"Probabilistic Semantics for Epistemic Modals" Moritz Schulz, New College oxford

The Thesis

It is prima facie plausible that “might” and “must” express a certain kind of
epistemic modality. We shall assume that “might” expresses some kind of epistemic possibility and that “must” can be used to express a corresponding kind
of epistemic necessity. Evidence for the view that “might” and “must” provide
an interdefinable pair of modals can be gained from examples like the following:

(1) They might be away.
(2) No, they must be at home.

It seems to be intuitively plausible that (2) is the negation of (1). In what
follows, we shall thus assume that “might” and “must” are interdefinable (in
their epistemic usages). In our informal discussion, we shall be mainly concerned
with “might” because it seems to be less ambiguous than “must”. Besides, we
shall only deal with indicative or present tense usages of “might”.
A good starting point for our investigation into the semantics of “might” is
the following observation:
(Basic Observation)
We are prepared to assert “It might be that X” iff our credence in X
is positive, i.e. iff C(X) > 0.
If one is not sure that they are away, one is in a position to say “They might
be at home”. And if one is sure that they are away, one should reject that they
might be at home. Moreover, the basic observation seems to provide a good
explanation why it would be an odd thing to say
(3) They might be at home, but I am certain that they are away.
Now, assuming that the basic observation gets the assertability conditions of
“might”-statements about right, we can go on and ask what our credence in a
“might”-statement should be. Since assertability goes by high credence, it follows
from the basic observation that our credence in a “might”-statement should be
high iff our credence in the embedded statement is non-zero. Actually, I would
like to argue for a more definite thesis:
(MIGHT)
Our credence in “It might be that X” should be 1 iff our credence
in X is positive and it should be 0 iff our credence in X is 0. Thus
C(Might X) = 1 iff C(X) > 0 and C(Might X) = 0 iff C(X) = 0 for
C being any reasonable credence function.

My main reason for proposing MIGHT stems from the observation that beliefs
in “might”-statements do not come in degrees. Rather, it seems to be an all or-
nothing matter. For instance, we usually do not qualify a “might”-statement
with a phrase such as “probably” which can be used to compare the likelihood
of statements. It is rather odd to say
(?) Probably they might be at home.
Also, we do not say that one “might”-statement is more likely than another:
(?) It is more likely that they might be at home than it is that they might be
away.
All this is evidence for allowing a “might”-statement to receive only two values.
In addition, it is hard to see what kind of further evidence (over and above nonzero
credence) would be needed in order to be certain about a “might”-statement.
It seems to be enough to give the embedded statement some (subjective) chance
of being true. So, it seems that one can make a good case for MIGHT.
Given our assumption about the interdefinability of “might” and “must”, our
thesis has a natural counterpart.
(MUST)
Our credence in “It must be that X” should be 1 iff our credence
in X is 1 and it should be zero otherwise. Thus C(Must X) = 1 iff
C(X) = 1 and C (Must X) = 0 iff C(X) < 1 for C being any reasonable
credence function.

Two envelope paradox solved. Jonny Blamey 22.6.2007

ALF:

Here is a well known paradox for decision theory. I offer you a choice of two envelopes. Let’s call them A and B. A contains twice as much money as B, but we don’t know how much money is in either. It is not possible to tell by looking which envelope is A and which envelope is B. You select one of the two envelopes. Let us call this envelope E1. We still don’t know whether E1 is A or B, but we can safely assume that it is equally likely to be either. So the probability that E1 is A = the probability that E1 is B = ½.
Now I offer you a choice. You can either keep E1 and take the money inside, let us call this option KEEP. Or you can exchange E1 for E2 and take the money in E2. Let us call this SWAP. Which do you choose and why? Remember, you want to end up with envelope A since it has twice as much money in it as envelope B.

BETH

Frankly I am indifferent. I therefore choose KEEP since it requires less effort. I am indifferent because I was initially equally likely to have chosen A as B, and by swapping I am just as likely to end up with A as B. We can express this by multiplying the expected utilities by their probabilities. For KEEP we get whatever value E1 contains, which is either the higher sum A, or the lower sum B. We multiply A by the probability that E1 contains A, and multiply B by the probability that E1 contains B. For SWAP we get whatever is contained in E2. so we multiply the probability that E1 contains A by B and the probability that E1 contains B by A.

PAYOFF KEEP
P(E1 =A)(A) = ½ A = B
+
P (E1 = B)(B) = ½ B = ¼ A
=
1 ½ B = ¾ A

PAYOFF SWAP
P(E1 =A)(B) = ½ A = B
+
P (E1 = B)(A) = ½ B = ¼ A
=
1 ½ B = ¾ A


In fact, probability notions aren’t even necessary since SWAP is no different from choosing E2 in the first place, and this option was open to me. SWAP is just a dithering form of KEEP.

ALF

Ok, I agree, it seems like INDIFFERENCE is the most rational attitude to have towards the choice between SWAP and KEEP. But suppose you were allowed to open E1 before deciding whether to KEEP or SWAP. E1 contains had a specific amount of money in it, let’s call it x. Now we can deduce that E2 must either contain either 2x or ½ x.

