Why a Ratio of Medians?
A GWAS produces association statistics across many genetic variants. A question that arises when reviewing these results is whether the statistics, considered across the genome, are centered on the scale expected under the null hypothesis of no association. The genomic inflation factor, usually written \(\lambda\) or \(\lambda_{GC}\), summarizes this comparison using medians.
The median is compared with a theoretical reference from the null distribution, not with zero. A correctly calibrated one-degree-of-freedom chi-square statistic is positive even under the null, and its median is about 0.455. Thus a ratio near one indicates that the observed median is near this null reference. The ratio alone does not explain why an observed median differs from the reference.
Deriving the Null Reference
Under the null hypothesis and the usual large-sample conditions for a one-degree-of-freedom association test, its test statistic has an approximate chi-square distribution with one degree of freedom. This is the standard asymptotic reference for such statistics as squared \(Z\) statistics and many one-degree-of-freedom Wald statistics. The derivation below explains why the median of this reference distribution, rather than its mean, appears in the definition.
If \(Z\sim N(0,1)\), then \(Z^2\) has a chi-square distribution with one degree of freedom.
Let \(c\) be the median of \(Z^2\). By the definition of a median for this continuous distribution, \(P(Z^2\leq c)=0.5\).
For \(c\geq 0\), the event \(Z^2\leq c\) is equivalent to \(-\sqrt{c}\leq Z\leq \sqrt{c}\). By symmetry of the standard normal distribution, \(P(Z^2\leq c)=2\Phi(\sqrt{c})-1\), where \(\Phi\) is the standard normal cumulative distribution function.
Setting the final expression equal to 0.5 gives the null median:
This calculation also shows why the null median is not one. The mean of a chi-square distribution with one degree of freedom is one, but the distribution is right-skewed; its median is lower. Using one as the denominator would therefore define a different ratio and would not be the usual genomic inflation factor.
Calculating the Factor From Summary Statistics
For a set of \(m\) variants, first identify the association statistics that are appropriate for this calculation. They should be on a comparable one-degree-of-freedom chi-square scale and should represent the same analysis or a clearly defined set of analyses. If the results provide one-degree-of-freedom \(Z\) statistics rather than chi-square statistics, square each \(Z\) statistic to obtain its chi-square statistic. Then find the sample median and divide it by 0.454936.
For an odd number of values, the sample median is the middle value after sorting. For an even number, it is conventionally the average of the two middle values. In a large GWAS, the same calculation is applied to the selected set of variant-level statistics in the summary results. The calculation itself does not require the statistics to be independent. However, correlation among variants, for example through linkage disequilibrium, matters when assessing how much the sample median might vary across repeated data sets.
Worked Example: An Odd Number of Chi-Square Statistics
Suppose a hypothetical GWAS summary contains the following one-degree-of-freedom chi-square statistics for nine variants: \(0.08,\ 0.17,\ 0.29,\ 0.41,\ 0.58,\ 0.72,\ 0.95,\ 1.20,\ 1.80\). Assume the statistics come from comparable tests whose null reference is approximately chi-square with one degree of freedom.
The values are already in increasing order, so the fifth value is the median: \(0.58\). Therefore,
The observed median is about 1.275 times the one-degree-of-freedom null median. This summarizes an upward shift in the median statistic for this hypothetical set; it does not identify the reason for that shift.
Worked Example: An Even Number of Statistics
In a second hypothetical summary, eight comparable one-degree-of-freedom chi-square statistics, in increasing order, are \(0.10,\ 0.22,\ 0.31,\ 0.39,\ 0.47,\ 0.63,\ 0.88,\ 1.15\). Because there are an even number of observations, average the fourth and fifth values:
The genomic inflation factor is
The sample median is modestly below the theoretical null median, giving a ratio below one. With only eight statistics, this calculation is particularly sensitive to the specific values observed; it is a descriptive ratio, not by itself evidence that the association test is systematically conservative.
Worked Example: Starting With Z Statistics
A hypothetical association summary reports absolute one-degree-of-freedom \(Z\) statistics of \(0.40,\ 0.70,\ 0.90,\ 1.20,\) and \(1.50\). Assume each test has the standard normal null reference, so that squaring its \(Z\) statistic gives a one-degree-of-freedom chi-square statistic. The squared values, in order, are \(0.16,\ 0.49,\ 0.81,\ 1.44,\) and \(2.25\).
