# Samples of Thoughts

## Poisson Regression

### Oceanic Tools

A binomial distriution with many trials (that is $n$ large) and a small probability of an event ($p$ small) approaches a Poisson distribution where both the mean and the variance are equal:

y <- rbinom(1e5, 1000, 1/1000)
c(mean(y), var(y))
[1] 0.996960 1.000841

A Poisson model allows us to model binomial events for which the number of trials $n$ is unknown.

We work with the Kline data, a dataset about Oceanic societies and the number of found tools. We will analyse the number of found tools in dependance of the population size and contact rate. The hypothesis is, that a society with a large population has a larger number of tools and further that a high contact rate to other societies increases the effect of population size.

library(rethinking)
data(Kline)
d <- Kline
knitr::kable(d)
culture population contact total_tools mean_TU
Malekula 1100 low 13 3.2
Tikopia 1500 low 22 4.7
Santa Cruz 3600 low 24 4.0
Yap 4791 high 43 5.0
Lau Fiji 7400 high 33 5.0
Trobriand 8000 high 19 4.0
Chuuk 9200 high 40 3.8
Manus 13000 low 28 6.6
Tonga 17500 high 55 5.4
Hawaii 275000 low 71 6.6

The data looks rather small, it has only 10 rows.

d$log_pop <- log(d$population)
d$contact_high <- ifelse( d$contact == "high", 1, 0)

We will use the following model: \begin{align*} T_i &\sim \text{Poisson}(\lambda_i) \\ \log \lambda_i &= \alpha + \beta_P \log P_i + \beta_C C_i + \beta_{PC}C_i \log P_i \\ \alpha &\sim \text{Normal}(0, 100) \\ \beta_P &\sim \text{Normal}(0, 1) \\ \beta_C &\sim \text{Normal}(0, 1) \\ \beta_{PC} &\sim \text{Normal}(0, 1) \end{align*} where $P$ is population, $C$ is contact_high, and the $T$ is total_tools.

m10.10 <- map(
alist(
total_tools ~ dpois( lambda ),
log(lambda) <- a + bp*log_pop +
bc*contact_high + bpc*contact_high*log_pop,
a ~ dnorm( 0, 100),
c(bp, bc, bpc) ~ dnorm(0, 1)
), data=d
)
precis(m10.10, corr=TRUE)
           mean         sd       5.5%     94.5%
a    0.94422292 0.36008794  0.3687328 1.5197130
bp   0.26401583 0.03466759  0.2086103 0.3194213
bc  -0.09002292 0.84140405 -1.4347491 1.2547033
bpc  0.04254207 0.09227171 -0.1049259 0.1900101
plot( precis( m10.10 ))

It appears as if the contact rate, as well as its interaction with population, has no influence on the number of tools. But the table is misleading. Let’s compute some counterfactual predictions. Consider two islands, both with log-population of 8 but one with high and the other with low contact rate.

post <- extract.samples(m10.10 )
lambda_high <- exp( post$a + post$bc + (post$bp + post$bpc)*8 )
lambda_low <- exp( post$a + post$bp*8 )
diff <- lambda_high - lambda_low
sum(diff > 0) / length(diff)
[1] 0.9532

There is a 95% probability that the high-contact island has more tools than the island with low-contact.

par(mfrow=c(1,2))
dens( diff , col="steelblue", lwd=2,
xlab="lambda_high - lambda_low")
abline(v = 0, lty=2)
abline(h=0, col="grey", lwd=1)

pr <- as.data.frame(precis(m10.10, prob=0.95))
bc <- post$bc[1:500] bpc <- post$bpc[1:500]
plot( bc, bpc, col=densCols(bc, bpc), pch=16, cex=0.7 )
lines( pr["bc", 3:4], rep(min(bpc), 2))
lines( rep(min(bc), 2),  pr["bpc", 3:4])
points(c( pr["bc", 1], min(bc) ), c(min(bpc), pr["bpc", 1] ),
pch=16, cex=0.7 )

The reason for this behaviour lies in the strong correlation in the uncertainty of the two parameters. It is important to not just inspect the marginal uncertainty in each parameter, but also the joint uncertainty.

