Has the behavior of log_sum_exp changed?

Developers
1

Hi,

At least when using stan 2.11 on Windows 7, I was able to run the following program and estimate q.

  • model.stan:
data {
  int N;
  int<lower=0, upper=1> Y[N];
}

parameters {
  real<lower=0, upper=1> q;
}

model {
  for (n in 1:N)
    target += log_sum_exp(
      log(0.5) + bernoulli_lpmf(Y[n] | q),
      log(0.5) + bernoulli_lpmf(Y[n] | 1)
    );
}
  • run-model.R:
library(rstan)

Y <- c(1,0,1,0,0,1,1,0,1,1,0,1,0)
N <- length(Y)
data <- list(N=N, Y=Y)

fit <- stan(file='model.stan', data=data, seed=1234)

However, if I use the current version (stan 2.17) on Windows 7, the following error message will appear.

SAMPLING FOR MODEL 'model' NOW (CHAIN 1).
Rejecting initial value:
  Log probability evaluates to log(0), i.e. negative infinity.
  Stan can't start sampling from this initial value.
...
[1] "Error in sampler$call_sampler(args_list[[i]]) : Initialization failed."
[1] "error occurred during calling the sampler; sampling not done"

The reason is that log(0.5) + bernoulli_lpmf(Y[n] | 1) will be negative infinity when Y[n] = 0. Does the current version of log_sum_exp evaluate all cases first? Is this expected behavior?

If so, do I need to write as follows?

data {
  int N;
  int<lower=0, upper=1> Y[N];
}

parameters {
  real<lower=0, upper=1> q;
}

model {
  for (n in 1:N) {
    if (Y[n] == 0) {
      target += bernoulli_lpmf(Y[n] | q);
    } else {
      target += log_sum_exp(
        log(0.5) + bernoulli_lpmf(Y[n] | q),
        log(0.5) + bernoulli_lpmf(Y[n] | 1)
      );
    }
  }
}

Regards,
Kentaro

2

Your first version works for me on Linux. The second version should be fine as well. Also, log_sum_exp should work when one of its inputs is negative infinity. You might try with the log_mix function. Basically, I don’t understand why the initialization is failing for you.

3

Thank you for the reply! log_mix also produces the same error on Windows 7… Does anyone reproduce it on Windows?

4

Yes, I see this on Windows 7. Still doesn’t make much sense.