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Introduction

I want to look at simple EB/variational/Bayesian approach to estimating a binary vector that lives approximately in a given subspace. You are given an \(nxn\), rank \(k\), projection matrix P and you want to find a vector \(v\) such that \(\rho(v) :=v'Pv/v'v\) is large. If \(v\) lies entirely in the subspace defined by \(P\) then \(\rho(v) = v'v/v'v = 1\).

My derivations here suggest using a log-likelihood \[l(v) \approx -0.5(n-k-2b)\log(1-\rho(v)),\] where \(b\) is a hyperparameter. For now we make the approximation that \(k,b\) are small so \(l(v) \approx -0.5n \log(1-\rho(v))\). We also make the approximation \(E_q(l(v)) = -0.5n \log(1-\rho(q))\) where \(\rho(q) = \bar{v}'P\bar{v}/<v'v>\). Here \(\bar{v},<v'v>\) denote the expectations of \(v,v'v\) under \(q\).

The variational EB approach then iterates: \[q \leftarrow EBNM\left( \frac{\mathbf{P}\bar{v}}{\rho(q)}, \frac{(1-\rho(q)) <v^Tv>}{n\rho(q)} \right)\]

For the Rademacher prior we have \(<v'v>=n\) and the update can be written entirely in terms of the posterior mean: \[\bar{v} = tanh(P\bar{v}/(1-\rho))\] While the idea is to use \(\rho=v'Pv/n\), I will also try fixing \(\rho\) in these updates to see what happens. Note that this approach is not really EB because the prior is fixed Rademacher. TODO: we should try estimating a generalized (asymmetric) Rademacher prior, so this becomes a genuinely EB approach.

Three group simulation (low noise)

Here I simulate 3 groups, with 20 members each.

K=3
p = 1000
n = 100
set.seed(1)
L = matrix(-1,nrow=n,ncol=K)
for(i in 1:K){L[sample(1:n,20),i]=1}
FF = matrix(rnorm(p*K), nrow = p, ncol=K)

X = L %*% t(FF) + rnorm(n*p,0,1)

The column space of \(X\) approximately contains three binary vectors (columns of \(L\)). Here I find the projection matrix corresponding to the column space of \(X\) (approximated using a rank 3 matrix, which helps remove the noise.) We can see the values of rho for the true L are very high.

U = svd(X)$u[,1:3] 
P = U %*% t(U) 
true_rho = diag(t(L) %*% P %*% L/n)
true_rho
[1] 0.9989868 0.9989723 0.9989671

Now, initializing from a random vector. If we set rho=0 then the iterates converge to something close to 0 (as expected, because tanh is a contraction). However, interestingly it does find a binary source!

set.seed(1)
v.init = sample(c(-1,1),n,replace=TRUE)
maxiter = 10000

rho = 0
v = v.init
for(i in 1:maxiter){
  v = tanh(P %*% v/(1-rho))
}
t(v) %*% P %*% v/n # print out rho
             [,1]
[1,] 0.0001436779
plot(v)

Version Author Date
d112e19 Matthew Stephens 2026-06-21 
cor(v,L) 
           [,1]        [,2]       [,3]
[1,] -0.9990229 -0.01279823 0.07729109

Try instead setting rho=0.5. It converges to a binary source again, but this time close to v in (-1,1).

rho = 0.5
v = v.init
for(i in 1:maxiter){
  v = tanh(P %*% v/(1-rho))
}
t(v) %*% P %*% v/n # print out rho
          [,1]
[1,] 0.9152038
plot(v)

Version Author Date
d112e19 Matthew Stephens 2026-06-21 
cor(v,L)
           [,1]          [,2]       [,3]
[1,] -0.9999768 -0.0002258368 0.06234581

This time we estimate rho. It converges to a binary source again, but this time close to v in (-1,1).

rho = 0
v = v.init
for(i in 1:maxiter){
  rho = as.numeric(t(v) %*% P %*% v/n) # update rho
  v = tanh(P %*% v/(1-rho))
}
t(v) %*% P %*% v/n # print out rho
          [,1]
[1,] 0.9989868
plot(v)

