How to corrected test statistics based on the LDSC intercept?

Question

[Isaac]

I find that some article will "correct the test statistics" in GWAS when the summary data's LD intercept is too high(e.g. >1.05). But I want to know how to achieve this? Is it dividing the z-score of all SNPS by the LD intercept?

I will be so grateful if anyone could help.

[LChart]

some article will

Can you provide a reference please?

[Isaac]

Sure, the article is "Multi-ancestry genome-wide association study of major depression aids locus discovery, fine mapping, gene prioritization and causal inference", and the part I mentioned is "To account for the minor inflation found in the Hispanic/Latin American samples (lambda1,000 = 1.002 and linkage disequilibrium score regression (LDSC) intercept 1.051; Supplementary Table 2), we corrected test statistics for this analysis based on the LDSC intercept."

Answer 1

[LChart]

The approach the authors are using is Genomic Control (Roeder, 1999). The history of the use of GC in GWAS was (and is) that, in many cases, the inflation of the quantile-quantile plot is due to population stratification rather than bona-fide genetic association, so the statistic values can be adjusted to "correct" for that baseline level of inflation. The correction is simply multiplying the standard error of each variant by sqrt(inflation). However, one issue is that the observed inflation is really a combination of polygenic risk (e.g., quantitative genetic model) plus any population stratification; so subsequent refinements used other approaches besides the overall genomic inflation to estimate the contribution of population stratification.

In LD Score Regression, it turns out that the intercept of the model is an estimator of the contribution to association of pure population stratification, and thus can be used as a plug-in value for genomic correction.

In terms of how to achieved this: Multiply the standard errors by sqrt(lambda) prior to computing statistics and p-values.

[Isaac]

I sincerely appreciate your detailed answer! I would like to confirm if using LDSC intercept for correction also multiplies the standard error by sqrt (intercept)? Or multiply it directly by the intercept? After all, lambda will expand when the sample size is large.

[LChart]

It gets multiplied by sqrt(lambda) because the adjusted variance should be lambda * sigma^2, and if X ~ N(0, s^2) then tX ~ N(0, t^2s^2); hence sqrt(lambda)X ~ N(0, lambda s^2).

[Isaac]

Thank you very much!