BETH

That’s right, because if E1 were A, then E2 would contain ½ x, whereas if E1 were B then E2 would contain 2x and that exhausts the possibilities.

ALF

In fact we don’t even have to open E1 since the same is true for all values x.

BETH

I suppose so.

ALF

If I elect to KEEP, I will end up with x for certain, but if I SWAP I will end up with either 2x or 1/2 x. Therefore it is rational to SWAP since I’ve ½ chance of ending up with ½ x and a ½ chance of ending up with 2x, making the expected utility of SWAP 1¼x. Therefore I should SWAP, and what is more I should pay up to ¼ x to SWAP.

BETH

That’s absurd since then you should also pay up to ¼ x to swap back on the same reasoning. It is absurd because you end up with an intransitive preference. You have reasoned that you should prefer E1 to E2 and E2 to E1.Your mistake is in assigning a single value to x. E1 contains a sum of money and you have elected to call that sum x. If we call the higher sum “A” and the lower sum “B”, then p(x = A) = ½ and p(x = B) = ½ . So we can calculate the utility of KEEP as being ½ A + ½ B, which is ¾ A or 1 ½ B. We don’t know which x is, but we know that x is either A or B. So this gives us 2 possibilities.
1. p (E1 = A and x = A). = ½
The Pay off for KEEP is (A = ½ B = x)
The Pay off for SWAP (½ A= B = 1/2x)
2. p (E1 = B and x =B). = ½
The Pay off for KEEP is (½ A= B = x)
The Pay off for SWAP (A = ½ B = 2x)

So the total Pay off for KEEP is:
½ (A = ½ B) + ½ ( ½ A = B ) = (3/4 A = 1 ½ B )
If x = A, the pay off is ¾ x; if x = B the pay off is 1 ½ x
And the total pay off for SWAP is:
½ (½ A= B ) + ½ (A = ½ B ) = (3/4 A = 1 ½ B)
If x = A, the pay off is ¾ x; if x = B the pay off is 1 ½ x

As you can see, if you are clear about the value of x and the probability that x has that value, the expected utility comes out the same whether you SWAP or KEEP.

ALF

Very clever, but suppose I actually open E1 and count the cash inside. Suppose it comes out at for example £12. I reason that if I KEEP I get £12; whereas if I SWAP, I get either £6 or £24. Since I was just as likely to have chosen A or B, then the expected utility of SWAP is £3 + £12 = £15. Here I am being completely clear that x = £12 and that the probability that x has this value is 1 since I know it to have this value having opened the envelope. Furthermore, the same reasoning applies however much money is in E1. I should always SWAP.

BETH

Your mistake here is in assuming that prob (E2 contains £6) = prob (E2 contains £24) = ½. Why do you assume that? The correct assignments are prob (E2 contains £6) = 2/3 and prob (E2 contains £24) = 1/3.


ALF

I assume that prob (E2 contains £6) = prob (E2 contains £24) = ½ because it is obvious. It is obvious to everyone who has written about this paradox and it is obvious to me. If E1 contained £12 and E1 contained the lesser sum then E2 contains £24. If E1 contained the greater sum then E2 contains £6. There is a equal chance that E1 contained the greater sum (A) or the lesser sum (B), so there is an equal chance that E2 contains £24 or £6.

BETH

The reasoning is seductive, but it you refer back to the utility calculations there is only a chance of getting 2x when x = B and there is only a chance of getting ½ x when x = A. So it is not clear that you reasoning is valid. However it is difficult to explain why your reasoning is wrong, so instead I will demonstrate why the probability that E2 contains 2x is 1/3 and the probability that E2 contains ½ x is 2/3 on the assumption that E1 contains x.

Frank Ramsey developed a measurement for a subjects degree of belief that p given indifference between the options
1. A for certain
2. B if p and C if ~p.
In these conditions the subjects degree of belief that p is equal to
(A – C)/(B – C)
This quantity can be shown to be a probability in that it should conform to the axioms of probability calculus if the subject doesn’t want to be victim to a Dutch book.

This fits rather well with the two envelope paradox. Let proposition p1 be that E2 contains twice the sum of money in E1. Let’s call this sum of money “x”.
So the two options open to us in the envelope problem are
KEEP: x for certain
SWAP: 2x if p1 and ½ x if not p1.
A subject who is indifferent between these options has a degree of belief in p1 equal to:
(x – ½ x)/(2x – ½ x)
= 1/3
Let p2 be that E2 contains half the sum of money in E1.
KEEP: x for certain
SWAP ½x if p2 and 2x if not p2
A subject who is indifferent between these options has a degree of belief in p2 equal to:
(x – 2x)/(½ x – 2x)

= 2/3

So a subject who is indifferent between the options KEEP and SWAP has a degree of belief 1/3 that E2 contains twice the amount in E1 and degree of belief 2/3 that E2 contains half the amount in E1.
We started off agreeing that we should be indifferent between the options KEEP and SWAP. If we should be indifferent, then our degree of belief should be what the formula says it is when we are indifferent. So our degree of belief that E2 contains twice the amount in E1 should be 1/3 and our degree of belief that E2 contains half the amount in E1 is 2/3.

