This is based on Example 5.1 (pages 143-5) – Figures 5.1 (page 144) and 5.2 (page 146) of Gamerman and Lopes (2006) which in turn is based on an example of Carlin, Gelfand and Smith (1992)

Essentially we consider count data where: $$Y_i ~ Poisson(\lambda_i)$$ subject to

$\lambda_i = \lambda_b$ if year $<$ changepoint, and $\lambda_i = \lambda_a$ otherwise. Assume $\lambda_b ~ Gamma(\alpha, \beta)$ and $\lambda_a ~ Gamma(\gamma, \delta)$. I think we have discrete uniform prior on $Pr[Change = Year]$. The posterior density is given by:

$$f(\lambda_b, \lambda_a, changepoint) \propto f(y_1, \ldots, y_n | \lambda_b, \lambda_a, changepoint)p(\lambda_b, \lambda_a, changepoint)$.

Setting the two sufficient statistics $s_b=\sum_{i=1}^{changepoint} y_i$ and $s_a = sum_{changepoint+1}^n y_i$ we have the posterior as:

$$f(\lambda_b, \lambda_a, changepoint) \propto \lambda_b^{\alpha + s_m - 1} e^{-(\beta+changepoint)\lambda_b} \lambda_a^{\gamma+s_a-1} e^{-(\delta s_a+n-changepoint)\lambda_a}$$

This has three simple conditional densities which are specified later.

In total, there are 112 observations of $y_i$ from 1851 to 1962.

In [3]:

```
from numpy import array
%load_ext rmagic
```

In [11]:

```
disasters = ( 4, 5, 4, 0, 1, 4, 3, 4, 0, 6, 3, 3, 4, 0, 2, 6,
3, 3, 5, 4, 5, 3, 1, 4, 4, 1, 5, 5, 3, 4, 2, 5,
2, 2, 3, 4, 2, 1, 3, 2, 2, 1, 1, 1, 1, 3, 0, 0,
1, 0, 1, 1, 0, 0, 3, 1, 0, 3, 2, 2, 0, 1, 1, 1,
0, 1, 0, 1, 0, 0, 0, 2, 1, 0, 0, 0, 1, 1, 0, 2,
3, 3, 1, 1, 2, 1, 1, 1, 1, 2, 4, 2, 0, 0, 1, 4,
0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1)
years = range(1851,1962)
len(disasters)
idx = range(111)
```

Out[11]:

In [12]:

```
%Rpush years disasters
%R plot(years, disasters)
```

In [13]:

```
import numpy as np
idxl = np.tril_indices(111)
idxu = np.triu_indices(111)
##create blank matrices of zeros
a=np.zeros(shape=(111,111))
b=np.zeros(shape=(111,111))
## Fill the upper triangle with ones
a[idxl] = 1
b[idxu] = 1
## Now get cumulative sums
cumsumafter = np.dot(b, disasters)
cumsumbefore = np.dot(a, disasters)
## Now append zero at start/end as appropriate.
cumsumafter = np.append(cumsumafter, 0)
cumsumbefore = np.append(0, cumsumbefore)
```

In [16]:

```
%Rpush years cumsumbefore
%R plot(years, cumsumbefore[-1])
```

In [18]:

```
from scipy import random, exp
#Provide values for the priors
alphaprior = 0.001
betaprior = 0.001
gammaprior = 0.001
deltaprior = 0.001
```

We need to do the last thing first and define a function to calculate the messiest full conditional; namelt:

$$f(changepoint) = \frac{\lambda_b^{\alpha + s_m - 1} e^{-(\beta+changepoint)\lambda_b} \lambda_a^{\gamma+s_a-1} e^{-(\delta s_a+n-changepoint)\lambda_a}}{\sum_{i=1}^n \lambda_b^{\alpha + s_m - 1} e^{-(\beta+changepoint)\lambda_b} \lambda_a^{\gamma+s_a-1} e^{-(\delta s_a+n-changepoint)\lambda_a}}$$

- This ought perhaps be calculated as logs first (computational overflow looks like a risk here)
- This is computed below by a loop which is inefficient
- But first, a small function to calculate the numerator (and at least make sure this works OK)

Having computed these values, we can sample (random.choice in scipy) a changepoint value.

In [19]:

```
def mprob(lb, la, ap, bp, gp, dp, m, sm, sl):
return (lb**(ap+sm-1))*(exp(-1*((bp)+m)*lb))*(la**(gp+sl-1))*(exp(-1*((dp)+(111-m))*la))
```

The easier full conditionals to simulate from are the two Poisson means.

$$f(\lambda_b) = Gamma(\alpha + s_m, \beta+changepoint)$$ and $$f(\lambda_b) = Gamma(\gamma + s_l, \delta + (n-changepoint))$$

This requires only the sufficient statistics $s_m$, $s_l$ and the current value of changepoint.

The Gibbs sampler proceeds in three stages:

- Simulate $\lambda_b$ given current sufficient statistics and parameter value
- Simulate $\lambda_a$ given current (updated) sufficient statistics and parameter values,
- Simulate $changepoint$ given current (updated) sufficient statistics and parameter values. Repeat.

In [24]:

```
outlambdabefore=np.zeros(5000)
outlambdaafter=np.zeros(5000)
outm=np.zeros(5000)
outlambdabefore[0]=12
outlambdaafter[0]=3
outm[0]=60
for iter in range(1,5000):
outlambdabefore[iter] = random.gamma(alphaprior+cumsumbefore[outm[iter-1]], 1.0/(betaprior+outm[iter-1]))
outlambdaafter[iter] = random.gamma(gammaprior+cumsumafter[outm[iter-1]], 1.0/(deltaprior+111-outm[iter-1]))
blah=np.zeros(111)
for i in range(111):
blah[i] = mprob(outlambdabefore[iter], outlambdaafter[iter], alphaprior, betaprior, gammaprior, deltaprior, i, cumsumbefore[i], cumsumafter[i])
condalpha = blah / sum(blah)
outm[iter]=random.choice(range(0,111), size=1, replace = 0, p=condalpha)
```

In [32]:

```
%Rpush outlambdabefore outlambdaafter outm
%R plot(c(1:5000), outlambdabefore, type = "l")
%R par(mfrow = c(2,2)); hist(outlambdabefore[c(1000:5000)], xlim = c(0,5)); hist(outlambdaafter[c(1000:5000)], xlim =c(0,5))
%R plot(xtabs(~outm[c(1000:5000)]))
%R acf(outlambdabefore[c(1000:5000)]) ; acf(outlambdaafter[c(1000:5000)])
```

Out[32]:

- Note we are throwing away the first 1000 values as a burn in when plotting the posterior simulations
- Note also that the change point is defined in terms of the index rather than years. Add 1851 to these values (the posterior mode should be around 1890)

This was a quick sketch of a Gibbs sampler for a rather famous change point problem. The first thing to note is that analytic values are available and so the answers can be checked.

As a programming exercise there is still a little work: the way the posterior value for the change point is calculated currently uses a for loop which could be vectorised. There are a few other efficiency tweaks that should be made. But it appears that on the whole this works.