Last updated: 2019-05-10
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Rmd | a713b11 | Matthew Stephens | 2019-05-10 | workflowr::wflow_publish(“simple_transform_simulation.Rmd”) |
Let’s try a simple simulation to test t test on log-tranformed count data.
Specifically I simulate \(X_i | s_i \sim Poi(s_i \lambda_i)\) for two groups where the library size \(s_i\) differs between the groups (by a factor of 10, so quite extreme) but distribution of \(\lambda_i\) is the same in each group.
Indeed, here i just fix the \(\lambda_i\) to be all equal, to a value such that the data have mean 1 in one group and mean 10 in the other group.
Then do transform \(Y_i = \log(X_i/(s_i/median(s_i)) + 1)\)
set.seed(1)
n = 100
s = c(rep(10^5,n), rep(10^4,n))
l = rep(1/10^4,2*n)
niter = 1000
pval = rep(0,niter)
for(i in 1:niter){
x = rpois(2*n, s*l)
y = log(x/(s/median(s))+1)
pval[i] = t.test(y[1:100],y[101:200])$p.value
}
hist(pval)
So we see the t test p values are very non-uniform. One can see why one might worry about this….
Plot one example:
plot(y)
mean(y[1:100])
[1] 1.883674
mean(y[101:200])
[1] 1.390502
Try the same thing but with only a factor 2 in library size
set.seed(1)
n = 100
s = c(rep(10^5,n), rep(0.5*10^5,n))
l = rep(1/(0.5*10^5),2*n)
niter = 1000
pval = rep(0,niter)
for(i in 1:niter){
x = rpois(2*n, s*l)
y = log(x/(s/median(s))+1)
pval[i] = t.test(y[1:100],y[101:200])$p.value
}
hist(pval)
(Not sure this is 100% correct… needs checking)
According to Taylor series expansion https://users.rcc.uchicago.edu/~aksarkar/singlecell-modes/transforms.html the bias should be V(x)/2(E(x) + 1)^2 where in the first simulation x is x/(1.8) or x/0.18 in the two groups (because this is s/median(s))
(10/1.8^2) /(2*(10/1.8+1)^2) - (1/0.18^2) / (2*(1/0.18+1)^2)
[1] -0.323183
So if that is right, to second order, the difference in mean of y between two groups should be 0.32 in our first simulation…so try correcting for this….
set.seed(1)
n = 100
s = c(rep(10^5,n), rep(10^4,n))
l = rep(1/10^4,2*n)
niter = 1000
pval = rep(0,niter)
for(i in 1:niter){
x = rpois(2*n, s*l)
y = log(x/(s/median(s))+1)
pval[i] = t.test(y[1:100],y[101:200]+0.32)$p.value
}
hist(pval)
sessionInfo()
R version 3.5.2 (2018-12-20)
Platform: x86_64-apple-darwin15.6.0 (64-bit)
Running under: macOS Mojave 10.14.4
Matrix products: default
BLAS: /Library/Frameworks/R.framework/Versions/3.5/Resources/lib/libRblas.0.dylib
LAPACK: /Library/Frameworks/R.framework/Versions/3.5/Resources/lib/libRlapack.dylib
locale:
[1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8
attached base packages:
[1] stats graphics grDevices utils datasets methods base
loaded via a namespace (and not attached):
[1] workflowr_1.2.0 Rcpp_1.0.1 digest_0.6.18 rprojroot_1.3-2
[5] backports_1.1.3 git2r_0.24.0 magrittr_1.5 evaluate_0.12
[9] stringi_1.2.4 fs_1.2.6 whisker_0.3-2 rmarkdown_1.11
[13] tools_3.5.2 stringr_1.3.1 glue_1.3.0 xfun_0.4
[17] yaml_2.2.0 compiler_3.5.2 htmltools_0.3.6 knitr_1.21