On Entailment and SupportAuthor(s): Frank JacksonSource: Noûs, Vol. 3, No. 3 (Sep., 1969), pp. 345-349Published by: WileyStable URL: http://www.jstor.org/stable/2214555 .

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On Entailment and Support

FRANK JACKSON

LA TROBE UNIVERSITY

It is still too widely supposed that the principle: "if p entails q and if e supports p, then e supports q at least as much as it supports p," is true.

Let's make a start on showing this principle is false by con- sidering what it is for one proposition to support another proposi- tion. I suggest that e supports p if and only if the probability of p given e is greater than the probability of p not given e. Thus, in the familiar notation:

(1) e supports p - P(p/e) > P(p).

The obvious competitors to (1) are (2) and (3).

(2) e supports p P(p/e) > 0.

(3) e supports p P(p/e) > 0.5.

Two counter-examples show (2) and (3) are inadequate. Against (2) and (3) we note that the probability that the Japanese attacked Pearl Harbour given that I am sitting down is greater than nought and greater than one-half but the fact I am sitting down does not support the proposition that the Japanese attacked Pearl Harbour. Against (3) we note that the fact Hyperion is being ridden by the best jockey may support that Hyperion will win without its being the case that it is more likely than not that Hyperion will win. As well as showing (2) and (3) are false these examples support (1). Why is it that my sitting down does not support that the Japanese attacked Pearl Harbour? Clearly because' my sitting down is irrele- vant to the attack and so does not increase the probability that it occurred. Why is it that being ridden by the best jockey generally supports the proposition that a horse will win? Clearly because,

345

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346 NOS

generally, being ridden by the best jockey increases the probability of winning.

There are also formal defects in (2) and (3). The following two principles are intuitively clear.

(4) if e supports p then W-e does not support p.

(5) if e supports p then e does not support Up.

Thus any adequate account of support must guarantee (4) and (5); (3) guarantees (5) as "P(p/e) > 0.5" entails "P(Qp/e) < 0.5" be- cause P(p) = 1 - P(-p). However (3) does not ensure (4); it is possible for both P(p/e) and P(p/-- e) to be greater than one-half (take p very probable and e irrelevant). (2) ensures neither (4) or (5).

It must now be shown that (1) guarantees (4) and (5). "P(p/e) > P(p)" entails "1 - P(Q-p/e) > 1 - P(Q-'p)", which gives "P(~p) > P(Qp/e)", thus (1) ensures (5). Likewise (1) ensures (4).

P(p/e) = P(p) P(e/p) and replacing 'e' by we' we get P(e)

5

P(p) * P(Qe/p) P(p/ e) = P(e)

P(p) [1 - P(e/p)]

1 - P(e)

Taking e to be possible this gives that P(p) is between P(p/e) and P(p/~e) and so that 'P(p/e) - P(p)' and 'P(p/~e) - P(p)' have opposite sign.

There is another account of support which is intuitively ap- pealing and satisfies (4) and (5), namely

(6) e supports p -- P(p/e) > P(p/,- e).

However there are decisive counter-examples to this. I shall give just one. Given a minimal amount of back-ground information, 'Hyperion will win' is not supported by 'Something exists', however it would be according to (6) as the probability of 'Hyperion will win' given 'Nothing exists' is zero.

This completes the case for (1). However (1) is deficient in an important respect. The support a proposition gives another proposi- tion is a function of the background information. 'Hyperion will win' is supported by 'Hyperion will be ridden by the best jockey' in most

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ENTAILMENT AND SUPPORT 347

cases. However, if one also knows that Hyperion's trainer dislikes the best jockey it may well be rational to lower one's estimate of the probability of Hyperion's winning in view of the fact that he is being ridden by the best jockey.

To take another example. The support 'Every marble so far drawn has been red' gives to 'Every marble is red' is clearly a func- tion of 'Every draw has been a random one'. Thus (1) applies only in the limit case where there are no other relevant facts, in general (1) should be replaced by (7).

(7) e supports p given h -: P(p/e * h) > P(p/h).

Commonly the background information is taken for granted and 'h' is suppressed, and for simplicity I shall follow this practice. Thus we will work from (1) "carrying in our head" in actual examples the relevant background information.