Version Author Date
d112e19 Matthew Stephens 2026-06-21 
cor(v,L)
     [,1]          [,2]   [,3]
[1,]   -1 -5.724587e-17 0.0625

Three group simulation (higher noise)

Here I repeat the above, with more noise:

K=3
p = 1000
n = 100
set.seed(1)
L = matrix(-1,nrow=n,ncol=K)
for(i in 1:K){L[sample(1:n,20),i]=1}
FF = matrix(rnorm(p*K), nrow = p, ncol=K)

X = L %*% t(FF) + rnorm(n*p,0,10)

The column space of \(X\) approximately contains three binary vectors (columns of \(L\)). The true value of rho is near 0.8.

U = svd(X)$u[,1:3] 
P = U %*% t(U) 
true_rho = diag(t(L) %*% P %*% L/n)
true_rho
[1] 0.807348 0.790952 0.829372

Now, initializing from a random vector. If we set rho=0 then the iterates converge to something close to 0 (as expected, because tanh is a contraction). This time it does not really find a binary source!

set.seed(1)
v.init.1 = sample(c(-1,1),n,replace=TRUE)
maxiter = 10000

rho = 0
v = v.init.1
for(i in 1:maxiter){
  v = tanh(P %*% v/(1-rho))
}
t(v) %*% P %*% v/n # print out rho
             [,1]
[1,] 8.746538e-05
plot(v)

Version Author Date
d112e19 Matthew Stephens 2026-06-21 
cor(v,L) 
          [,1]       [,2]       [,3]
[1,] -0.428421 -0.7348204 0.09921965

Try instead setting rho=0.5. It is getting closer to a binary source.

rho = 0.5
v = v.init.1
for(i in 1:maxiter){
  v = tanh(P %*% v/(1-rho))
}
t(v) %*% P %*% v/n # print out rho
          [,1]
[1,] 0.6826304
plot(v)

Version Author Date
d112e19 Matthew Stephens 2026-06-21 
cor(v,L)
           [,1]       [,2]        [,3]
[1,] -0.1350534 -0.8809243 -0.08754405

This time we estimate rho. It converges to something closer to a binary source but does not get it exactly right.

rho = 0
v = v.init.1
for(i in 1:maxiter){
  rho = as.numeric(t(v) %*% P %*% v/n) # update rho
  v = tanh(P %*% v/(1-rho))
}
t(v) %*% P %*% v/n # print out rho
          [,1]
[1,] 0.8272076
plot(v)

Version Author Date
d112e19 Matthew Stephens 2026-06-21 
cor(v,L)
            [,1]       [,2]        [,3]
[1,] -0.07294291 -0.8723976 -0.07514556

Try a different random start: this solution is closer to a true source (TODO: check the objective function).

set.seed(3)
v.init.3 = sample(c(-1,1),n,replace=TRUE)
rho = 0
v = v.init.3
for(i in 1:maxiter){
  rho = as.numeric(t(v) %*% P %*% v/n) # update rho
  v = tanh(P %*% v/(1-rho))
}
t(v) %*% P %*% v/n # print out rho
          [,1]
[1,] 0.8254599
plot(v)

Version Author Date
d112e19 Matthew Stephens 2026-06-21 
cor(v,L)
           [,1]        [,2]       [,3]
[1,] 0.02204825 -0.07501538 -0.9648287

Try EBICA

set.seed(3)
v.init.3 = sample(c(-1,1),n,replace=TRUE)
rho = 0
v = v.init.3

for(i in 1:maxiter){
  v = tanh(n*P %*% v/as.numeric(sqrt(n * t(v) %*% P %*% v)))
}
t(v) %*% P %*% v/n # print out rho
         [,1]
[1,] 0.487831
plot(v)

Version Author Date
d112e19 Matthew Stephens 2026-06-21 
cor(v,L)
           [,1]       [,2]      [,3]
[1,] 0.01389254 -0.1671942 -0.906508

Try centering X

In the above I did not center X. So here I do so. This reduces the “true rho” substantially (below 0.5). The issue is that the original binary vectors are no longer in the subspace spanned by columns of Xcenter - the centering shifts the +-1 vectors so they are no longer +-1.

