Lower bound on each outcome error is vector not constant error

here is some toy data. My outcome variable has a lower bound which is outcome >= -x where x is a predictor.

This seems to be a truncated data situation so I looked on page 99 of the stan manual it goes over adding a constraint on the outcome: http://www.uvm.edu/~bbeckage/Teaching/DataAnalysis/Manuals/stan-reference-2.8.0.pdf

The examples define the contraint as a single number “real U”. In my case the boundary is not a single number U but a vector of numbers so I declared “real U[N]”. When I run the code I get this error:

ERROR - Lower bounds in truncated distributions must be univariate.

It seems I’m very close to a solution here if a vector bound was available. Is there a way to incorporate a lower bound that is a vector? Thank you.

Here is the toy code:

x =   rep(seq(-10,10,1),each=5) 
y =  rep(0,length(x) )

for(i in 1:length(x)){
y[i] = sample( seq(-x[i],-x[i]+ ifelse(x[i]<0 ,
sample(seq(5,40,.1),1) ),1)


weights = sample( seq(1,1000,1) ,length(x))
groups = rep( letters[1:5], times =length(x)/5 )

# create a numeric vector to indicate the categorical groups
dat$GROUP_ID  = match(  dat$group, levels( dat$group   )  ) 
dat$BOUND = -dat$x
standat <- list(
  N = nrow(dat),
  y = dat$y,
  x = dat$x,
  nGROUPS = 5,
  L = dat$BOUND, 

stanmodelcode = '
data {
int<lower=1> N; 
int nGROUPS;
real y[N];                                 
real x[N];   
int<lower=1, upper=nGROUPS> GROUP_ID[N];
real L[N];

transformed data{


parameters {
real intercept;  
vector[nGROUPS] intercept_x2;
real beta;                   
real<lower=0> sigma;    
real<lower=0>  sigma_2;

transformed parameters {  // none needed

model {
real mu;

// priors
intercept~ normal(0,10);
intercept_x2 ~ normal(0,sigma_2);
beta ~ normal(0,10);
sigma ~ normal(0,10);
sigma_2 ~ normal(0,10);

// likelihood

for(i in 1:N){
mu = intercept + intercept_x2[ GROUP_ID[i] ] +  beta*x[i];
y[i] ~ normal(mu, sigma) T[L,];


fit22 = stan(model_code=stanmodelcode, data=standat, iter=2000, warmup=500, chains = 1)

With T[L[i],]

ugh yes of course I forgot about the loop. thank you!