ANÁLISIS DE LA INGESTA

Recall from the beginning of Section 1.4 we stated the following: Given a set of variablesV, if for everyX, Y _∈Vwe draw an edge fromXtoY if and only ifX

is a direct cause ofY relative to V, we call the resultant DAG acausal DAG. Given the manipulation definition of causation offered earlier, by ‘X being a

M

E

A

L

O

T

Figure 1.15: The actionAhas a causal arc into the system other than through

M.

F

D

G

Figure 1.16: The causal relationships if F had a causal influence onG other than throughD.

direct cause ofY relative toV’ we mean that a manipulation ofX changes the probability distribution ofY, and that there is no subsetW_⊆V_−{X, Y_}such that if we instantiate the variables inWa manipulation ofXno longer changes the probability distribution ofY. When constructing a causal DAG containing a set of variablesV, we callV‘our set of observed variables.’ Recall further from the beginning of Section 1.4 we said we would illustrate why we feel the joint probability (relative frequency) distribution of the variables in a causal DAG often satisfies the Markov condition with that DAG. We do that first; then we state the causal Markov Assumption.

Why Causal DAGs Often Satisfy the Markov Condition

Considerfirst the situation concerningfinasteride, DHT, and hair regrowth dis- cussed in Section 1.4.1. In this case, our set of observed variablesVis_{F, D, G_}. We learned that finasteride level has a causal influence on DHT level. So we placed an edge fromF to Din Figure 1.14. We learned that DHT level has a causal influence on hair regrowth. So we placed an edge fromDtoGin Figure 1.14. We suspected that the causal effect finasteride has on hair regrowth is only through the lowering of DHT levels. So we did not place an edge from

1.4. CREATING BAYESIAN NETWORKS USING CAUSAL EDGES 53

H

X

Z

Y

Figure 1.17: XandY are not independent if they have a hidden common cause

H.

F affectedG by some means other than by decreasing DHT levels), we would also place an edge from F to G as shown in Figure 1.16. Assuming the only causal connection between F and G is as indicated in Fig 1.14, we would feel that F and G are conditionally independent given D because, once we knew the value of D, we would have a probability distribution of G based on this known value, and, since the value ofF cannot change the known value ofDand there is no other connection betweenF andG, it cannot change the probability distribution ofG. Manipulation experiments have substantiated this intuition. That is, there have been experiments in which it was established thatXcauses

Y, Y causesZ,X andZ are not probabilistically independent, and X and Z

are conditionally independent given Y. See [Lugg et al, 1995] for an example. In general, when all causal paths from X toY contain at least one variable in our set of observed variables V, X and Y do not have a common cause, there are no causal paths fromY back toX, and we do not have selection bias, then we feelXandY are independent if we condition on a set of variables including at least one variable in each of the causal paths fromX toY. Since the set of all parents of Y is such a set, we feel the Markov condition is satisfied relative to XandY.

We sayX andY have a common causeif there is some variable that has causal paths into bothXandY. IfXandY have a common causeC, there is often a dependency between them through this common cause (But this is not necessarily the case. See Exercise 2.34.). However, if we condition onY’s parent in the path fromCtoY, we feel we break this dependency for the same reasons discussed above. So, as long as all common causes are in our set of observed variablesV, we can still break the dependency betweenXand Y (assuming as above there are no causal paths from Y to X) by conditioning on the set of parents of Y, which means the Markov condition is still satisfied relative toX

observed variablesV. Such a common cause is called ahidden variable. If two variables had a hidden common cause, then there would often be a dependency between them, which the Markov condition would identify as an independency. For example, consider the DAG in Figure 1.17. If we only identified the variables

X, Y, and Z, and the causal relationships that X and Y each caused Z, we would draw edges from each of X and Y to Z. The Markov condition would entailX andY are independent. But ifX andY had a hidden common cause

H, they would not ordinarily be independent. So, for us to assume the Markov condition is satisfied, either no two variables in the set of observed variablesV can have a hidden common cause, or, if they do, it must have the same unknown value for every unit in the population under consideration. When this is the case, we say the set iscausally sufficient.

Another violation of the Markov condition, similar to the failure to include a hidden common cause, is when there is selection bias present. Recall that, in the beginning of Section 1.4.1, we noted that iffinasteride use (F) and apprehension about lack of hair regrowth (G) are both causes of hypertension (Y), and we happen to be observing individuals hospitalized for treatment of hypertension,

we would observe a probabilistic dependence betweenF andGdue to selection bias. This situation is depicted in Figure 1.12 (e). Note that in this situation our set of observed variablesVis _{F, G_}. That is, Y is not observed. So if neither

F nor Gcaused each other and they did not have a hidden common cause, a causal DAG containing only the two variables (i.e. one with no edges) would still not satisfy the Markov condition with the observed probability distribution, because the Markov condition saysF andGare independent when indeed they are not for this population.

