Monday 12 February 2018

An Improved Method for Solving Hybrid Influence Diagrams

Most decisions are made in the face of uncertain factors and outcomes. In a typical decision problem, uncertainties involve both continuous factors (e.g. amount of profit) and discrete factors (e.g. presence of a small number of risk events). Tools such as decision trees and influence diagrams are used to cope with uncertainty regarding decisions, but most implementations of these tools can only deal with discrete or discretized factors and ignore continuous factors and their distributions.

A paper just published in the International Journal of Approximate Reasoning presents a novel method that overcomes a number of these limitations. The method is able to solve decision problems with both discrete and continuous factors in a fully automated way. The method requires that the decision problem is modelled as a Hybrid Influence Diagrams, which is an extension of influence diagrams containing both discrete and continuous nodes, and solves it by using a state-of-the-art inference algorithm called Dynamic Discretization. The optimal policies calculated by the method are presented in a simplified decision tree.

The full reference is:

Yet, B., Neil, M., Fenton, N., Dementiev, E., & Constantinou, A. (2018). "An Improved Method for Solving Hybrid Influence Diagrams". International Journal of Approximate Reasoning. DOI: 10.1016/j.ijar.2018.01.006  Preprint (open access) available here.
UPDATE (22 Feb 2018): The full published version the paper is available online for free for 50 days here: https://authors.elsevier.com/c/1Wc6D,KD6ZG8y-

Acknowledgements: Part of this work was performed under the auspices of EU project ERC-2013-AdG339182-BAYES_KNOWLEDGE

Saturday 10 February 2018

Decision-making under uncertainty: computing "Value of Information"

Information gathering is a crucial part of decision making under uncertainty. Whether to collect additional information or not, and how much to invest for such information are vital questions for successful decision making. For example, before making a treatment decision, a physician has to evaluate the benefits and risks of additional imaging or laboratory tests and decide whether to ask for them. Value of Information (VoI) is a quantitative decision analysis technique for answering such questions based on a decision model. It is used to prioritise the parts of a decision model where additional information is expected to be useful for decision making.

However, computing VoI in decision models is challenging especially when the problem involves both discrete and continuous variables. A new paper in the IEEE Access journal illustrates a simple and practical approach that can calculate VoI using Influence Diagram models that contain both discrete and continuous variables. The proposed method can be applied to a wide variety of decision problems as most decisions can be modelled as an influence diagram, and many decision modelling tools, including Decision Trees and Markov models, can be converted to an influence diagram.

The full reference is:

Yet, B., Constantinou, A., Fenton, N., & Neil, M. (2018). Expected Value of Partial Perfect Information in Hybrid Models using Dynamic Discretization.  IEEE Access. DOI: 10.1109/ACCESS.2018.2799527

Acknowledgements: Part of this work was performed under the auspices of EU project ERC-2013-AdG339182-BAYES_KNOWLEDGE, EPSRC project EP/P009964/1: PAMBAYESIAN, and ICRAF Contract No SD4/2012/214 issued to Agena.

Wednesday 7 February 2018

Lawnmower v terrorist risk: the saga continues

 Kim Kardashian's tweet comparing risk from lawnmowers v terrorists triggered the award and debate

Yesterday Significance Magazine (the magazine of the Royal Statistical Society and the American Statistical Association) published an article “Lawnmowers versus Terrorists”  with the strapline:
The Royal Statistical Society’s first ‘International Statistic of the Year’ sparked plenty of online discussion. Here, Norman Fenton and Martin Neil argue against the choice of winner, while Nick Thieme writes in support.
Our case, titled “A highly misleading view of risk”, was an edited version of a paper  previously publicised in a blog post that itself followed up on original concerns raised by Nicholas Nassim Taleb about the RSS citation and the way it had been publicised. The ‘opposing’ case made by Nick Thieme was essentially a critique of our paper.

We have today published a response to Nick’s critique.

Monday 5 February 2018

Revisiting a Classic Probability Puzzle: the Two Envelopes Problem

Many people have heard about the Monty Hall problem. A similar (but less well known and more mathematically interesting) problem is the two envelopes problem, which Wikipedia describes as follows:
“You are given two indistinguishable envelopes, each containing money, one contains twice as much as the other. You may pick one envelope and keep the money it contains. Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch?”
The problem has been around in various forms since 1953 and has been extensively discussed (see, for example Gerville-Réache for a comprehensive analysis and set of references)  although I was not aware of this until recently.

We actually gave this problem (using boxes instead of envelopes) as an exercise in the supplementary material for our Book, after Prof John Barrow of University of Cambridge first alerted us to it. The ‘standard solution’ (as in the Monty Hall problem) says that you should always switch. This is based on the following argument:
If the envelope you choose contains \$100 then there is an evens chance the other envelope contains \$50 and an evens chance it contains \$200. If you do not switch you have won \$100. If you do switch you are just as likely to decrease the amount you win as increase it. However, if you win the amount increases by \$100 and if you lose it only decreases by \$50. So your expected gain is positive (rather than neutral). Formally, if the envelope contains S then the expected amount in the other envelope is 5/4 times X (i.e. 25% more).
In fact (as pointed out by a reader Hugh Panton), the problem with the above argument is that it equally applies to the ‘other envelope’ thereby suggesting we have a genuine paradox. In fact, it turns out that the above argument only really works if you actually open the first envelope (which was explicitly not allowed in the problem statement) and discover it contains S. As Gerville-Réache shows, if the first envelope is not opened, the only probabilistic reasoning that does not use supplementary information leads to estimating expectations as infinite amounts of each envelope. Bayesian reasoning can be used to show that there is no benefit in switching, but that is not what I want to describe here.

What I found interesting is that I could not find - in any of the discussions about the problem - a solution for the case where we assume there is a finite maximum prize, even if we allow that maximum to be as large as we like. With this assumption it turns out that we can prove (without dispute) that there is no benefit to be gained if you stick or switch. See this short paper for the details:
Fenton N E, "Revisiting a Classic Probability Puzzle: the Two Envelopes Problem" 2018, DOI10.13140/RG.2.2.24641.04960