Skip to main content

Probabilistic Inferences

Recall that XYX\Rightarrow Y means that XX is the premise and YY is the conclusion. Of course, most arguments have more than one premise. In this section, think of XX as the conjunction of all the premises of an argument. For example, an instance of Modus Ponens is the argument with premises PP and PQP\rightarrow Q, and conclusion QQ. In this notation, we represent this argument as P(PQ)QP\wedge (P\rightarrow Q)\Rightarrow Q. Sometimes it is important to list the premises separately. We use a comma to separate the premises. For instance, we might write Modus Ponens as P,PQQP,P\rightarrow Q\Rightarrow Q.

Consider an argument XYX\Rightarrow Y. We have introduced three ways to evaluate how XX supports YY:

  1. XYX\Rightarrow Y is (deductively) valid: There is no truth value assignment that makes XX true and YY false.
  2. Given a stochastic truth table, XX evidentially supports YY: Pr(YX)>0.5Pr(Y\mid X)>0.5.
  3. Given a stochastic truth table, XX is positively relevant to YY: Pr(YX)>Pr(Y)Pr(Y\mid X)>Pr(Y).

There is an important difference between item 1 and items 2 and 3. There is only one truth table for an argument XYX\Rightarrow Y, but there are infinitely many stochastic truth tables. Whether XX evidentially supports (resp. is positively relevant to) YY depends on the stochastic truth table. One consequence of this is that for atomic propositions PP and QQ, while P⊭QP\not\models Q (the argument PQP\Rightarrow Q is not valid), there are stochastic truth tables in which PP evidentially supports (resp. is positively relevant to) QQ. For instance, use the following interactive stochastic truth table to find the following:

  1. A stochastic truth table where PP evidentially supports QQ and PP is positively relevant to QQ.
  2. A stochastic truth table where PP does not evidentially support QQ and PP is not positively relevant to QQ.
  3. A stochastic truth table where PP evidentially supports QQ and PP is not positively relevant to QQ.
  4. A stochastic truth table where PP does not evidentially support QQ and PP is positively relevant to QQ.
PPQQ
T\mathsf{T}T\mathsf{T}
T\mathsf{T}F\mathsf{F}
F\mathsf{F}T\mathsf{T}
F\mathsf{F}F\mathsf{F}
The sum of the numbers assigned to each row must be 1.
Pr(P)Pr(P)  = \ =\ undefinedPr(Q)Pr(Q)  = \ =\ undefined
Pr(QP)Pr(Q \mid P) = \ =\ Pr(QP)Pr(Q\wedge P) // Pr(P)Pr(P)  = \ =\ undefined / undefined = \ =\ undefined
PP does not evidentially support QQ
PP is not positively relevant to QQ

In the remainder of this section, we identify some interesting differences between valid arguments and inductively strong arguments.

Non-Monotonicity#

We have already observed that valid arguments are monotonic:

For all X,YX, Y, and ZZ, if XYX\models Y, then X,ZYX,Z\models Y.

That is, any valid argument remains valid with the addition of any premise. It is not hard to find examples of seemingly strong arguments that are no longer as strong with the addition of certain premises. For instance, the following is a good argument:

Tweety is a bird. So, Tweety flies.

While this is not a valid argument, it is an inductively strong argument. However, if you learn that Tweety is a penguin, then the conclusion that Tweety flies no longer follows from the premises. This type of non-monotonicity is typical of probabilistic reasoning. The first observation is that evidential support is not monotonic.

Observation

There are formulas XX, YY and ZZ and a stochastic truth table such that XX evidentially supports YY, but conditional on ZZ, XX does not evidentially supports YY.

To illustrate this phenomenon, find a stochastic truth table for PP, QQ and RR such that:

  1. Pr(QP)>0.5Pr(Q\mid P) > 0.5 (so PP evidentially supports QQ)
  2. Pr(QPR)<0.5Pr(Q\mid P\wedge R)< 0.5 (so PRP\wedge R does not evidentially support QQ)
PPQQRR
T\mathsf{T}T\mathsf{T}T\mathsf{T}
T\mathsf{T}T\mathsf{T}F\mathsf{F}
T\mathsf{T}F\mathsf{F}T\mathsf{T}
T\mathsf{T}F\mathsf{F}F\mathsf{F}
F\mathsf{F}T\mathsf{T}T\mathsf{T}
F\mathsf{F}T\mathsf{T}F\mathsf{F}
F\mathsf{F}F\mathsf{F}T\mathsf{T}
F\mathsf{F}F\mathsf{F}F\mathsf{F}
Pr(QP)Pr(Q \mid P)  = \ =\ Pr(QP)Pr(Q\wedge P)  / \ /\ Pr(P)Pr(P)  = \ =\ 0.300 / 0.500 = \ =\ 0.600
Pr(QPR)Pr(Q \mid P\wedge R)  = \ =\ Pr(QPR)Pr(Q\wedge P\wedge R)  / \ /\ Pr(PR)Pr(P\wedge R)  = \ =\ 0.200 / 0.300 = \ =\ 0.667
This is not an example of a failure of monotonicity for evidential support:
Pr(QP)>0.5Pr(Q \mid P) > 0.5 and Pr(QPR)>0.5Pr(Q \mid P\wedge R) > 0.5

