Is it possible to test the hypothesis that the significant variants are in a set of SNPs?

[garyzhubc]

Is it possible to test the hypothesis that the significant variants are in a set of SNPs?

[garyzhubc]

Not just sets from the susie_get_cs()

[stephens999]

sorry I don't think we understand the question.

[garyzhubc]

Can I test the hypothesis like "beta_2,beta_3\ne0, beta_1,beta_4=0"?

[stephens999]

perhaps the simplest way would be to sample from the (approximate) posterior distribution on beta, and then use that sample to compute the probability of any particular combination
of 0/non-zero beta values.

[garyzhubc]

Something like this (test the hypothesis that significant variants are in the middle)?

posterior<-unlist(lapply(1:1000, function(i) (max(samp$b[40:60,i])>0 | min(samp$b[40:60,i])<0) & (samp$b[1,1:39] == 0 & samp$b[1,61:101] == 0)))

result in:

> mean(posterior)
[1] 0.9756098

[stephens999]

I think the code doesn't look quite right, but something like that, yes

[garyzhubc]

I made an edit on the code. Could you tell me why it doesn't look right?

[stephens999]

you have samp$b[1,1:39]
but samp$b[40:60,i]

so the indices of b don't look consistent through your code

[garyzhubc]

Thanks for spotting this mistake. I'm now using:

> posterior<-unlist(lapply(1:num_samples, function(i) (max(samp$b[40:60,i])>0 | min(samp$b[40:60,i])<0) & all(c(samp$b[1:39,i] == 0, samp$b[61:101,i] == 0))))
> mean(posterior)
[1] 0
> posterior<-unlist(lapply(1:num_samples, function(i) (max(samp$b[1:39,i])>0 | min(samp$b[1:39,i])<0) & all(samp$b[40:101,i] == 0)))
> mean(posterior)
[1] 0
> posterior<-unlist(lapply(1:num_samples, function(i) (max(samp$b[61:101,i])>0 | min(samp$b[61:101,i])<0) & all(samp$b[1:60,i] == 0)))
> mean(posterior)
[1] 0

which I believe is consistent. However, it looks like these all have very small probabilities. Do you recommend instead of testing all(samp$b[1:60,i] == 0) try something like all(abs(samp$b[1:60,i]) < episilon), or do you think what I'm doing is okay?

[garyzhubc]

Can anyone respond to this issue please? @pcarbo

[pcarbo]

The issue of working with very small probabilities is a common issue and there are some ways to help with this. If you can share with us a reproducible example illustrating exactly what you are trying to do, I might be able to help you.

As a general piece of advice, I recommend starting with an example that is as simple as possible, e.g., an example with exactly 4 variables X.

[garyzhubc]

An example of four variables b1, b2, b3, b4: suppose I want to calculate P(b1=0 and b4 =0 and (b2!=0 or b3!=0)) as the probability that the causal variant is in {b2,b3}, shall I run this program below to get the probability?

samp<-susie_get_posterior_samples(res_, num_samples)
posterior23<-unlist(lapply(1:num_samples, function(i) (max(samp$b[2:3,i])>0 | min(samp$b[2:3,i])<0) & all(c(samp$b[1,i] == 0, samp$b[4,i] == 0))))
mean(posterior23)

If I rank posterior1, posterior2, posterior3, posterior12, posterior23, posterior13, posterior123 and select the one with probability greater than 0.95, will I get the same credible interval (default coverage = 0.95):

susie_get_cs(res_)

[pcarbo]

@garyzhubc This is not a reproducible example because some variables in your code (e.g., res_) are not defined.

[stephens999]

However, it looks like these all have very small probabilities. Do you recommend instead of testing all(samp$b[1:60,i] == 0) try something like all(abs(samp$b[1:60,i]) < episilon), or do you think what I'm doing is okay?

I think what you are doing looks OK, and you are just getting the answer that there is a very small probability of the event you are looking at. You could also try the epsilon approach you suggested.

[garyzhubc]

I also tried using PIP directly instead of sampling.

prod(1-res$pip[1:34])*(1-prod(1-res$pip[35:68]))*prod(1-res$pip[69:101])

still gives zero probabilities. See #203 (comment)

[garyzhubc]

@garyzhubc This is not a reproducible example because some variables in your code (e.g., res_) are not defined.

I could do the same on this example https://stephenslab.github.io/susieR/articles/sparse_susie_eval.html:

create_sparsity_mat = function(sparsity, n, p) {
  nonzero          <- round(n*p*(1-sparsity))
  nonzero.idx      <- sample(n*p, nonzero)
  mat              <- numeric(n*p)
  mat[nonzero.idx] <- 1
  mat              <- matrix(mat, nrow=n, ncol=p)
}
n <- 1000
p <- 1000
beta <- rep(0,p)
beta[c(1,300,400,1000)] <- 10 
X.dense  <- create_sparsity_mat(0.99,n,p)
X.sparse <- as(X.dense,"CsparseMatrix")
y <- c(X.dense %*% beta + rnorm(n))
susie.sparse <- susie(X.sparse,y)

Using Monte Carlo sample from posterior:

num_samples<-10000
samp<-susie_get_posterior_samples(susie.sparse, num_samples)
posterior<-unlist(lapply(1:num_samples, function(i) (max(samp$b[341:680,i])>0 | min(samp$b[341:680,i])<0) & all(c(samp$b[1:340,i] == 0, samp$b[681:1000,i] == 0))))
mean(posterior)

Using PIP:

prod(1-susie.sparse$pip[1:340])*(1-prod(1-susie.sparse$pip[341:680]))*prod(1-susie.sparse$pip[681:1000])

Both gives probability zero.

[pcarbo]

@garyzhubc I think the issue is that in your example all the inclusion probabilities are either 1 or very, very small, so it may be tricky to use a naive Monte Carlo sampling approach:

hist(log10(susie.sparse$alpha),n = 64)

One idea that comes to mind is importance sampling, but you might want to start with an example where the probabilities are less extreme.