Finally, we must also make certain that ifXhas a causal influence onY, then

Y does not have a causal influenceX. In this way we guarantee that the identi- fied causal edges will indeed yield a DAG. Causal feedback loops (e.g. the situa- tion identified in Figure 1.12 (c)) are discussed in [Richardson and Spirtes, 1999]. Before closing, we note that if we mistakenly draw an edge from X to Y

in a case where X’s causal influence on Y is only through other variables in the model, we have not done anything to thwart the Markov condition being satisfied. For example, consider again the variables in Figure 1.14. IfF’s only influence onGwas throughD, we would not thwart the Markov condition by drawing an edge fromF toG. That is, this does not result in the structure of the DAG entailing any conditional independencies that are not there. Indeed, the opposite has happened. That is, the DAG fails to entail a conditional in- dependency (namelyI(_{F_},_{G_}|{D_})) that is there. This is a violation of the faithfulness condition (discussed in Chapter 2), not the Markov condition. In general, we would not want to do this because it makes the DAG less informative and unnecessarily increases the size of the instance (which is important because, as we shall see in Section 3.6, the problem of doing inference in Bayesian net- works is#P-complete). However, a few mistakes of this sort are not that serious as we can still expect the Markov condition to be satisfied.

1.4. CREATING BAYESIAN NETWORKS USING CAUSAL EDGES 55 The Causal Markov Assumption

We’ve offered a definition of causation based on manipulation, and we’ve argued that, given this definition of causation, a causal DAG often satisfies the Markov condition with the probability distribution of the variables, which means we can construct a Bayesian network by creating a causal DAG. In general, given any definitions of ‘causation’ and ‘direct causal influence,’ if we create a causal DAG _G = (V,E) and assume the probability distribution of the variables in V satisfies the Markov condition with _G, we say we are making the causal Markov assumption.

As discussed above, if the following three conditions are satisfied the causal Markov assumption is ordinarily warranted: 1) there must be no hidden common causes; 2) selection bias must not be present; and 3) there must be no causal feedback loops. In general, when constructing a Bayesian network using identi- fied causal influences, one must take care that the causal Markov assumptions holds.

Often we identify causes using methods other than manipulation. For exam- ple, most of us believe smoking causes lung cancer. Yet we have not manipulated individuals by making them smoke. We believe in this causal influence because smoking and lung cancer are correlated, the smoking precedes the cancer in time (a common assumption is that an effect cannot precede a cause), and there are biochemical changes associated with smoking. All of this could possibly be ex- plained by a hidden common cause (Perhaps a genetic defect causes both.), but domain experts essentially rule out this possibility. When we identify causes by any means whatsoever, ordinarily we feel they are ones that could be identified by manipulation if we were to perform a RCE, and we make the causal Markov assumption as long as we are confident exceptions such as conditions (1), (2) and (3) in the preceding paragraph are not present.

An example of constructing a causal DAG follows.

Example 1.34 Suppose we have identified the following causal influences by some means: A history of smoking (H) has a causal effect both on bronchitis (B) and on lung cancer (L). Furthermore, each of these variables can cause fatigue (F). Lung Cancer (L) can cause a positive chest X-ray (C). Then the DAG in Figure 1.10 represents our identified causal relationships among these variables. If we believe 1) these are the only causal influences among the variables; 2) there are no hidden common causes; and 3) selection bias is not present, it seems reasonable to make the causal Markov assumption. Then if the conditional distributions specified in Figure 1.10 are our estimates of the conditional relative frequencies, that DAG along with those specified conditional distributions constitute a Bayesian network which represents our beliefs.

Before closing we mention an objection to the causal Markov condition. That is, unless we abandon the ‘locality principle’ the condition seems to be violated in some quantum mechanical experiments. See [Spirtes et al, 1993, 2000] for a discussion of this matter.

C V S C V S

(a) (b)

Figure 1.18: CandS are not independent in (a), but the instantiation ofV in (b) renders them independent.

The Markov Condition Without Causation

Using causal edges is just one way to develop a DAG and a probability dis- tribution that satisfy the Markov condition. In Example 1.25 we showed the joint distribution of V (value), S (shape), andC (color) satisfied the Markov condition with the DAG in Figure 1.3 (a), but we would not say that the color of an object has a causal influence on its shape. The Markov condition is simply a property of the probabilistic relationships among the variables. Furthermore, if the DAG in Figure 1.3 (a) did capture the causal relationships among some causally sufficient set of variables and there was no selection bias present, the Markov condition would be satisfied not only with that DAG but also with the DAGS in Figures 1.3 (b) and (c). Yet we certainly would not say the edges in these latter two DAGs represent causal influence.