The middle squared value is \(0.81\). Thus,
The ratio is about 1.780 for this five-variant set. This example illustrates the conversion from \(Z\) statistics to the required chi-square scale; it does not mean that the \(Z\) statistics themselves should be divided by 0.454936.
What the Ratio Measures
The numerator is an empirical quantity: the median of the selected association chi-square statistics. The denominator is a theoretical reference: the median under the one-degree-of-freedom null distribution. Their ratio describes the location of the observed median relative to that reference. In a very large set of well-calibrated null tests, one expects the observed median to be near 0.454936, so the ratio is expected to be near one, though a finite sample need not give exactly one.
| Value of \(\lambda_{GC}\) | Direct interpretation | What it does not establish by itself |
|---|---|---|
| Near 1 | The observed median is near the one-degree-of-freedom null median. | That every test is calibrated or that there are no unusual signals in the tail. |
| Above 1 | The observed median exceeds the null reference. | The cause of the excess, or whether it reflects any particular source. |
| Below 1 | The observed median is below the null reference. | That the analysis is definitively conservative; sampling variation can also affect the ratio. |
Because a median describes the middle of a distribution, \(\lambda_{GC}\) is not a summary of the largest association statistics alone. A handful of very large statistics can be important for discovery, but they do not determine the median unless they are numerous enough to move its middle. Conversely, a ratio near one does not guarantee that the most extreme results have the correct behavior.
The ratio is also not a count of associated variants, a measure of an individual variant's effect, or a direct estimate of a particular source of bias. Several features of an analysis can affect the distribution of observed test statistics, and the ratio by itself cannot distinguish among them. A QQ plot can display how observed statistics compare with their null expectations across the distribution; the genomic inflation factor reduces one aspect of that comparison to a single number.
Scope and Assumptions
The denominator 0.454936 applies specifically to a chi-square distribution with one degree of freedom. Before calculating the usual \(\lambda_{GC}\), check that the statistics being summarized are one-degree-of-freedom association statistics or have been validly converted to that scale. A test with a different number of degrees of freedom has a different null distribution and therefore a different null median. Applying 0.454936 to such statistics would not be a like-for-like comparison.
The null reference also relies on the association statistics being calibrated under their stated null models, at least approximately. For instance, the usual large-sample chi-square approximation is an assumption about the test and data conditions, not a consequence of taking a median. The ratio can still be computed as a descriptive summary if the reference is questionable, but its interpretation as a comparison with the null median is then limited.
Variants are often correlated, so the many statistics in a GWAS should not automatically be treated as independent observations when reasoning about uncertainty in \(\lambda_{GC}\). Correlation does not change the arithmetic definition: sort the selected statistics, find the median, and divide. It can, however, affect the variability of that median. A point estimate of the ratio does not provide an uncertainty interval or a test of whether it differs from one.
Common Mistake
A common mistake is to reason, “\(\lambda_{GC}=1.2\), so the study must have a particular amount of population stratification,” or “the study must have 20% more causal variants.” Neither conclusion follows from the ratio. The calculation says only that the observed median chi-square statistic is 1.2 times the one-degree-of-freedom null median. It does not assign a cause, quantify the number of causal variants, or say that each statistic is inflated by the same amount.
Another frequent error is dividing the median by one because the null chi-square mean is one. The definition uses the null median, not the null mean. For one degree of freedom, that denominator is approximately 0.454936, as derived above.
Check Your Understanding
Answer each question using the definition and assumptions given above.
- Why is the denominator for the usual one-degree-of-freedom genomic inflation factor about 0.455 rather than one?
- A hypothetical set has a median chi-square statistic of 0.60. Calculate \(\lambda_{GC}\) and state what the ratio describes.
- For eight sorted statistics, which values determine the sample median, and how are they combined?
- What transformation is needed before using reported one-degree-of-freedom \(Z\) statistics in this calculation?
- Why does a value above one not, by itself, identify the cause of the observed upward shift?