Xcenter = scale(X,scale=FALSE)
Ucenter = svd(Xcenter)$u[,1:3] 
Pcenter = Ucenter %*% t(Ucenter) 
diag(t(L) %*% Pcenter %*% L/n)
[1] 0.4262193 0.3058118 0.4351670

The method now fails to find them (not suprisingly):

set.seed(3)
v.init.3 = sample(c(-1,1),n,replace=TRUE)
rho = 0
v = v.init.3
for(i in 1:maxiter){
  rho = as.numeric(t(v) %*% Pcenter %*% v/n) # update rho
  v = tanh(Pcenter %*% v/(1-rho))
}
t(v) %*% Pcenter %*% v/n # print out rho
          [,1]
[1,] 0.4034526
plot(v)

Version Author Date
d112e19 Matthew Stephens 2026-06-21 
cor(v,L)
          [,1]       [,2]      [,3]
[1,] 0.4690021 0.06030247 0.5967123

I tried adding an intercept to P (so now the intercept is in the space spanned by \(P\)). But now the method finds the trivial solution of all 1s. (When I ran this about five times with different seeds it did eventually find one of the sources. But the trivial solution has the largest rho, so it tends to win.) TODO: we should try generalizing the Rademacher prior to try to solve this problem.

Ucenter = cbind(rep(1/sqrt(n),n),Ucenter) # add an intercept
Pcenter = Ucenter %*% t(Ucenter) 
diag(t(L) %*% Pcenter %*% L/n) #rho is back to 0.8 or so
[1] 0.7862193 0.6658118 0.7951670
set.seed(3)
v.init.3 = sample(c(-1,1),n,replace=TRUE)
rho = 0
v = v.init.3
v = sample(c(-1,1),n,replace=TRUE)
for(i in 1:maxiter){
  rho = as.numeric(t(v) %*% Pcenter %*% v/n) # update rho
  v = tanh(Pcenter %*% v/(1-rho))
}
t(v) %*% Pcenter %*% v/n # print out rho
     [,1]
[1,]    1
plot(v)

Version Author Date
d112e19 Matthew Stephens 2026-06-21 
cor(v,L)
Warning in cor(v, L): the standard deviation is zero
     [,1] [,2] [,3]
[1,]   NA   NA   NA

sessionInfo()
R version 4.4.2 (2024-10-31)
Platform: aarch64-apple-darwin20
Running under: macOS Sequoia 15.6.1

Matrix products: default
BLAS:   /System/Library/Frameworks/Accelerate.framework/Versions/A/Frameworks/vecLib.framework/Versions/A/libBLAS.dylib 
LAPACK: /Library/Frameworks/R.framework/Versions/4.4-arm64/Resources/lib/libRlapack.dylib;  LAPACK version 3.12.0

locale:
[1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8

time zone: America/Chicago
tzcode source: internal

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

loaded via a namespace (and not attached):
 [1] vctrs_0.7.2       cli_3.6.5         knitr_1.51        rlang_1.1.7      
 [5] xfun_0.56         stringi_1.8.7     otel_0.2.0        promises_1.5.0   
 [9] jsonlite_2.0.0    workflowr_1.7.2   glue_1.8.0        rprojroot_2.1.1  
[13] git2r_0.36.2      htmltools_0.5.9   httpuv_1.6.16     sass_0.4.10      
[17] rmarkdown_2.30    jquerylib_0.1.4   evaluate_1.0.5    tibble_3.3.1     
[21] fastmap_1.2.0     yaml_2.3.12       lifecycle_1.0.5   whisker_0.4.1    
[25] stringr_1.6.0     compiler_4.4.2    fs_1.6.6          Rcpp_1.1.1       
[29] pkgconfig_2.0.3   rstudioapi_0.18.0 later_1.4.6       digest_0.6.39    
[33] R6_2.6.1          pillar_1.11.1     magrittr_2.0.4    bslib_0.10.0     
[37] tools_4.4.2       cachem_1.1.0