Not only verbs but also believes can be conjugated

Following on from last week, where I presented a simple example of a Bayesian network with discrete probabilities to predict the number of claims for a motor insurance customer, I will look at continuos probability distributions today. Here I follow example 16.17 in Loss Models: From Data to Decisions [1].

Suppose there is a class of risks that incurs random losses following an exponential distribution (density \(f(x) = \Theta {e}^{- \Theta x}\)) with mean \(1/\Theta\). Further, I believe that \(\Theta\) varies according to a gamma distribution (density \(f(x)= \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha \,-\, 1} e^{- \beta x } \)) with shape \(\alpha=4\) and rate \(\beta=1000\).

In the same way as I had good and bad driver in my previous post, here I have clients with different characteristics, reflected by the gamma distribution. I shall call the gamma distribution with the above parameters my prior parameter distribution and the exponential distribution the prior predictive distribution.

The textbook tells me that the unconditional mixed distribution of an exponential distribution with parameter \(\Theta\), whereby \(\Theta\) has a gamma distribution, is a Pareto II distribution (density \(f(x) = \frac{\alpha \beta^\alpha}{(x+\beta)^{\alpha+1}}\)) with parameters \(\alpha,\, \beta\). Its k-th moment is given in the general case by
\[
E[X^k] = \frac{\beta^k\Gamma(k+1)\Gamma(\alpha - k)}{\Gamma(\alpha)},\; -1 < k < \alpha. \] Thus, I can calculate the prior expected loss (\(k=1\)) as \(\frac{\beta}{\alpha-1}=\,\)333.33.
Now suppose I have three independent observations, namely losses of $100, $950 and $450 over the last 3 years. The mean loss is $500, which is higher than the $333.33 of my model.

Question: How should I update my belief about the client's risk profile to predict the expected loss cost for year 4 given those 3 observations?

Visually I can regard this scenario as a graph, with evidence set for years 1 to 3 that I want to propagate through to year 4.

Read more �

Not only verbs but also believes can be conjugated

Following on from last week, where I presented a simple example of a Bayesian network with discrete probabilities to predict the number of claims for a motor insurance customer, I will look at continuos probability distributions today. Here I follow example 16.17 in Loss Models: From Data to Decisions [1].

Suppose there is a class of risks that incurs random losses following an exponential distribution (density \(f(x) = \Theta {e}^{- \Theta x}\)) with mean \(1/\Theta\). Further, I believe that \(\Theta\) varies according to a gamma distribution (density \(f(x)= \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha \,-\, 1} e^{- \beta x } \)) with shape \(\alpha=4\) and rate \(\beta=1000\).

In the same way as I had good and bad driver in my previous post, here I have clients with different characteristics, reflected by the gamma distribution. I shall call the gamma distribution with the above parameters my prior parameter distribution and the exponential distribution the prior predictive distribution.

The textbook tells me that the unconditional mixed distribution of an exponential distribution with parameter \(\Theta\), whereby \(\Theta\) has a gamma distribution, is a Pareto II distribution (density \(f(x) = \frac{\alpha \beta^\alpha}{(x+\beta)^{\alpha+1}}\)) with parameters \(\alpha,\, \beta\). Its k-th moment is given in the general case by
\[
E[X^k] = \frac{\beta^k\Gamma(k+1)\Gamma(\alpha - k)}{\Gamma(\alpha)},\; -1 < k < \alpha. \] Thus, I can calculate the prior expected loss (\(k=1\)) as \(\frac{\beta}{\alpha-1}=\,\)333.33.
Now suppose I have three independent observations, namely losses of $100, $950 and $450 over the last 3 years. The mean loss is $500, which is higher than the $333.33 of my model.

Question: How should I update my belief about the client's risk profile to predict the expected loss cost for year 4 given those 3 observations?

Visually I can regard this scenario as a graph, with evidence set for years 1 to 3 that I want to propagate through to year 4.

Read more �

Predicting claims with a Bayesian network

Here is a little Bayesian Network to predict the claims for two different types of drivers over the next year, see also example 16.15 in [1].

Let's assume there are good and bad drivers. The probabilities that a good driver will have 0, 1 or 2 claims in any given year are set to 70%, 20% and 10%, while for bad drivers the probabilities are 50%, 30% and 20% respectively.

Further I assume that 75% of all drivers are good drivers and only 25% would be classified as bad drivers. Therefore the average number of claims per policyholder across the whole customer base would be:

0.75*(0*0.7 + 1*0.2 + 2*0.1) + 0.25*(0*0.5 + 1*0.3 + 2*0.2) = 0.475

Now a customer of two years asks for his renewal. Suppose he had no claims in the first year and one claim last year. How many claims should I predict for next year? Or in other words, how much credibility should I give him?

