Elsewhere we’ve described how we selected a “primary” \(h^2_g\) result for each of the unique phenotypes in the UKB GWAS. These phenotypes range widely in terms of sample size and polygenicity, among other factors. These features can affect our confidence in the stability of the LD Score Regression results. For example we previously noted concerns about bias at low effective sample sizes.

We document here out evaluation of confidence in each of the LDSR \(h^2_g\) results. This is separate from our evaulation of statistical significance for these results. Here we look at:

We conclude with a summary of our current confidence classifications.

Minimum sample size

After we’ve defined a “primary” \(h^2_g\) result for each of the 4178 unique phenotypes in the Round 2 GWAS, we know that the sample size (or effective sample size) for many of these phenotypes is quite low. For example, case counts are quite small for most ICD codes and job codes. To avoid unnecessary multiple testing for the significance of the \(h^2_g\) results we therefore aim to identify the minimum (effective) sample size required to provide sufficient power to make LDSR analysis viable. [NB: We focus first on a bare minimum in terms of power, we’ll return to the question of bias later.]

Goal: determine the minimum (effective) sample size where we’d expect to have some useful level of power to detect a reasonable \(h^2_g\).

Question 1: What’s the relationship between sample size and precision for LDSR?

To evaluate the effective sample size required to have power in LDSR analyses, we consider the relationship between \(N_{eff}\) and the observed standard error (SE) for the LDSR \(h^2_g\) estimate for the 4178 Round 2 phenotypes.

Takeaway: There’s a clear inverse relationship between \(N_{eff}\) and the SE of the \(h^2_g\) estimate. The SE is also very large (e.g. wider than the 0-1 range for \(h^2\)) at small effective sample sizes.

For both visualization and modelling, it’s useful to look at this relationship in terms of the inverse variance (\(1/SE^2\)):