ALF: That is absurd and I can tell you why. If the reasoning was valid then it would apply equally to E2. This would make the probability that E2 contains twice the amount in E1 = 1/3 and the probability that E1 contains twice the amount in E2 also 1/3. But this exhausts the possibilities so the numbers should add up to 1. Worse still the probability the E2 contains half amount in E1 = 2/3, but so does the probability that E1 contains half the amount in E2, that means that according to your reasoning either the disjunction has a probability higher than 1, or there is a probability of at least 1/3 that both amounts are lower than the other. And the worst subjectivist crime of all, your degree of belief not only does, but should change according to how the case is described. For you have only a 1/3 degree of belief that E2 has half the amount in E1 but 2/3 degree of belief that E1 has twice the amount in E2. But surely this is the same state of affairs.

BETH: Go back and read your Kripke. “The amount in E1” can be a rigid or non-rigid designator, as can “The amount in E2”. Let us rigidly designate the amount in E1 as x and the amount in E2 as y. For non rigid designation we will use E1 and E2. If we rigidly designate (or simply find out) the value of x, but we don’t know the amount in E2, then prob (E2 = 2x) = 1/3 and prob (E2 = ½ x) = 2/3. On the other hand if we rigidly designate (or know) the value of y but don’t know the value of E1, then prob (E1= 2y) = 1/3 and prob (E2 = 1/2y) = 2/3.

To picture this suppose E1 = £4 and E2 = £2. Let call the sum of E2 and E1: T for total. A for the lesser sum and B for the greater sum


Possible values for E2 and T given E1 = 4 = x

1.
2. E2 = 1/2x = B
3.
4. E1 = x = A or B
5.
6. T = 1 ½ x = A + ½ B
7.
8. E2 = 2x = A
9.
10.
11.
12. T = 3x = A + B.

Possible values for E1 and T given E2 = 2 =y

1. E1 = ½ y = B
2. E2 = y = A or B
3. T = 1 ½ y = A + B
4. E1 = 2y = A
5.
6. T = 3x = A + B



So when I say that the probability that the amount in E2 is half the amount in E1 = 2/3 I mean the probability that E2 = 2 = 2/3. This is because “the amount in E1” rigidly designates x or 4. So the probability that E1 = 4 is 1. But when I say the probability that the amount in E1 is twice the amount in E2 I mean the probability that (E1 = 4) = 1/3. This time it is “the amount in E2” which is the rigid designator. When a term is rigidly designated it is assumed to have a probability 1.

Incidentally, don’t assume that we are talking about Bayesian or Kolmogorov conditionalization here. The prior or unconditional probabilities of E2 = 2 and E1 = 4 are either zero or undefined. Here is how Kolmogorov defines conditional probabilities

P(E2=2 E1=4)
=
P(E2=2 ∩E1=4)/ P(E1=4)

provided P(b) > 0.

But I take it that we don’t know any of the unconditional probabilities.

ALF:
Perhaps we could think of the total amount of money in both envelopes as defining the logical space of probability. A probability can be expressed as a proper fraction. If we take the total amount in both envelopes to be the denominator and the amount in each envelope to be numerators we get the result that envelope A contains 2/3 of the total and envelope B contains 1/3 of the total. When we open E1 we discover that there is £x inside. What we are interested in is whether we have 1/3 of the total or 2/3 of the total. If we have 1/3 of the total then E2 contains 2x and the denominator is 3x. If we have 2/3 of the total then E2 contains ½ x and the denominator is 1½ x.
In effect we have a 1/3 share of the 2x space of probability and a 2/3 share of the ½ x space of probability.
BETH:
Yes, here’s how I look at it. When we open E1 and find a sum of money in there, (call it x) we know that the total possible amount of money is 3x. Given 3x, we can make a pair of envelopes with x and 2x respectively; or we can make 2 pairs of envelopes with x and ½ x. Therefore given that E1 contains x, the probability that E2 contains ½ x is twice that of the probability that E2 contains 2x. Given that this exhausts the possibilities the probability that E2 contains 2x is 1/3 and the probability that E1 contains ½ x is 2/3.
ALF:
So contra Casper J. Albers, Barteld P. Kooi and Willem Schaafsma in Synthese 2005 145: 89–109, we can, and have, resolved the two envelope problem. To sum up: Indifference between KEEP and SWAP is the rational attitude. The correct degree of belief to have that the second envelope contains twice the amount contained in the first envelope is 1/3 and the correct degree of belief that the second envelope contains half the amount in the first envelope is 2/3. Therefore the expected utility of both SWAP and KEEP are the same. An explanation for the counter intuitive probability assignments is that 3x = 2(1 ½ x) so the lower pair of envelopes is twice as likely as the higher pair. Perhaps in slogan form: for every pair of socks there are two odd socks.
BETH That’ll explain why you wear odd socks half the time then Alf, or should that be 2/3s of the time!