Can (Should) I center the observations of an exponential distribution?

Modeling
1

Let’s think of a simple exponential model (which works well and fast)

data {
  int <lower = 0> N;     // number of observations
  int <lower = 0> K;     // number of parameters 
  row_vector[K] X[N];
  real y[N];
}

parameters {
  vector[K] beta;
}

model {
  beta ~ normal(0, 1);
  for(t in 1:N) 
   y[t] ~ exponential(exp(X[t]*beta)); 
}

My model is actually more complicated and the chains don’t mix. I am trying to reparameterize the model using the fact that if X \sim Exp(\lambda) then X \times \lambda \sim Exp(1)

data {
  int <lower = 0> N;     // number of observations
  int <lower = 0> K;     // number of parameters 
  row_vector[K] X[N];
  real y[N];
}

parameters {
  vector[K] beta;
}

transformed parameters{
  real y_std [N];
  for(t in 1:N)
    y_std[t] = y[t]* exp(X[t]*beta); 
}

model {
  beta ~ normal(0, 1);
  for(t in 1:N)
    y_std[t] ~ exponential(1); 
}

But this second model is not right (does not recover the parameters in synthetic data). Does anyone know why? Thanks!

Edit: see code in response below

1 Like
2

I think the reparametrization introduces a change of variables and you’ll need the Jacobian adjustment.

4 Likes
3

Thanks, that was it. The following model code produces posterior parameters distributions that match the first code (and the truth). It is is 50% slower for this simple example.

model {
  real y_std [N];
  for(t in 1:N)
    y_std[t] = y[t]* exp(X[t]*beta);
  beta ~ normal(0, 1);
  for(t in 1:N){
    y_std[t] ~ exponential(1);
    target += X[t]*beta; 
  }
}

So that was it. Thanks.

4 Likes
4

Happy to help! You should mark your own answer as the solution. It has the code. I merely gave a hint.

2 Likes
5

Thanks. I did now.