# Other count regressions

Corrie November 11, 2018

## Multinomial Regression

A multinomial regression is used when more than two things can happen. As an example, suppose we are modelling career choices for some young adults. Let’s assume there are three career choices one can take and expected income is one of the predictors. One option to model the career choices would be the explicit multinomial model which uses the multinomial logit. The multinomial logit uses the multinomial distribution which is an extension of the binomial distribution to the case with events.

### Explicit Multinomial Model

In the explicit multinomial model we need linear models for different events. In our example, we’d thus have two different linear models.

We first simulate the career choices:

```
library(rethinking)
set.seed(2405)
N <- 1000 # number of individuals
income <- 1:3 # expected income of each career
score <- 0.5*1:3 # scores for each career, based on income
# convert scores to probabilities
(p <- softmax(score[1], score[2], score[3]) )
```

```
[1] 0.1863237 0.3071959 0.5064804
```

Next, we simulate the career choice an individual made.

```
career <- sample(1:3, N, prob=p, replace=TRUE)
```

To fit the model, we use the `dcategorical`

likelihood.

```
m10.16 <- map(
alist(
career ~ dcategorical( softmax( 0, s2, s3 )),
s2 <- b*2, # linear model for event type 2
s3 <- b*3, # linear model for event type 3
b ~ dnorm(0, 5)
), data=list(career=career)
)
```

```
precis(m10.16)
```

```
mean sd 5.5% 94.5%
b 0.3499246 0.02952974 0.3027303 0.3971188
```

So what does this estimate mean? From the parameter `b`

we can compute
the scores `s2 =`

0.62 and `s3 =`

0.93. The probabilities are then
computed from these scores via `softmax()`

. However, these score are
very difficult to interpret. In this example, we assigned a fixed score
of 0 to the first event, but we also could have assigned a fixed score
to the second or third event. The score would be very different in that
case. The probabilities obtained through `softmax()`

will stay the same
though, so it’s absolutely important to always convert these scores to
probabilities.

```
post <- extract.samples(m10.16)
post$s2 <- post$b * 2
post$s3 <- post$b * 3
probs <- apply(post, 1, function(x) softmax(0, x[2], x[3]))
(pr <- apply(probs, 1, mean) )
```

```
[1] 0.1705797 0.3428060 0.4866143
```

The probabilities are reasonably close to the one we used to simulate our data.

For comparison, the proportions of our simulated data:

```
table(career) / length(career)
```

```
career
1 2 3
0.180 0.314 0.506
```

Note again, the scores we obtain from the model are very different from the scores we used for the simulation, only the probabilities match!

Let’s have a look at the densities for each probability:

```
dens(probs[1,], main="Probability for Event 1")
```

```
dens(probs[2,], main="Probability for Event 2")
```

```
dens(probs[3,], main="Probability for Event 3")
```

The distribution for the probability of the second event is heavily skewed. I assume this is since the three probabilities are not linearly independent of each other: They need to sum up to 1. Furthermore, we constructed the non-fixed scores as multiples of each other.

**Further notes**

There are various ways how to put the linear models and its scores in the softmax. Here we used

```
s1 <- 0
s2 <- b*2
s3 <- b*3
```

which translates to trying to find a
such
that

But actually, even without the error term
, it’s not guaranteed that such a
exists.
In this case, such a
does
not exist, so the best
is the
one that minimizes the error term. The solution to this optimization
problem is indeed very close to the result we got for `b`

.

There are also other possibilities in how we specify our model. One option for example would be to use different multiplicands for the scores:

```
s1 <- 0
s2 <- b
s3 <- b*2
```

which would indeed get better estimates in this case, since there is a that almost solves the resulting equations exactly. For such a model, we would get the following estimate for

```
mean sd 5.5% 94.5%
b 0.5098878 0.04125471 0.4439548 0.5758208
```

and the resulting probability estimates would be

```
[1] 0.1840538 0.3060746 0.5098716
```

For comparison, the proportions in our simulated data:

```
career
1 2 3
0.180 0.314 0.506
```

These are quite close!

We don’t need to restrict ourselves to just one parameter for the scores and its multiples, we an also introduce another parameter for each score. For example, we could specify the models as follows:

```
s1 <- 0
s2 <- b2
s3 <- b2 + b3
```

This also gives better estimates than the original model but also
introduces correlation between `b2`

and `b3`

.

```
mean sd 5.5% 94.5%
b2 0.5563025 0.0934674 0.4069236 0.7056815
b3 0.4771160 0.0718306 0.3623168 0.5919151
```

The resulting probability estimates are

```
[1] 0.1800222 0.3139969 0.5059809
```

Interestingly, if we use a model, where each score gets its own parameter as follow

```
s1 <- 0
s2 <- b2
s3 <- b3
```

the correlation of the estimates get worse:

```
mean sd 5.5% 94.5%
b2 0.5560173 0.09345729 0.4066545 0.7053801
b3 1.0331494 0.08675576 0.8944969 1.1718018
```

The resulting probabilities:

```
[1] 0.1800629 0.3139782 0.5059589
```

In total, which model specification to pick in an explicit multinomial is not quite straightforward.