A better way to assess whether a predictor is expected to improve prediction is to use model comparison. We will compare 5 models.

A model that omits the interaction:

# no interaction
m10.11 <- map(
alist(
total_tools ~ dpois( lambda ),
log(lambda) <- a + bp*log_pop + bc*contact_high,
a ~ dnorm(0, 100),
c(bp, bc) ~ dnorm(0, 1)
), data=d
)

Two more models, each with only one of the predictors:

# one predictor
m10.12 <- map(
alist(
total_tools ~ dpois( lambda ),
log(lambda) <- a + bp*log_pop,
a ~ dnorm(0, 100),
bp ~ dnorm(0, 1)
), data=d
)
m10.13 <- map(
alist(
total_tools ~ dpois( lambda ),
log(lambda) <- a + bc*contact_high,
a ~ dnorm(0, 100),
bc ~ dnorm( 0, 1)
), data=d
)

A “null” model with only the intercept:

# no predictor
m10.14 <- map(
alist(
total_tools ~ dpois( lambda ),
log(lambda) <- a,
a ~ dnorm(0, 100)
), data=d
)

We compare all models using WAIC:

( islands.compare <- compare( m10.10, m10.11, m10.12, m10.13, m10.14, n=1e4) )
            WAIC        SE      dWAIC       dSE     pWAIC       weight
m10.11  79.22960 11.185306  0.0000000        NA  4.290270 5.805228e-01
m10.10  80.09441 11.382020  0.8648086  1.191099  4.822664 3.767285e-01
m10.12  84.44679  8.819275  5.2171839  8.306973  3.757409 4.274863e-02
m10.14 141.30179 31.414794 62.0721813 34.262213  8.212397 1.927598e-14
m10.13 150.67501 44.862135 71.4454091 46.903474 16.906044 1.776830e-16
plot( islands.compare )

The top two models include both predictors but the top model excludes the interaction. However, a lot of model weight is assigned to both models. This suggests that the interactionis probably overfit but the model set is decent evidence that contact rate matters.

Let’s plot some counterfactual predictions.

# different plot markers for high/low contact
pch <- ifelse( d$contact_high==1, 16, 1) plot( d$log_pop, d$total_tools, col="steelblue", pch=pch, xlab="log-population", ylab="total tools") # sequence of log-population sizes to compute over log_pop.seq <- seq(from=6, to=13, length.out = 30) # compute trend for high contact islands d.pred <- data.frame( log_pop = log_pop.seq, contact_high = 1 ) lambda.pred.h <- ensemble( m10.10, m10.11, m10.12, data=d.pred) lambda.med <- apply( lambda.pred.h$link, 2, median )
lambda.PI <- apply( lambda.pred.h$link, 2, PI ) # plot predicted trend for high contact islands lines( log_pop.seq, lambda.med, col="steelblue") shade( lambda.PI, log_pop.seq, col=col.alpha("steelblue", 0.2)) # compute trend for low contact islands d.pred <- data.frame( log_pop = log_pop.seq, contact_high = 0 ) lambda.pred.l <- ensemble( m10.10, m10.11, m10.12, data=d.pred) lambda.med <- apply( lambda.pred.l$link, 2, median )
lambda.PI <- apply( lambda.pred.l$link, 2, PI ) # plot predicted trend for high contact islands lines( log_pop.seq, lambda.med, lty=2) shade( lambda.PI, log_pop.seq, col=col.alpha("black", 0.1 )) Next, we’ll check if the MAP estimates accurately describe the shape of the posterior. m10.10stan <- map2stan( m10.10, iter=3000, warmup=1000, chains=4) precis( m10.10stan)  mean sd 5.5% 94.5% n_eff Rhat4 a 0.94146861 0.36526139 0.3440976 1.5085371 2722.643 1.001720 bp 0.26359976 0.03523350 0.2091810 0.3204490 2713.974 1.001768 bc -0.08922339 0.84032207 -1.4562229 1.2267930 2786.084 1.001813 bpc 0.04271506 0.09228461 -0.1012971 0.1929941 2842.538 1.001923 The estimates and intervals are the same as before. However, if we look at the pairs plot pairs(m10.10stan) we see that there is a very high correlation between the parameters a and bp and between bc and bpc. Hamiltonian Monte Carlo can handle these correlations, however, it would still be better to avoid them. We can center the predictors to reduce the correlation: d$log_pop_c <- d$log_pop - mean(d$log_pop)

