A problem regarding "Log probability evaluates to log(0)"

Modeling
1

Hello Stan users and exports,

I have a problem as the tittle describes.
My model is a little complex so I paste the parts that cause error.

data{
  int<lower=0> y[3,18];
  real log_max; // log(1000)
//etc...
}

parameters{
  vector<lower = 0>[2] sigma;
  vector[2] U;
  matrix[2,2] B;
  matrix<upper=log_max>[2,18]log_y_2;
//etc...
}

transformed parameters{
simplex[3] theta[18];
matrix<upper=log_max>[3,18]log_y;

  for(t in 1:18){
    log_y[1,t]=log_y_2[1,t];
    log_y[2,t]=log_y_2[2,t];
    if ((exp(log_max)-exp(log_y_2[1,t])-exp(log_y_2[2,t]))<25)
    log_y[3,t]=log(25);
    else 
    log_y[3,t]=log(exp(log_max)-exp(log_y_2[1,t])-exp(log_y_2[2,t]));
  }
    for (t in 1:18){
    theta[t]=softmax(exp(log_y[,t]));
  }
}

model{
  for (i in 1:18) {
        y[,i] ~ multinomial(theta[i,]);
       }
  for(t in 2:18){
  target+= multi_normal_lpdf(log_y_2[,t] |U+B*log_y_2[,t-1],diag_matrix(sigma)); 
//etc... 
  }

After many attempts, I found that log_y_3[3,t] in transformed parameters is the problem. If I write

    if ((exp(log_max)-exp(log_y_2[1,t])-exp(log_y_2[2,t]))<25)
    log_y[3,t]=log(25);
    else 
    log_y[3,t]=log(exp(log_max)-exp(log_y_2[1,t])-exp(log_y_2[2,t]));

there would be a log(0) error.
However, if I write

    if ((exp(log_max)-exp(log_y_2[1,t])-exp(log_y_2[2,t]))<25)
    log_y[3,t]=log(25);
    else 
    log_y[3,t]=log(50);

It will work.
But if change 50 to 1000 or 500, it will report an error once again.

Actually, I need log_y[3,t]=log(exp(log_max)-exp(log_y_2[1,t])-exp(log_y_2[2,t]));. Because log_y is a matrix that elements in each column follow multinomial distribution. How can I successfully achieve this goal?

Thank you in advance

2

Hi,
just going through unanswered questions. I think this problem did get eventually handled (or is in the process of being handled) in https://discourse.mc-stan.org/t/how-do-i-limit-the-sum-of-the-elements-in-a-vector-to-less-than-a-constant is that correct? Or is there something else you need?

Best of luck!

3

Yes, this question is in the process of being handled in that link. These were originally two different questions, but they are now being solved together. Thank you for the concern.