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**Solution Assignment 6.2**

Hello,

I was going through the exercises in order to prepare for the exam and noticed that the Decision Network (Problem 6.2.1) is not contained in the official solutions. Is this an accident? It would be helpful to have some example solution of how the network is ought to look like. I had some correction in my official solution from a tutor, but a visual representation gives much more insight I guess.

Kind Regards,

Chris.

No, it’s primarily a „screw tikz, I’ll do this later somehow“ coupled with a lagging memory on my part. Sorry about that. I’ll take care of it once I’m back in germany.

It’s pretty easy to do in dot

It’s also pretty easy to do in tikz. Except that in either case I need time to figure out how and to actually do it, and i the case of tikz I actually learn something useful in the process

**Question regarding 6.2.1**

Hi, I am also a bit confused regarding the solution of Problem 6.2.1.

The Expected Utility of doing a Biopsy is:

EU(B) = P(Liv|B)U(Liv) + P(¬Liv|B)U(¬Liv) − 50

Q: Why do we subtract 50 only once in the end and not for each conditional probability independently like:

EU(B) = P(Liv|B)*(U(Liv) − 50) + P(¬Liv|B)*(U(¬Liv) − 50)

I thought our utility gets influenced by both Biopsy and Liv/¬Liv at the same time.

Thank you!

@Petra: I think both is mathematically the same.

EU(B) = P(Liv|B)U(Liv) + P(¬Liv|B)U(¬Liv) − 50 = 100a - 10b -50

and

EU(B) = P(Liv|B)*(U(Liv) − 50) + P(¬Liv|B)*(U(¬Liv) − 50) = 50a - 60b

b = 1-a so

100a - 10b -50 = 100a -10 +10a -50 = 110a -60 = 50a + 60a -60 = 50a - 60(1 - a) = 50a + 60b

means

P(Liv|B)*(U(Liv) − 50) + P(¬Liv|B)*(U(¬Liv) − 50) = P(Liv|B)U(Liv) + P(¬Liv|B)U(¬Liv) − 50

Thank you very much for this explanation.

This was exactly the hint I was looking for

Bump

*6.2.1 Solution still missing

I also have a question.

Is there and edge between LC → B?

If there is; isnt the probability P(B|LC) missing in the calculation?

I don’t think this makes much sense because what you are saying is that given LC, what is the probability of doing a biopsy. But with the Biopsy you want to find out whether you have LC or not. Correct me if I’m wrong

B is a decision node - decision nodes never have incoming arrows, since they represent actions the agent has to *decide* to undertake or not. They only have outgoing arrows, representing that they influence the probabilities of *other* nodes.

Sorry it took so long

One *could* argue that given the stated probabilities there should be an arrow from B to Liv, since B occurs as a condition in the probabilities stated for Liv. But quite obviously from the given probabilities, Liv and B are conditionally independent given C.

Shouldn’t there be an arrow from B to Liv since doing a biopsy can kill a SMAIB patient?

Sorry, reading helps

What about Exam WS17/18 Problem 2.3 ?

or

Exam SS18 Problem 2.2?

or

Russel & Norvig Page 635

Fair point, and I should have checked other examples before I answered, if only to avoid confusion.

But note that the arrows going into the decision notes do not represent a “set of parents” representing conditional probabilities. So I should rephrase: one *can* have ingoing arrows in decision networks that represent observable variables of the network - i.e. those pieces of information which are actually used for making a decision - but consequently, they have a different semantics than the arrows in a (normal) bayesian network.

To be precise: Nodes with an arrow into a decision node represent *evidence nodes*, the actions automatically represent query variables. This is different to bayesian networks, where the network itself does not distinguish between query variables, hidden variables and evidence. Hence, if you think about decision networks as “bayesian networks augmented by decision nodes and a utility node” as we introduced them, action nodes shouldn’t have any ingoing arrows at all.

It does make absolute sense to mark the observables as having-arrows-into-the-decision-nodes though, of course, so yes, your objection is still valid