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Rmd 7f572ae Matthew Stephens 2023-11-06 workflowr::wflow_publish("flashier_log1p.Rmd")
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Rmd 38c2d39 Matthew Stephens 2023-10-21 wflow_publish("flashier_log1p.Rmd") 
library(flashier)
Loading required package: ebnm
Loading required package: magrittr
Loading required package: ggplot2

Introduction

When applied to (log1p-transformed) Poisson data, flashier often overestimates the rank (too many factors). However this is usually in simulations where the underlying low-rank structure is the log-mean (or maybe the mean). In practice we don’t know if the low-rank structure is in the log mean or some other function of the mean - for example, in the log1p(mean). Motivated by this, I look at how (nonnegative) flashier behaves on log1p transformed data when the underlying log1p(mu) is itself low rank with non-negative factors.

log1p mean and variance

I want to start by studying the mean and variance of log(1+X)

By Taylor series of \(f(x) = log(1+x)\) we have \[\log(1+X) \approx \log(1+\mu) + (X-\mu)f'(\mu) + 0.5 (X-\mu)^2 f''(\mu)\] so \[E(\log(1+X)) = \log(1+\mu) - 0.5 \mu/(1+\mu)^2\] For small \(\mu\) this is approximately \(\mu/2\) and for large \(\mu\) it is \(\log(1+\mu)\).

Rearranging the above gives \[(log(1+X) - log(1+\mu))^2 \approx (X-\mu)^2 f'(\mu)^2\] so \[var(log(1+X)) \approx \mu/(1+\mu)^2.\]

We can check the accuracy by simulation. It matches pretty well. Also, I note that the standard deviation is not that variable for mu in the range exp(-5) to exp(3). (a factor of just under 10 variation in standard deviation, so not entirely negligible, but it could be worse). This might suggest we might get away without taking account of variation in standard error, which is practically very convenient.

mm = exp(seq(-5,3,length=9))
m = v =mm
for(i in 1:length(mm)){
  x = rpois(100000,mm[i])
  m[i] = mean(log(1+x))
  v[i] = var(log(1+x))
}
plot(mm,m)
lines(mm, log(1+mm) - 0.5*mm/(1+mm)^2 )

Version Author Date
342338e Matthew Stephens 2023-10-21 
plot(mm,sqrt(v))
lines(mm, sqrt(mm)/(1+mm))
abline(v=1)

Version Author Date
342338e Matthew Stephens 2023-10-21

log(c+X)

We can extend this to \(log(c+X)\). By Taylor series we have: \[E(log(c+X)) = log(c+\mu) - 0.5 \mu/(c+\mu)^2\] and \[Var(log(c+X)) \approx \mu/(c+\mu)^2.\]

c=0.1
mm = exp(seq(-5,3,length=9))
mc = vc =mm
for(i in 1:length(mm)){
  x = rpois(100000,mm[i])
  mc[i] = mean(log(c+x))
  vc[i] = var(log(c+x))
}

plot(mm, log(c+mm) - 0.5*mm/(c+mm)^2,type="l" )
points(mm,mc)

Version Author Date
342338e Matthew Stephens 2023-10-21 
plot(mm,sqrt(vc))
lines(mm, sqrt(mm)/(c+mm))

Version Author Date
342338e Matthew Stephens 2023-10-21

Interesting that (unless I made a mistake) the mean is quite off for small mu and the variance is also off. I need to understand this better. However, I do note again that the standard deviation doesn’t vary too much with mm.

log(1+cX)

Here I instead extend this to \(log(1+cX)\) (which is sparse). By Taylor series we have: \[E(log(1+cX)) = log(1+c\mu) - 0.5 c^2 \mu /(1+c\mu)^2\] and \[Var(log(c+X)) \approx \mu c^2/(1+c\mu)^2.\]

c=10
mm = exp(seq(-5,3,length=9))
mc = vc =mm
for(i in 1:length(mm)){
  x = rpois(100000,mm[i])
  mc[i] = mean(log(1+c*x))
  vc[i] = var(log(1+c*x))
}

plot(mm, log(1+c*mm) - 0.5*c^2 * mm/(1+c*mm)^2,type="l" )
points(mm,mc)