Conjunction Fallacy#

We start with a famous reasoning experiment from the Nobel prize winning psychologists Amos Tversky and Dan Kahneman. Suppose that you are told the following about Linda:

Linda is 31 years old, single, outspoken, and very bright. She majored in philosophy. As a student, she was deeply concerned with issues of discrimination and social justice, and also participated in anti-nuclear demonstrations.

Which is more probable?

  1. Linda is a bank teller
  2. Linda is a bank teller and is active in the feminist movement.

Most people respond that 2 is more probable than 1. Let BB mean that "Linda is a bank teller" and FF mean "Linda is active in the feminist movement". Then, anyone who reports that 2 is more probable than 1, is saying that Pr(BF)>Pr(B)Pr(B\wedge F) > Pr(B). However, this contradicts the laws of probability: Since (BF)B(B\wedge F)\rightarrow B is a tautology, in any stochastic truth table Pr(B)Pr(BF)Pr(B)\ge Pr(B\wedge F). In fact, people that respond that 2 is more probable than 1 seem to be violating the following stronger principle of probabilistic reasoning:

Observation

For all XX, YY and ZZ and all stochastic truth tables, if ZYZ\rightarrow Y is a tautology, then

Pr(ZX)Pr(YX)Pr(Z\mid X)\le Pr(Y\mid X)

Let EE represent the evidence that you are told about Linda (e.g., she was a philosophy major, deeply concerned with issues of discrimination and social justice, etc.). Then, according to the above observation, Pr(BFE)Pr(BE)Pr(B\wedge F\mid E)\le Pr(B\mid E). So if EE evidentially supports BFB\wedge F, then EE must evidentially support BB. More generally, we have the following:

Observation

For all formulas XX, YY and ZZ and stochastic truth tables, if ZYZ\rightarrow Y is a tautology and XX evidentially supports ZZ, then XX evidentially supports YY.

One explanation for why many people say that 2 is more probable than 1 is that they are evaluating the relevance of the evidence to the two possibilities rather than measuring the evidential support of the evidence to the BB and BFB\wedge F. Unlike evidential support, positive relevance is not preserved under tautological implications.

Observation

There are formulas XX, YY and ZZ and a stochastic truth table such that ZYZ\rightarrow Y is a tautology, XX is positively relevant to ZZ, but XX is not positively relevant to YY.

To illustrate, finish filling in the following stochastic truth table so that:

  1. Pr(BFE)>Pr(BF)Pr(B\wedge F\mid E) > Pr(B\wedge F) (EE is positively relevant to BFB\wedge F)
  2. Pr(BE)<Pr(B)Pr(B \mid E) < Pr(B) (EE is negatively relevant to BB)

Sure-Thing Principle#

In order to show that something follows from a disjunction you must show that it follows from each disjunct. That is, the following is an important principle of valid inferences:

Observation

For all formulas XX, YY and ZZ, if XZX\models Z and YZY\models Z, then X\veeYZX\veeY\models Z.

A special case of the above observation is the following: If XZX\models Z and ¬XZ\neg X\models Z, then ZZ must be a tautology. To see why this is true, suppose that XZX\models Z and ¬XZ\neg X\models Z. Then, every truth value assignment falls into one of two categories:

  1. The truth value assignment makes XX true. Then, since XZX\models Z, the truth value assignment makes ZZ true

  2. The truth value assignment makes XX false. Then, since ¬XZ\neg X\models Z, the truth value assignment makes ZZ true.

Since ZZ is true in every truth value assignment, ZZ is a tautology. Can we adapt the above reasoning to show evidential support and positive relevance?

The first observation is that the above reasoning can be used to show evidential support between two formulas. We start with the following preliminary observation:

Observation

If XX evidentially supports YY and ¬X\neg X evidentially supports YY, then YY is probable: For all formulas XX and YY and all stochastic truth tables, if Pr(YX)>0.5Pr(Y\mid X)>0.5 and Pr(Y¬X)>0.5Pr(Y\mid \neg X)>0.5, then Pr(Y)>0.5Pr(Y)>0.5.

