Hello.

I’m using the `loo`

package to compare linear regression models, when the outcome variable has measurement errors. My problem is that I’m not sure of the expression of likelihood required by `loo`

.

Given a model like this:

y_i \sim \mathrm{normal}(u_i, s_i)

u_i \sim \mathrm{normal}(\beta_0 + \beta_1 x_i, \sigma)

The values y_i, s_i and x_i are observed. s_i is the standard deviation of the measurement error of each point. The value u_i is not observed.

`loo`

requires the value of the likelihood of each point. In this case I’m not sure if I have to use the joint likelihood of y_i, u_i (a bivariate normal) like this:

```
generated quantities {
vector[N] log_like;
for (i in 1:N) {
log_like[i] = normal_lpdf(y[i] | u[i], sy[i]) +
normal_lpdf(u[i] | beta0 + beta1*x[i], sigma);
}
}
```

or the marginal likelihood of y_i alone (a univariate normal) like this:

```
generated quantities {
vector[N] log_like;
for (i in 1:N) {
log_like[i] = normal_lpdf(y[i] | beta0 + beta1*x[i], sqrt(sigma**2 + sy[i]**2));
}
}
```

I have read the papers, but I’m not a statistician so I’m not sure which likelihood (if any) is correct in this case.