Showing posts with label gRain. Show all posts
Showing posts with label gRain. Show all posts
Hit and run. Think Bayes!
At the R in Insurance conference Arthur Charpentier gave a great keynote talk on Bayesian modelling in R. Bayes' theorem on conditional probabilities is strikingly simple, yet incredibly thought provoking. Here is an example from Daniel Kahneman to test your intuition. But first I have to start with Bayes' theorem.
\[
P(H \cap D) = P(H|D) \, P(D) = P(D|H) \, P(H)
\]
Or if I rearrange it:
\[
P(H|D) = \dfrac{P(D|H) \, P(H)}{P(D)}
\]
Imagine \(H\) is short for hypothesis and \(D\) is short for data, or evidence. Then Bayes' theorem states that the probability of a hypothesis given data is the same as the likelihood that we observe the data given the hypothesis, weighted by the prior belief of the hypothesis, normalised by the probability that we observe the data regardless of the hypothesis.
The tricky bit in real life is often to figure out what the hypothesis and data are.
What is here the data and what is here the hypothesis? Intuitively you may think that the proportion of Blue and Green cabs is the data at hand and the witness accusation that a Blue cab was involved in the accident is the hypothesis. However, after some thought I found the following assignment much more helpful, as then \(P(H|D)\) matches the above question:
\(H =\) Accident caused by Blue cab. \(D =\) Witness said the cab was Blue.
With this it is straightforward to get the probabilities of \(P(H)=15\%\) and \(P(D|H)=80\%\). But what is \(P(D)\)? Well, when would the witness say that the cab was Blue? Either, when the cab was Blue and so the witness is right, or when the cab was actually Green and the witness is incorrect. Thus, following the law of total probability:
$$\begin{align}
P(D) & = P(D|H) P(H) + P(D | \bar{H}) P(\bar{H})\\
& = 0.8 \cdot 0.15 + 0.2 \cdot 0.85 = 0.29
\end{align}$$Therefore I get \(P(H|D)=41\%\). Thus, even if the witness states that the cab involved in the accident was Blue, the probability of this being true is only \(41\%\).
An alternative way to think about this problem is via a Bayesian Network. The colour of the cab will influence the statement of the witness. In R I can specify such a network using the gRain package [2], which I discussed in an earlier post. Here I provide the distribution of the cabs and the conditional probabilities of the witness as an input. After I compile the network, I can again read off the probabilities that a Blue cab was involved, when the witness said so.
Bayes' theorem
Bayes' theorem states that given two events \(D\) and \(H\), the probability of \(D\) and \(H\) happening at the same time is the same as the probability of \(D\) occurring, given \(H\), weighted by the probability that \(H\) occurs; or the other way round. As a formula it can be written as:\[
P(H \cap D) = P(H|D) \, P(D) = P(D|H) \, P(H)
\]
Or if I rearrange it:
\[
P(H|D) = \dfrac{P(D|H) \, P(H)}{P(D)}
\]
Imagine \(H\) is short for hypothesis and \(D\) is short for data, or evidence. Then Bayes' theorem states that the probability of a hypothesis given data is the same as the likelihood that we observe the data given the hypothesis, weighted by the prior belief of the hypothesis, normalised by the probability that we observe the data regardless of the hypothesis.
The tricky bit in real life is often to figure out what the hypothesis and data are.
Hit and run accident
This example is taken from Daniel Kahneman's book Thinking, fast and slow [1].A cab was involved in a hit and run accident at night. Two cab companies, the Green and the Blue, operate in the city. 85% of the cabs in the city are Green and 15% are Blue. A witness identified the cab as Blue. The court tested the reliability of the witness under the same circumstances that existed on the night of the accident and concluded that the witness correctly identified each one of the two colours 80% of the time and failed 20% of the time.
What is the probability that the cab involved in the accident was Blue rather than Green knowing that this witness identified it as Blue?
What is here the data and what is here the hypothesis? Intuitively you may think that the proportion of Blue and Green cabs is the data at hand and the witness accusation that a Blue cab was involved in the accident is the hypothesis. However, after some thought I found the following assignment much more helpful, as then \(P(H|D)\) matches the above question:
\(H =\) Accident caused by Blue cab. \(D =\) Witness said the cab was Blue.
With this it is straightforward to get the probabilities of \(P(H)=15\%\) and \(P(D|H)=80\%\). But what is \(P(D)\)? Well, when would the witness say that the cab was Blue? Either, when the cab was Blue and so the witness is right, or when the cab was actually Green and the witness is incorrect. Thus, following the law of total probability:
$$\begin{align}
P(D) & = P(D|H) P(H) + P(D | \bar{H}) P(\bar{H})\\
& = 0.8 \cdot 0.15 + 0.2 \cdot 0.85 = 0.29
\end{align}$$Therefore I get \(P(H|D)=41\%\). Thus, even if the witness states that the cab involved in the accident was Blue, the probability of this being true is only \(41\%\).
An alternative way to think about this problem is via a Bayesian Network. The colour of the cab will influence the statement of the witness. In R I can specify such a network using the gRain package [2], which I discussed in an earlier post. Here I provide the distribution of the cabs and the conditional probabilities of the witness as an input. After I compile the network, I can again read off the probabilities that a Blue cab was involved, when the witness said so.
R code
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:
To answer the above question I present the data here as a Bayesian Network using the
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.
[2] Søren Højsgaard (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/
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
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] Søren Højsgaard (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