m10.10stan.c <- map2stan(
alist(
total_tools ~ dpois( lambda ),
log(lambda) <- a + bp*log_pop_c + bc*contact_high +
bcp*log_pop_c*contact_high,
a ~ dnorm(0, 10),
bp ~ dnorm(0, 1),
bc ~ dnorm(0, 1),
bcp ~ dnorm(0, 1)
),
data=d, iter=3000, warmup=1000, chains=4
)
precis(m10.10stan.c)
          mean         sd       5.5%     94.5%    n_eff    Rhat4
a   3.30922186 0.08957579  3.1615992 3.4495053 3510.400 1.000697
bp  0.26316573 0.03504921  0.2083146 0.3200965 4450.243 1.000443
bc  0.28669798 0.11713837  0.0971141 0.4766120 3667.258 1.000495
bcp 0.06443262 0.17111533 -0.2113402 0.3405277 5206.212 1.000072

The estimates look different because of the centering, but the predictions remain the same. If we look at the pairs plot now, we see that the strong correlations are gone:

pairs(m10.10stan.c)

The Markov Chain has also become more efficient which we can see at the increased number of effective samples.

### Examples with different exposure Time

When the length of observation varies, the counts we observe also vary. We can handle this by adding an offset to the linear model.

As in the previous example, we have a monastery that completes about 1.5 manuscripts per day, that is $\lambda=1.5$.

num_days <- 30
y <- rpois( num_days, 1.5)

Now y holds 30 days of simulated counts of completed manuscripts. Next, we consider to buy another monastery which unfortunately doesn’t keep daily records but weekly records instead. Let’s assume this monastery has a real daily rate of $\lambda = 0.5$. To simulate data on a weekly basis, we just multiply this average by 7, the exposure:

num_weeks <- 4
y_new <- rpois( num_weeks, 0.5*7 )

So y_new holds four weeks of counts of completed manuscripts.

y_all <- c(y, y_new)
exposure <- c( rep(1, 30), rep(7, 4) )
monastery <- c(rep(0, 30), rep(1, 4))
d <- data.frame(y=y_all, days=exposure, monastery=monastery)
knitr::kable( head(d) )
y days monastery
3 1 0
1 1 0
1 1 0
3 1 0
2 1 0
2 1 0

Next, we fit a model to estimate the rate of manuscript production at each monastery.

d$log_days <- log(d$days)

m10.15 <- map(
alist(
y ~ dpois( lambda ),
log(lambda) <- log_days + a + b*monastery,
a ~ dnorm(0, 100),
b ~ dnorm(0, 1)
), data=d
)
post <- extract.samples(m10.15 )
lambda_old <- exp( post$a ) lambda_new <- exp( post$a + post\$b)
precis( data.frame( lambda_old, lambda_new))
                mean        sd      5.5%     94.5%    histogram
lambda_old 1.8471127 0.2495663 1.4672893 2.2651459  ▁▁▁▃▇▇▃▂▁▁▁
lambda_new 0.5934118 0.1447132 0.3898765 0.8506193 ▁▂▅▇▅▃▁▁▁▁▁▁

The estimates are indeed very close to the real values.