More generally, we can show that XX evidentially supports YY by demonstrating that both of the following are true:

  1. XX evidentially supports YY conditional on ZZ; and
  2. XX evidentially supports YY conditional on ¬Z\neg Z.

The first statement means that Pr(YXZ)>0.5Pr(Y\mid X\wedge Z) > 0.5 and the second statements means that Pr(YX¬Z)>0.5Pr(Y\mid X\wedge\neg Z) > 0.5. So, the following justifies the above claim.

Observation

For all formulas XX, YY and ZZ and all stochastic truth tables,
if Pr(XYZ)>0.5Pr(X\mid Y\wedge Z)>0.5 and Pr(XY¬Z)>0.5Pr(X\mid Y \wedge \neg Z)>0.5,
then Pr(XY)>0.5Pr(X\mid Y)>0.5.

The situation is more complicated with positive relevance. We want to show that if

  1. YY is positively relevant to XX conditional on ZZ and
  2. YY is positively relevant to XX conditional on ¬Z\neg Z,

then YY is positively relevant to XX.

The first complication is that "YY is positively relevant to XX conditional on ZZ" can be interpreted in two ways:

  1. Pr(XYZ)>Pr(X)Pr(X\mid Y\wedge Z) > Pr(X)
  2. Pr(XYZ)>Pr(XZ)Pr(X\mid Y\wedge Z) > Pr(X\mid Z)

Using the second interpretation, we can adapt the above reasoning to positive relevance, called the conditional sure-thing principle.

Conditional Sure-Thing Principle

For all formulas XX, YY and ZZ and all stochastic truth tables,
if Pr(XYZ)>Pr(X)Pr(X\mid Y\wedge Z)>Pr(X) and Pr(XY¬Z)>Pr(X)Pr(X\mid Y \wedge \neg Z)>Pr(X),
then Pr(XY)>Pr(X)Pr(X\mid Y)>Pr(X).

Surprisingly, there is no analogous result when "YY is positively relevant to XX conditional on ZZ" uses the first interpretation:

Failure of the Unconditional Sure-Thing Principle

There are formulas XX, YY and ZZ and a stochastic truth table, such that

  1. Pr(XYZ)>Pr(XZ)Pr(X\mid Y\wedge Z)>Pr(X\mid Z),
  2. Pr(XY¬Z)>Pr(X¬Z)Pr(X\mid Y \wedge \neg Z)>Pr(X\mid \neg Z), but
  3. Pr(XY)Pr(X)Pr(X\mid Y)\le Pr(X).

The failure of the unconditional sure-thing principle is related to a well-known issue in statistics known as Simpson's Paradox. Suppose that a University is hiring in Philosophy and Mathematics, and that 13 men and 13 women in total apply for jobs.

  1. In the Mathematics department, 5 men and 8 women apply for jobs. Suppose that 1 of the men is hired and 2 of the women are hired. The success rate favors women over men:

    15<28\frac{1}{5} < \frac{2}{8}
  2. In the Philosophy department, 8 men and 5 women apply for jobs. Suppose that 6 of the men are hired and 4 of the women are hired. The success rate favors women over men:

    68<45\frac{6}{8} < \frac{4}{5}

So, in both departments, the success rate favors women over men. However, when we determine the overall success rate, the inequalities reverse: Overall, 7 of the 13 men were hired and 6 out of the 13 women were hired. This means that the overall success rate for men is better than for women:

713>613\frac{7}{13} > \frac{6}{13}

How can it be that each Department favors women applicants, and yet overall men fare better than women?

See this video for a discussion of this.

Practice Questions#

  1. Is PQ¬QP\Rightarrow Q\vee \neg Q valid? Does PP evidentially supports (Q¬Q)(Q\vee\neg Q)? Is PP positively relevant to QQ?
  1. Is P¬PQP\wedge\neg P\Rightarrow Q valid? Does P¬PP\wedge\neg P evidentially supports QQ? Is P¬PP\wedge\neg P positively relevant to QQ?
  1. Is PQPP\wedge Q\Rightarrow P valid? Does PQP\wedge Q evidentially supports PP? Is PQP\wedge Q positively relevant to QQ?
  1. Can you find a stochastic truth table for PP, QQ and RR such that PP evidentially supports QQ, but PP does not evidentially support QRQ\vee R?
  1. Can you find a stochastic truth table for PP, QQ and RR such that PP is positively relevant to QQ, PP is positively relevant to RR, but PP is not positively relevant to QRQ\vee R?