posterior odds value when shat2 is infinite

[MapleLin-Bio]

Hi there,

I am confused about the way to deal with log(po) in the single_effect_regression() function.

# log(po) = log(BF * prior) for each SNP
    lbf = dnorm(betahat,0,sqrt(V + shat2),log = TRUE) -
      dnorm(betahat,0,sqrt(shat2),log = TRUE)
    lpo = lbf + log(prior_weights + sqrt(.Machine$double.eps))

    # Deal with special case of infinite shat2 (e.g., happens if X does
    # not vary).
    lbf[is.infinite(shat2)] = 0
    lpo[is.infinite(shat2)] = 0
    maxlpo = max(lpo)

I'm not sure if it's my misunderstanding. In the special case of infinite shat2, why lpo[is.infinite(shat2)] is set to 0, not log(prior_weights + sqrt(.Machine$double.eps))? If we specify null_weight , the corresponding column of X will always meet this case, but lpo[is.infinite(shat2)] = 0 implies that the prior weight is fixed as 1.

Thank you!

[pcarbo]

@MapleLin-Bio I think you are right, this is a mistake in the code.