Some Final Examples

To solidify the notion that the Markov condition is often satisfied by a causal DAG, we close with three simple examples. We present these examples using an intuitive approach, which shows how humans reason qualitatively with the dependencies and conditional independencies among variables. In accordance with this approach, we again use the name of a variable to stand also for a value. For example, in modeling whether an individual has a cold, we use a variableC

whose value ispresent or absent, and we also useC to represent that a cold is present.

Example 1.35 If Alice’s husband Ralph was planning a surprise birthday party for Alice with a caterer (C), this may cause him to visit the caterer’s store (V). The act of visiting that store could cause him to be seen (S) visiting the store. So the causal relationships among the variables are the ones shown in Figure 1.18 (a). There is no direct path from C toS because planning the party with the caterer could only cause him to be seen visiting the store if it caused him to actually visit the store. If Alice’s friend Trixie reported to her that she had seen Ralph visiting the caterer’s store today, Alice would conclude that he may be planning a surprise birthday party because she would feel there is a good chance Trixie really did see Ralph visiting the store, and, if this actually was the case, there is a chance he may be planning a surprise birthday party. SoC

1.4. CREATING BAYESIAN NETWORKS USING CAUSAL EDGES 57 (a) (c) R S C (b) R S C R S C H (d) R S C H

Figure 1.19: IfC is the only common cause ofRand S (a), we need to instan- tiate onlyC (b) to render them independent. If they have exactly two common causes, C andH (c), we need to instantiate both C andH (d) to render them independent.

and S are not independent. If, however, Alice had already witnessed this same act of Ralph visiting the caterer’s store, she would already suspect Ralph may be planning a surprise birthday party. Trixie’s testimony would not affect here belief concerning Ralph’s visiting the store and therefore would have no affect on her belief concerning his planning a party. SoCandSare conditionally independent givenV, as the Markov condition entails for the DAG in Figure 1.18 (a). The instantiation of V, which rendersC and S independent, is depicted in Figure 1.18 (b) by placing a cross throughV.

Example 1.36 A cold (C) can cause both sneezing (S) and a runny nose (R). Assume neither of these manifestations causes the other and, for the moment, also assume there are no hidden common causes (That is, this set of variables is causally sufficient.). The causal relationships among the variables are then the ones depicted in Figure 1.19 (a). Suppose now that Professor Patel walks into the classroom with a runny nose. You would fear she has a cold, and, if so, the cold may make her sneeze. So you back off from her to avoid the possible sneeze. We see then thatS and R are not independent. Suppose next that Professor Patel calls school in the morning to announce she has a cold which will make her late for class. When she finally does arrive, you back off

B (a) (b) F A F B A

Figure 1.20: B andF are independent in (a), but the instantiation ofAin (b) renders them dependent.

her nose is running, this has no affect on your belief concerning her sneezing because the runny nose no longer makes the cold more probable (You know she has a cold.). SoS andR are conditionally independent givenC, as the Markov condition entails for the DAG in Figure 1.19 (a). The instantiation of C is depicted in Figure 1.19 (b).

There actually is at least one other common cause of sneezing and a runny nose, namely hay fever (H). Suppose this is the only common cause missing from Figure 1.19 (a). The causal relationships among the variables would then be as depicted in Figure 1.19 (c). Given this, conditioning onC is not sufficient to renderR andS independent, becauseRcould still make S more probable by makingH more probable. So we must condition on both C and H to renderR and S independent. The instantiation of C and H is depicted in Figure 1.19 (d).

Example 1.37 Antonio has observed that his burglar alarm (A) has sometimes gone offwhen a freight truck (F) was making a delivery to the Home Depot in back of his house. So he feels a freight truck can trigger the alarm. However, he also believes a burglar (B) can trigger the alarm. He does not feel that the appearance of a burglar might cause a freight truck to make a delivery or vice versa. Therefore, he feels that the causal relationships among the variables are the ones depicted in Figure 1.20 (a). Suppose Antonio sees a freight truck making a delivery in back of his house. This does not make him feel a burglar is more probable. SoF andB are independent, as the Markov condition entails for the DAG in Figure 1.20 (a). Suppose next that Antonio is awakened at night by the sounding of his burglar alarm. This increases his belief that a burglar is present, and he begins fearing this is indeed the case. However, as he proceeds to investigate this possibility, he notices that a freight truck is making a delivery in back of his house. He reasons that this truck explains away the alarm, and therefore he believes a burglar probably is not present. So he relaxes a bit. Given the alarm has sounded, learning that a freight truck is present decreases the probability of a burglar. So the instantiation of A, as depicted in

EXERCISES 59

Figure 1.20 (b), rendersF andB conditionally dependent. As noted previously, the instantiation of a common effect creates a dependence between its causes because each explains away the occurrence of the effect, thereby making the other cause less likely.

Note that the Markov condition does not entail thatF andB are condition- ally dependent givenA. Indeed, a probability distribution can satisfy the Markov