{,}000\) independent standardized markers in \(N=50{,}000\) individuals. Assume 100 markers each have a marginal effect of magnitude \(0.01\), and the other 900 are null. Assume the single-marker approximation is appropriate and all null tests are calibrated.

For each affected marker,

\delta_j\approx \frac{50{,}000(0.01)^2}{1-(0.01)^2} = \frac{5}{0.9999} \approx 5.001.

Each affected test therefore has expected squared statistic about \(1+5.001=6.001\); each null test has expectation 1. The average expected squared statistic across all 1,000 tests is

\frac{900(1)+100(6.001)}{1{,}000} = \frac{1{,}500.1}{1{,}000} \approx 1.500.

This average is above 1 even though, by assumption, the 900 null tests are properly calibrated and there is no population confounding. The example is intentionally simplified: correlated markers and heterogeneous information change the distribution. Its conclusion is narrower and important: many genuine non-null tests can raise a genome-wide summary of test-statistic magnitude.

Linkage Disequilibrium Spreads Signal

The independent-marker calculation is not a literal description of most genomes. With LD, a tested marker can correlate with a causal variant and show association even if it has no direct effect on the phenotype. If standardized marker \(X_j\) has correlation \(r_{jk}\) with causal marker \(X_k\), and the phenotype contains the term \(\beta_k X_k\), then this term contributes \(r_{jk}\beta_k\) to the marginal association at \(X_j\). With several causal variants, the marginal effect is the sum of these contributions, \(\sum_k r_{jk}\beta_k\), under the linear model. LD can therefore spread genuine signal across multiple tests; those tests are correlated, not independent replications of distinct causal effects.

Worked Example: A Noncausal Marker Tags a Causal Effect

Suppose two hypothetical standardized genotypes \(X_1\) and \(X_2\) have correlation \(r_{12}=0.5\). Only \(X_1\) has a direct effect, with standardized effect \(\beta_1=0.02\), and the phenotype is standardized with independent noise. In the large-sample marginal model, the association at \(X_2\) contributed by this causal effect is

a_2=r_{12}\beta_1=(0.5)(0.02)=0.01.

For \(N=20{,}000\), the approximate noncentralities at the causal marker and its correlated tag are, respectively,

\delta_1\approx\frac{20{,}000(0.02)^2}{1-0.02^2} =\frac{8}{0.9996}\approx 8.003, \qquad \delta_2\approx\frac{20{,}000(0.01)^2}{1-0.01^2} =\frac{2}{0.9999}\approx 2.000.

Both tests can contribute noncentral signal, although only \(X_1\) has a direct effect in this model. Their dependence through LD means that counting the two elevated statistics as two independent causal effects would be incorrect.

Conditions, Diagnostics, and Limits

The argument assumes that the null distribution is correctly specified after accounting for the study design and relevant covariance among individuals. It also uses large-sample approximations and, in the numerical examples, standardized variables and a simple single-marker test. A mixed-model score or Wald test generally uses model-specific information; its noncentrality should be derived from that test rather than copied mechanically from the simple formula. Unequal sample sizes, imputation quality, allele frequencies, and variance-component estimation can also make information differ across markers.

A QQ plot or a genome-wide summary can reveal that the observed statistics depart from a central-null reference, but elevation alone does not identify its cause. It can reflect genuine polygenic signal, residual confounding, technical problems, or a combination. The preceding calculations establish that polygenicity is one possible source even when confounding is controlled; they do not establish that a particular observed scan is free of confounding. Nor should a mean-based calculation be substituted for the median used by \(\lambda\).

Several checks help avoid overinterpretation:

Check the null calibration premise. The polygenic explanation presumes that truly null tests have the intended null distribution under the fitted model. It does not excuse a misspecified covariance model or an unmodeled design problem.
Account for dependence. LD makes nearby association statistics correlated. A large number of elevated markers need not represent the same number of independent causal effects.
Separate scale from evidence at a locus. Genome-wide elevation is a property of a collection of tests; it does not by itself confirm any one variant’s causal role.
Keep the summary statistic in view. The expected mean of squared statistics and the median-based genomic inflation factor are different summaries, so one cannot be inferred exactly from the other.

A common mistake is to reason, “The test-statistic distribution is shifted above the null, so the mixed model must have failed to remove confounding.” That conclusion does not follow: under the idealized calculations above, genuine effects alone shift the statistics. The correction is to treat elevation as a diagnostic observation, then assess whether the null model is calibrated and consider genuine association signal as a separate contributor. Conversely, calling every elevation “polygenicity” is also unjustified without evidence; the shift alone cannot determine its source.

Key takeaway: Under a calibrated null, a test has positive noncentrality when the effect on the quantity it tests is nonzero. In a single-marker marginal test, effects at correlated variants can cancel, so a nonzero direct effect need not produce positive noncentrality at that marker. When many small effects contribute across the genome—and LD tags some of them—the genome-wide distribution can rise without population confounding. Inflation is a pattern to explain, not a diagnosis by itself.

Check Your Understanding

Use the assumptions stated in each question. The solutions follow the questions.

A signed association statistic is approximately \(N(\mu,1)\) with \(\mu=-1.5\). What are the noncentrality parameter of its square and the expected squared statistic?
In the independent-marker approximation, 50 of 500 tests have noncentrality 2 and the rest are null. What is the average expected squared statistic across the 500 tests?
A standardized causal genotype has effect \(0.03\). A second standardized genotype has correlation \(0.4\) with it and no direct effect. What marginal effect at the second genotype is contributed by the causal genotype?
Why does an elevated genome-wide statistic distribution not, by itself, prove that population confounding remains?
Why should a median-based inflation factor not be calculated by simply substituting the average expected chi-square value?

Solutions. (1) The noncentrality is \(\delta=\mu^2=(-1.5)^2=2.25\), and \(E(Z^2)=1+\delta=3.25\). The negative sign affects the signed statistic’s direction, not the noncentrality. (2) The average is \([450(1)+50(1+2)]/500=(450+150)/500=1.20\). (3) The tagged marginal effect is \(r\beta=(0.4)(0.03)=0.012\). (4) Genuine nonzero effects also yield noncentral tests, so a collection of genuine associations can shift the distribution even when null tests are calibrated. The observed shift alone cannot distinguish that explanation from confounding. (5) A median and a mean are different summaries; the median depends on the full distribution and is not determined by its average alone.