flashier_log1p.Rmd

---
title: "flashier_log1p"
author: "Matthew Stephens"
date: "2023-10-21"
output: workflowr::wflow_html
editor_options:
  chunk_output_type: console
---

```{r}
library(flashier)
```

## 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.

```{r}
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 )

plot(mm,sqrt(v))
lines(mm, sqrt(mm)/(1+mm))
abline(v=1)
```


### 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
$$Var(log(c+X)) \approx \mu/(c+\mu)^2.$$

```{r}
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)
    
plot(mm,sqrt(vc))
lines(mm, sqrt(mm)/(c+mm))
```

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.


## Simple simulation

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

```{r}
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)
```

Try flashier
```{r}
fit.nn.1 = flash(log(X+1),ebnm_fn = ebnm_point_exponential,var_type=2,backfit=TRUE)
```
The greedy approach fits 6 factors, but the backfitting ends up
with 5 factors. And one of the factors is capturing a single row.

```{r}
plot(fit.nn.1$F_pm[,4])
which.max(fit.nn.1$F_pm[,4])
plot(fit.nn.1$L_pm[,4])
plot(X[,54])
fit.nn.1$flash_fit$tau[54]
```

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).
```{r}
fit.nn.1 = flash(log(X+1),ebnm_fn = ebnm_point_exponential,var_type=2,backfit=TRUE, S=0.5)
```

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.