Version Author Date
342338e Matthew Stephens 2023-10-21 
plot(mm,sqrt(vc))
lines(mm, c *sqrt(mm)/(1+c*mm))

Simple simulation

I do a simple simulation where \(log(1+\mu) =LF'\) where \(L\) and \(F\) are non-negative and \(LF'\) is rank 4..

set.seed(1)
n= 1000
p = 200
K = 4
LL = matrix(runif(n*K),nrow=n)
FF = matrix(runif(p*K),nrow=p)
mu = matrix(exp(LL %*% t(FF))-1,ncol=p, nrow=n)
X= matrix(rpois(n*p,mu), ncol=p, nrow=n)
hist(mu)

mean(X>0)
[1] 0.744955

Try flashier

fit.nn.1 = flash(log(X+1),ebnm_fn = ebnm_point_exponential,var_type=2,backfit=TRUE)
Adding factor 1 to flash object...
Adding factor 2 to flash object...
Adding factor 3 to flash object...
Adding factor 4 to flash object...
Adding factor 5 to flash object...
Adding factor 6 to flash object...
Adding factor 7 to flash object...
Factor doesn't significantly increase objective and won't be added.
Wrapping up...
Done.
Backfitting 6 factors (tolerance: 2.98e-03)...
  Difference between iterations is within 1.0e+02...
  Difference between iterations is within 1.0e+01...
  --Estimate of factor 6 is numerically zero!
  --Estimate of factor 6 is numerically zero!
  --Estimate of factor 4 is numerically zero!
  Difference between iterations is within 1.0e+00...
  Difference between iterations is within 1.0e-01...
Wrapping up...
Done.
Nullchecking 6 factors...
  One factor is identically zero.
Wrapping up...
  Removed one factor.
Done.

The greedy approach fits 6 factors, but the backfitting ends up with 5 factors. And one of the factors is capturing a single row.

plot(fit.nn.1$F_pm[,4])

which.max(fit.nn.1$F_pm[,4])
[1] 55
plot(fit.nn.1$L_pm[,4])

plot(X[,54])

fit.nn.1$flash_fit$tau[54]
[1] 4.274929

So it seems the extra factor here is due to flash converging to a solution that picks out one column, and sets its variance close to 0 (tau is very big). And the null check does not work because the flash objective goes to infinity as tau goes to infinity with a perfect fit to the column. We could probably avoid this by setting a minimum tau somehow (or regularizing it).

After experimenting I found that S=0.5 fixed it (but smaller values of S do not).

fit.nn.1 = flash(log(X+1),ebnm_fn = ebnm_point_exponential,var_type=2,backfit=TRUE, S=0.5)
Adding factor 1 to flash object...
Adding factor 2 to flash object...
Adding factor 3 to flash object...
Adding factor 4 to flash object...
Adding factor 5 to flash object...
Factor doesn't significantly increase objective and won't be added.
Wrapping up...
Done.
Backfitting 4 factors (tolerance: 2.98e-03)...
  Difference between iterations is within 1.0e+02...
  Difference between iterations is within 1.0e+01...
  Difference between iterations is within 1.0e+00...
  Difference between iterations is within 1.0e-01...
  Difference between iterations is within 1.0e-02...
Wrapping up...
Done.
Nullchecking 4 factors...
Done.

We need more investigation, but maybe flash can do OK with estimating the rank if the true log(1+mu) is low rank and we fix the issue with outlying factors capturing a single column.


sessionInfo()
R version 4.2.1 (2022-06-23)
Platform: x86_64-apple-darwin17.0 (64-bit)
Running under: macOS Big Sur ... 10.16

Matrix products: default
BLAS:   /Library/Frameworks/R.framework/Versions/4.2/Resources/lib/libRblas.0.dylib
LAPACK: /Library/Frameworks/R.framework/Versions/4.2/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     

other attached packages:
[1] flashier_1.0.0 ggplot2_3.4.3  magrittr_2.0.3 ebnm_1.0-55   

loaded via a namespace (and not attached):
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[21] splines_4.2.1     stringr_1.5.0     munsell_0.5.0     mixsqp_0.3-48    
[25] compiler_4.2.1    httpuv_1.6.9      xfun_0.37         pkgconfig_2.0.3  
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