If (1) formulates the notion of support, (8) gives the notion of support to a degree:

(8) e supports p to degree d - [P(p/e) - P(p) = d],

and from (8)

(9) e supports q at least as much as p - [P(q/e) - P(q) > P(p/e) - P(p)].

Thus (10) is the principle (stated in English in the first paragraph) which I wish to show false.

(10) [p entails q] D [P(q/e) - P(q) > P(p/e) - P(p)].

The following shows (10) is false. Suppose I have ten red marbles and ten green marbles and

draw, with replacement, two marbles at random from the twenty; then there are four equiprobable cases. The first draw is red and the second draw is red, the first draw is red and the second green, the first is green and the second is green, and the first is green and the second red.

Thus the initial probability that both draws are the same colour is 0.5 (2 cases out of 4) and that both draws are red is 0.25 (1 case out of 4). Now the probability that both draws are red given that the first draw is red is 0.5, and thus 'The first draw is red' sup- ports 'Both draws are red', but the probability that both draws are the same colour given that the first draw is red is unchanged at 0.5; thus 'The first draw is red' does not support 'Both draws are the

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348 NONS

same colour'. But 'Both draws are red' entails 'Both draws are the same colour' so (10) is false.

To guard against misunderstanding perhaps I should say explicitly that this example does not, and I do not know what could, show that (11) is false,

(11) [p entails q] D [P(p/e) < P(q/e)]

It is perhaps also worth noting explicitly that though many devel- opments of Raven-type paradoxes appeal to (10), noting that (10) is false provides no real help in resolving the paradoxes because (12) follows from (11), and so is true,

(12) [p entails and is entailed by q] D [P(p/e) - P(p) = P(q/e) - P(q)].

Finally, I want to show why the argument that follows fails. The argument is quoted from Ian Hacking's Logic of Statistical Inference (Cambridge, 1965) pp. 33, 34. The explanation of the notation (from p. 32) is "h/d < i/e shall signify that e supports i at least as well as d supports h".

"But there is one possible defect in Koopman's logic which had better be recorded. His logic must be a trifle too strong, for it includes the principle of anti-synmnetry:

If h/e < i/d, then I.i/d < .' h/e.

Doubt is cast on the principle if we put e = d = 'a box con- tains 40 black balls and 60 green ones; a ball will shortly be drawn from the box'. Let i = 'the next ball to be drawn from the box will be black', and let j = 'the next ball to be drawn from the box will be green'. I take it that e supports both i and j; without e or some other piece of evidence, there would be no support for either of those propositions, but given e, there is some support. Now i and j are contraries, so / implies -i. By the thesis of implication quoted earlier, it follows that e sup- ports --i at least as well as j. Hence e furnishes support both for i and Hi.

Finally take h = 'there are fish in the Dead Sea'. I take it that e furnishes no support whatever for this proposition or its negation. Hence I conclude that

h/e <i/e, while also Thle < abi/e,

contrary to anti-symmetry."

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ENTAILMENT AND SUPPORT 349

This argument counts against what I have said on two counts: first, if e supports i and Hi, (5) must be rejected; second, the account of support given has the anti-symmetry principle as a consequence. This latter may be shown thus. On our account of support the principle is:

if P(h/e) - P(h) < P(i/d) - P(i), then

P(Qi/d) - P(-i) <? P(Q- h/e) - P(Qh)

which reduces to, as P(Q.'p) = 1 -P(p),

if P(h/e) - P(h) < P(i/d) - P(i), then

P(i) - P(i/d) < P(h) - P(h/e)

which is an instance of: if a < b then -b < -a. The short way with Hacking's argument is to say that it de-

pends on what he calls the thesis of implication, which is just (10), and so is false. In the circumstances this would be too short; how- ever, I think it is sufficient to note that since the probability of H.-i is one minus the probability of i, if e increases the probability of i it cannot also increase the probability of Hi. I think a confusion be- tween internal and external negation may have led Hacking to think that e supports -i, because e does support the internal nega- tion of i, that is, e does support 'the next ball to be drawn from the box will not be black', what e does not support is 'it is not the case that the next ball to be drawn from the box will be black'.*

* This paper has benefited from discussions with B. D. Ellis.

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