I am trying to understand the derivations from Sullivan et al. (2015) in the Supplementary Material. There, it is mentioned in the first page that the least squares estimate of the j-th SNP effect, considering the polygenicity linear model $φ=Xβ + ε$, is $\hat{β_j} := X^T_j φ/N$ . Normally, the least squares solution of $\hat{\beta}$ is $(X^TX)^{-1}X^T\phi$. By comparing the two equations with each other, this means that, in this case, $X^TX=N I_{M\times M}$. However, LD is assumed, therefore $E(X_{ij}X_{ik})\neq 0$, so the covariance matrix cannot be diagonal. Could someone explain to me how the derivation for the beta estimate was performed, please? What am I missing? I checked that the same thing is mentioned in more than one sources such as here.
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$\begingroup$ I still don't understand what you're asking. If you want to try and rephrase your question I can give it a shot, but I'm not sure where the confusion lies right now. $\endgroup$– user15286Commented Apr 13, 2022 at 3:11
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$\begingroup$ I edited the question, I hope now it is clearer what is the mathematical issue I am having trouble understanding. $\endgroup$– Vasilis LemonidisCommented Apr 14, 2022 at 10:13
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Not entirely clear on what you're asking but the beta estimates in LDSC come from GWAS summary statistics. As I understand it, they aren't "derived" so much as just set equivalent to what they linear model defines them as, which is their additive effects on the phenotype as implied by the model. So for each SNP j, it's effect size (beta) is whatever difference one copy of the allele makes on the phenotype (phi), divided by N, the number of SNPs.
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$\begingroup$ Sorry for being late, but what do you mean by the phrasing "equivalent to what the linear model defines them as"? If only the linear model solution was considered, then the estimated beta-s would be the pseudo-inverse of X times X' times phi, namely the least squares solution. $\endgroup$ Commented Apr 11, 2022 at 19:44
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$\begingroup$ Accepting your answer, I think I misunderstood the phrasing in the supplementary note, it is indeed only a way of defining a weighted effect on the explained output, nothing to do with least squares solution... $\endgroup$ Commented May 22, 2022 at 17:31