CypRiv
commented
1 year ago I noticed this thread while I was about to post similar questions regarding the Distortion test. Apart from the questions that have already been mentioned, I am adding another one related to the calculation of the expected bias.
In the paper, it is stated "The null distribution is generated by substituting nO variants detected as outliers by the MR-PRESSO outlier test with nE−2nO non-outliers, which are drawn with replacement from the entire set of non-outlier variants."
However, I believe that the behavior of the getRandomBias function is different:
-
The code starts by placing the outlier indices at the beginning of the indices array: indices <- c(refOutlier, replicate(nrow(data)-length(refOutlier), sample(setdiff(1:nrow(data), refOutlier))[1]))
-
Then it computes the linear regression with the first indices[len(indices)-len(outliers)] rows: data = data[indices[1:(length(indices) - length(refOutlier))], ]
-
This implies that the indices corresponding to the outliers are always included in the model. I think this contradicts the description in the paper, which specifies that the outlier variants should be replaced by randomly sampled non-outlier variants.
-
I believe that for the code to behave similarly to the paper's description, the indices array should be obtained by placing the outlier indices at the end: indices <- c(replicate(nrow(data)-length(refOutlier), sample(setdiff(1:nrow(data), refOutlier))[1]), refOutlier)
Thanks a lot for providing clarification on this!
Best, Cyprien
Dear authors,
Since the numerator of the empirical p-value of the article and the GitHub code is written differently, are the following H0 and Ha correct? Ho: causal effect estimates not distorted by horizontal pleiotropy Ha: causal effect estimates are distorted by horizontal pleiotropy
In the distortion test, how to draw with replacement nE − 2no from the entire set of non-outlier variants？
In the empirical p-value, how to calculate the expected distortion？
Thank you.
Best, Tina