# Using the same linear predictor for multiple responses in BRMS

I am wondering how best to code using the same linear predictor in multiple responses, in BRMS. As a simple example, imagine we are counting alive birds and dead birds 🐦. The mean total number of birds depends on some covariates:
\mu = \sum_i \beta_i x_i

The number of alive birds is:
b_a \sim \mathrm{Poisson}(p_a \mu)

The number of dead birds is:
b_d \sim \mathrm{Poisson}((1-p_a) \mu)

I have read about coding multiple responses in BRMS but am looking for an example of sharing the same linear predictor between the multiple responses in this way. Any pointers welcome 🙂

I think calling two likelihoods would work:

vector[N] mu = X * beta; // beta as vector and X as matrix

b_a ~ poisson(p_a * mu);
b_d ~ poisson((1-p_a) * mu);



Explicitly, I think what would happen is this:

\mu = \bf{X} \beta (Using matrix notation, so \mu is a vector, with entries i to N)

p(b_a | \lambda=p_a \mu) = \frac{(p_a \mu)^{b_a} e^{-p_a \mu}}{b_a !}

p(b_d | \lambda=(1-p_a) \mu) = \frac{((1-p_a) \mu)^{b_d} e^{-(1-p_a) \mu}}{b_a !}

Bringing over to log probability, which is what Stan works with:

log~p(b_a) = b_a log(p_a \mu) - p_a \mu - log(b_a !)

log~p(b_d) = b_d log((1-p_a) \mu) - (1-p_a) \mu - log(b_d !)

Adding the two together and dropping additive constants that don’t depend on parameters, we have:

log~p(b_a, b_d) = log[p_a^{b_a} (1-p_a)^{b_d} \mu ^{b_a+b_d}] - \mu

So explicitly I think you could call the joint likelihood as:

target += log(p_a^b_a * (1-p_a)^b_d * mu^(b_a + b_d)) - mu;


And it might help to pre-calculate some of this stuff

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… Thanks, have edited my question to make it clear that I am trying to figure out if this is possible within BRMS. As you point out, this is straightforward to code up in Stan, but my actual problem is quite a bit more complex. If possible, I would prefer to stay within BRMS, so I can take advantage of all the other great features of BRMS.

Sorry, my reply got sent prematurely.

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Okay, so judging from this vignette on modelling multiple response variables and this vignette on specifying a nonlinear model it seems like this should be straightforward:

prior1 <- prior(normal(0, 1), nlpar = "b1") + prior(normal(0, 1), nlpar = "b2") + ...

ba_bf <- bf(b_a ~ (b1*predictor_1 + ... + bk*predictor_k) * p_a)
bd_bf <- bf(b_d ~ (b1*predictor_1 + ... + bk*predictor_k) * (1-p_a))

fit <- brm(
ba_bf + bd_bf,
data = bird_data, chains = 2, cores = 2, family = poisson(), nl = TRUE, , prior = prior1
)


I think the trick might be to use nl = TRUE to specify some of the shared structure with p_a and (1-p_a). Might want to look a little more closely about how to tell brm() which variables are data and which are parameters

Just for clarity, here is complete example Stan code:

data {
int<lower=1> N;  // total number of observations
int alive[N];  // alive birds
vector[N] x;  // environmental variable
}

parameters {
real beta0;
real beta1;
real<lower=0, upper=1> p;
}

transformed parameters {
vector[N] mu;
mu = exp(beta0 + beta1*x);
}

model {
beta0 ~ normal(0, 3);
beta1 ~ normal(0, 3);
p ~ uniform(0, 1);

alive ~ poisson(p * mu);
dead ~ poisson((1 - p) * mu);
}



If this is saved in multiple-responses.stan, then it works with the following rstan code:

library(rstan)
set.seed(2546)
N = 100
x=rnorm(N, 0, 1)
bird_data = list(x=x,
N=N,
alive=rpois(N, lambda=0.1*exp(x)),
)

fit <- stan(file='multiple-responses.stan', data=bird_data)


I tried @ScottAlder’s suggestion, but it threw errors, and I haven’t yet figured out how to make a model like this within BRMS.

Sorry! I haven’t quite been able to pick my way through the non-linear and multiple response documentation to get a solution that works. The first example I have found where reasoning in Stan has been easy, but reasoning in BRMS, not so much!

Sorry for jumping in so late. brms does currently not allow for shared predictor terms in a multivariate because of some decision choices I regret making initially. This will be changed with brms 3.0, though.

Since both of your models are poisson, I still have a solution for you. Write the data in long format, that is both responses b_a and b_d under each other as the same variable. Let’s call it b. Define a variable z that indicates if the observations belong to b_a and b_d. The full specification then looks as follows.

library(brms)

df <- data.frame(b = 1:100, predictor_1 = rnorm(100), z = rep(0:1, each = 50))

form <- bf(
b ~ eta * p^z * (1 - p)^(1-z),
eta ~ predictor_1 + ...,
p ~ 1,
nl = TRUE
)

bprior <- prior(normal(0, 5), nlpar = eta) +
prior(normal(0, 5), nlpar = p, lb = 0, ub = 1)
make_stancode(form, df, poisson(), prior = bprior)
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Thanks for clarifying 👍 and for the example. Look forward to BRMS 3.0 ♥️