Quantifying the magnitude of a population decline with Bayesian time-series modelling
Population abundances tend to vary year to year. This variation can make
it make it hard detect a change and hard to quantify exactly what that
Bayesian time-series analysis can help us quantify a decline and put
uncertainty bounds on it too. Here I will use the R-INLA
package to fit a time-series model to a
For instance, take the pictured time-series. Quantifying change as the
difference between the first and last time-points is obviously
misleading. Doing so would imply that abundance has declined by 77% from
the historical value.
Another approach would be to compare the average of the first and last
decades. Doing so would yield a 72% decline.
A better way might be to model the population trend over time and then
estimate our change from the model. An advantage of a model is we can be
more specific (and transparent) about assumptions. Using Bayesian
time-series analysis we can also pick a model that is appropriate for
the distribution of the data. e.g. here I will use negative
errors, because the abundances are counts.
Simulating a time-series
Let’s simulate a simple time-series of counts that declines at a
constant (exponential) rate:
set.seed(1969) n <- 50 rho <- 0.8 prec <- 10 init <- 7 acoef <- c(rep(-0.01, n/2), rep(-0.03, n/2)) eta <- init + (acoef*(1:n)) y <- rnbinom(n, mu = exp(eta), size = 5) data <- list(y =y, z = 1:n)
The results are pictured above. In short we calculate
eta which is the
linear predictor. Its exponent determines the mean of the negative
binomial process, hence the exponential (not linear) decline. The
size = 5 means that the count data will be moderately over-dispersed –
that is quite noisy (the variance will be > than the mean).
Notice that I put a bit of a ‘kink’ in eta by having the rate of decline
acoef), increase halfway through the time-series like this:
We can calculate the magnitude of change like this:
1 - (y[n]/y) =
0.77 which can be interpreted as what fraction of the intial years
biomass has been lost?
What we want to do is put a smoother over time, so our estimate accounts
for short-term variation. We also want to get credibility intervals (95%
probability intervals) on the estimate of decline.
Fitting a time-series model with INLA
R-INLA is a pretty handy package. It is a very
way to fit generalized linear models and can also handle a huge range of
different types of random
effects. Here we will use
an auto-regressive lag one process. That is saying that abundance at
time t depends on abundance at time t-1 with some constant correlation
We could also include other covariates in the model, for instance,
abundance of a predator that might eat our organism of interest.
However, we won’t go into that here.
First up, we specify the formula in INLA’s notation:
library(INLA) f1 <- y ~ 1 + f(z, model = "ar1")
Which just says model
y as a function of an intercept term (constant
mean) + an autoregressive process that depends on
z (just an index for
the year). The
f() is INLA’s standard notation for modelling its
different random effects models (AKA latent models), here we chose the
Now the trick is to get INLA to compute the % difference between the
expectation for the first and last time points as it goes. That way
we not only get an estimate of the change but also we will get the full
posterior distribution, so we can get our CIs. Note I put emphasis on
expectation, because we won’t simply be calculating the difference
between the first and last points (we did that above), but will in fact
be calculating the difference between the model’s estimated mean values
for the first and last points.
To do that we use INLA’s linear
This feature let’s us compute a linear combination of the estimated
covariates. Here we will ask for the difference between the first and
last values of
z (the time-series).
inla will multiply the
values by the numbers we give below (here
1), then sum them
lc1 <- inla.make.lincomb(z = c(-1, rep(NA, n-2), 1)) names(lc1) = "lc1"
The row of
NA in the middle just says to ignore the middle points.
Note that we have also ignored the intercept. More on this in a moment.
Now we can fit the model, using negative binomial errors and specifying
our linear combination:
m1 <- inla(f1,family = "nbinomial", data = data, lincomb = lc1, control.inla = list(lincomb.derived.only = TRUE), control.predictor = list(compute = TRUE, link = 1) )
And the results summary can be obtained (not shown here):
Note the large estimate for
rho, which is the auto-correlation
parameter. That happens because we have such a strong decline.
We could have also fit just a linear trend to our data, however, because
the rate varies over time (and we wouldn’t know that beforehand if we
had real data, not data we made up), it is nice to use
ar1 which has a
kind of smoothing effect.
Also worth noting that modelling an trending (non-stationary)
time-series with an
ar1 is not technically correct if we want to
estimate rho. However, here we use
ar1 because it acts like a smoother
in a GAM [e.g. see
Examining the model’s predictions
It is pretty straightforward to plot INLA’s predictions for the
time-series. They are all stored in the
plot(data$z, data$y, col = 'grey', type = 'l', lwd = 2, xlab = "years", ylab = "Abundance") lines(data$z, m1$summary.fitted.values$mean, col = "tomato", lwd = 2) lines(data$z, m1$summary.fitted.values[,3], col = "tomato", lwd = 1, lty = 2) lines(data$z, m1$summary.fitted.values[,5], col = "tomato", lwd = 1, lty = 2) legend('topright', legend = c("observed", "expected", "95% CIs"), lty = c(1,1,2), col = c("grey", "tomato", "tomato"))
Now we can extract the change in population size from our linear
m1$summary.lincomb.derived$mean ##  -1.081196
However, this is not quite the number we want. It is the change in
or the linear predictor. Negative binomial models have a log-link, so
that is the change in log-space. We want the real change. If we
100 * (1 - exp(m1$summary.lincomb.derived$mean)) ##  66.08103
we get the change as a % from the historical mean The reason being is
that the linear combination is
log(eta[n]) - log(eta). Taking the
exponent gets the change and one minus that is the % loss.
It is easy now to get the CIs on the linear combination:
m1$summary.lincomb.derived ## ID mean sd 0.025quant 0.5quant 0.975quant mode kld ## lc1 1 -1.081196 0.309715 -1.695217 -1.08053 -0.4716486 -1.079157 0
We could also look at the marginal and plot the posterior distribution
of the % loss:
losses <- 100 * (1 - exp(m1$marginals.lincomb.derived[][,1])) dens <- m1$marginals.lincomb.derived[][,2] plot(losses, dens, type = 'l', xlim = c(0, 100), xlab = "% loss", ylab = "density", las = 1, main = "Posterior density for % loss") polygon(x = c(losses, rev(losses)), y = c(dens, rep(0, length(dens))), col = "turquoise3")