To answer the above question I present the data here as a Bayesian Network using the gRain package [2]. I start with the contingency probability tables for the driver type and the conditional probabilities for 0, 1 and 2 claims in year 1 and 2. As I assume independence between the years I set the same probabilities. I can now review my model as a mosaic plot (above) and as a graph (below) as well.

Next, I set the client's evidence (0 claims in year one and 1 claim in year two) and propagate these back through my network to estimate the probabilities that the customer is either a good (73.68%) or a bad (26.32%) driver. Knowing that a good driver has on overage 0.4 claims a year and a bad driver 0.7 claims I predict the number of claims for my customer with the given claims history as 0.4789.


Alternatively I could have added a third node for year 3 and queried the network for the probabilities of 0, 1 or 2 claims given that the customer had zero claims in year 1 and one claim in year 2. The sum product of the number of claims and probabilities gives me again an expected claims number of 0.4789.

References

[1] Klugman, S. A., Panjer, H. H. & Willmot, G. E. (2004), Loss Models: From Data to Decisions, Wiley Series in Proability and Statistics.

[2] S�ren H�jsgaard (2012). Graphical Independence Networks with the gRain Package for R. Journal of Statistical Software, 46(10), 1-26. URL http://www.jstatsoft.org/v46/i10/

Session Info

R version 3.0.2 (2013-09-25)
Platform: x86_64-apple-darwin10.8.0 (64-bit)

locale:
[1] en_GB.UTF-8/en_GB.UTF-8/en_GB.UTF-8/C/en_GB.UTF-8/en_GB.UTF-8

attached base packages:
[1] grid      stats     graphics  grDevices utils     datasets  methods  
[8] base     

other attached packages:
[1] Rgraphviz_2.6.0 gRain_1.2-2     gRbase_1.6-12   graph_1.40.0   

loaded via a namespace (and not attached):
[1] BiocGenerics_0.8.0 igraph_0.6.6       lattice_0.20-24    Matrix_1.1-0      
[5] parallel_3.0.2     RBGL_1.38.0        stats4_3.0.2       tools_3.0.2

Predicting claims with a Bayesian network

Here is a little Bayesian Network to predict the claims for two different types of drivers over the next year, see also example 16.15 in [1].

Let's assume there are good and bad drivers. The probabilities that a good driver will have 0, 1 or 2 claims in any given year are set to 70%, 20% and 10%, while for bad drivers the probabilities are 50%, 30% and 20% respectively.

Further I assume that 75% of all drivers are good drivers and only 25% would be classified as bad drivers. Therefore the average number of claims per policyholder across the whole customer base would be:

0.75*(0*0.7 + 1*0.2 + 2*0.1) + 0.25*(0*0.5 + 1*0.3 + 2*0.2) = 0.475

Now a customer of two years asks for his renewal. Suppose he had no claims in the first year and one claim last year. How many claims should I predict for next year? Or in other words, how much credibility should I give him?

To answer the above question I present the data here as a Bayesian Network using the gRain package [2]. I start with the contingency probability tables for the driver type and the conditional probabilities for 0, 1 and 2 claims in year 1 and 2. As I assume independence between the years I set the same probabilities. I can now review my model as a mosaic plot (above) and as a graph (below) as well.

Next, I set the client's evidence (0 claims in year one and 1 claim in year two) and propagate these back through my network to estimate the probabilities that the customer is either a good (73.68%) or a bad (26.32%) driver. Knowing that a good driver has on overage 0.4 claims a year and a bad driver 0.7 claims I predict the number of claims for my customer with the given claims history as 0.4789.


Alternatively I could have added a third node for year 3 and queried the network for the probabilities of 0, 1 or 2 claims given that the customer had zero claims in year 1 and one claim in year 2. The sum product of the number of claims and probabilities gives me again an expected claims number of 0.4789.

References

[1] Klugman, S. A., Panjer, H. H. & Willmot, G. E. (2004), Loss Models: From Data to Decisions, Wiley Series in Proability and Statistics.

[2] S�ren H�jsgaard (2012). Graphical Independence Networks with the gRain Package for R. Journal of Statistical Software, 46(10), 1-26. URL http://www.jstatsoft.org/v46/i10/

Session Info

R version 3.0.2 (2013-09-25)
Platform: x86_64-apple-darwin10.8.0 (64-bit)

locale:
[1] en_GB.UTF-8/en_GB.UTF-8/en_GB.UTF-8/C/en_GB.UTF-8/en_GB.UTF-8

attached base packages:
[1] grid      stats     graphics  grDevices utils     datasets  methods  
[8] base     

other attached packages:
[1] Rgraphviz_2.6.0 gRain_1.2-2     gRbase_1.6-12   graph_1.40.0   

loaded via a namespace (and not attached):
[1] BiocGenerics_0.8.0 igraph_0.6.6       lattice_0.20-24    Matrix_1.1-0      
[5] parallel_3.0.2     RBGL_1.38.0        stats4_3.0.2       tools_3.0.2