#### Including Predictor Variables

The model sofar isn’t very interesting yet. It doesn’t use any predictor variables and thus for every individual it simply outputs the same probabilities as above. Let’s get this a bit more interesting by including family income as a predictor variable into our model. We first simulate new data where the family income has an influence on the career taken by an individual:

```
N <- 100
# simulate family incomes for each individual
family_income <- runif(N)
# assign a unique coefficient for each type of event
b <- (1:-1)
career <- c()
for (i in 1:N) {
score <- 0.5*(1:3) + b * family_income[i]
p <- softmax(score[1], score[2], score[3])
career[i] <- sample(1:3, size=1, prob = p)
}
```

To include the predictor variable in our model, we include it for both linear models of event type 1 and type 2. It is important that the predictor variable has the same value in both models but we use unique parameters:

```
m10.17 <- map(
alist(
career ~ dcategorical( softmax( 0, s2, s3)),
s2 <- a2 + b2*family_income,
s3 <- a3 + b3*family_income,
c(a2, a3, b2, b3) ~ dnorm(0, 5)
),
data=list(career=career, family_income=family_income)
)
precis(m10.17)
```

```
mean sd 5.5% 94.5%
a2 -0.6593230 0.5301211 -1.5065589 0.1879128
a3 0.9576812 0.4229572 0.2817140 1.6336485
b2 1.0681466 0.8794221 -0.3373397 2.4736329
b3 -1.4571748 0.8330821 -2.7886009 -0.1257488
```

Again, we should look at the probabilities and not the scores.

First, we compute scores for each individual:

```
scores <- link(m10.17)
```

```
str(scores)
```

```
List of 2
$ s3: num [1:1000, 1:100] 0.754 0.475 1.308 0.646 0.896 ...
$ s2: num [1:1000, 1:100] -0.956 -0.882 -0.71 -0.935 -0.464 ...
```

This time, since the score depends on the family income, we get different scores for each individual. For the first individual, we would then get the following probability estimates:

```
apply(softmax(0, scores$s2[,1], scores$s3[,1]), 2, mean)
```

```
[1] 0.2655084 0.1576247 0.5768670
```

### Multinomial using Poisson

One way to fit a multinomial likelihood is to refactor it into a series of Poisson likelihoods. It is often computationally easier to use Poisson rather than multinomial likelihood. We first will take an example from earlier with only two events (that is, a binomial case) and model it using Poisson regression:

```
data("UCBadmit")
d <- UCBadmit
```

First the binomial model for comparison:

```
set.seed(2405)
m_binom <- map(
alist(
admit ~ dbinom(applications, p),
logit(p) <- a,
a ~ dnorm(0, 100)
), data=d
)
precis(m_binom)
```

```
mean sd 5.5% 94.5%
a -0.4567393 0.03050707 -0.5054955 -0.4079832
```

And then the Poisson model. To obtain the probability of admission, we model both the rate of admission and the rate of rejection using Poisson:

```
d$rej <- d$reject
m_pois <- map2stan(
alist(
admit ~ dpois(lambda1),
rej ~ dpois(lambda2),
log(lambda1) <- a1,
log(lambda2) <- a2,
c(a1, a2) ~ dnorm(0, 100)
), data=d, chains=3, cores=3
)
```

We will consider only the posterior means.

For the binomial model, the inferred probability of admission is:

```
logistic(coef(m_binom))
```

```
a
0.3877596
```

In the Poisson model, the probability of admission is given by:

That is:

```
a <- as.numeric(coef(m_pois))
exp(a[1]) / ( exp(a[1]) + exp(a[2]) )
```

```
[1] 0.3876155
```

#### Career Choices again, with Poisson

In our example from above, estimating probabilities for career choices, we would build the following model:

First, we need to reform the data set:

```
d <- data.frame(career,
career1= 1*(career==1),
career2= 1*(career==2),
career3= 1*(career==3))
```

```
m10.16_poisson <- map2stan(
alist(
career1 ~ dpois(lambda1),
career2 ~ dpois(lambda2),
career3 ~ dpois(lambda3),
log(lambda1) <- b1, # linear model for event type 1
log(lambda2) <- b2, # linear model for event type 2
log(lambda3) <- b3, # linear model for event type 3
c(b1, b2, b3) ~ dnorm(0, 5)
), data=d
)
precis(m10.16_poisson, corr = TRUE)
```

This then gives us the following probability estimates for the three career choices:

```
a <- as.numeric(coef(m10.16_poisson))
exp(a) / sum( exp(a) )
```

```
[1] 0.3008142 0.2691109 0.4300749
```

For comparison, the proportions in the data set:

```
career
1 2 3
0.30 0.27 0.43
```

## Geometric Model

The geometric distribution can be used if you have a count variable that
is counting the number of events up until something happens. Modelling
the probability of the terminating event is also known as *Event history
analysis* or *Survival analysis*. If the probability of that event
happening is constant through time and the units of time are discrete,
then the geometric distribution is a common likelihood function that has
maximum entropy under certain conditions.

```
# simulate
N <- 100
x <- runif(N)
y <- rgeom(N, prob=logistic(-1 + 2*x ))
# estimate
m10.18 <- map(
alist(
y ~ dgeom( p) ,
logit(p) <- a + b*x,
a ~ dnorm(0, 10),
b ~ dnorm(0, 1)
), data=list(y=y, x=x)
)
precis(m10.18)
```

```
mean sd 5.5% 94.5%
a -1.044255 0.2133367 -1.385209 -0.7033021
b 1.705896 0.4224419 1.030752 2.3810393
```