fold change in Limma is calculated

Question

[Learner]

I am trying to understand how limma calculates the fold change

Is it

log2(mean(groupA)- log2(mean(groupB)

[russhh]

could I ask why you removed your motivating example?

Answer 1

[russhh]

Shouldn't you be running mean(log2(groupB)) - mean(log2(groupA)) ?

 x <- data.frame(grp = rep(c("a", "b"), each = 3),
                   z = c(4.36e-3, 5.72e-3, 4.17e-3, 2.85e-2, 3.37e-2, 3.27e-2))

You'd have to log-transform these to use them in limma. Nonetheless:

 x
  grp       z
1   a 0.00436
2   a 0.00572
3   a 0.00417
4   b 0.02850
5   b 0.03370
6   b 0.03270

For simplicity we'll use lm

 lm (log2(z) ~ grp, data = x)

    Call:
lm(formula = log2(z) ~ grp, data = x)

Coefficients:
(Intercept)         grpb  
     -7.732        2.746

grpb coef is 2.746 as in limma

Consider both the log of the geometric mean and the log of the arithmetic mean within each group

y <- x %>% 
    group_by(grp) %>%
    summarise(
        log_gmean = mean(log2(z)),
        log_amean = log2(mean(z)))

> y
# A tibble: 2 x 3
     grp log_gmean log_amean
  <fctr>     <dbl>     <dbl>
1      a -7.732321 -7.717857
2      b -4.986189 -4.982411

# Differences in the log-means:
y[2, 2:3] - y[1, 2:3]
  log_gmean log_amean
1  2.746132  2.735446

Difference of the log-geometric-mean is the value you want, not the difference of the log-arithmetic-mean. I can't replicate your value of 2.77 for log(mu_B) - log(mu_A); sorry.