# Probabilistic Inferences

Recall that $X\Rightarrow Y$ means that $X$ is the premise and $Y$ is the conclusion. Of course, most arguments have more than one premise. In this section, think of $X$ as the conjunction of all the premises of an argument. For example, an instance of Modus Ponens is the argument with premises $P$ and $P\rightarrow Q$, and conclusion $Q$. In this notation, we represent this argument as $P\wedge (P\rightarrow Q)\Rightarrow Q$. 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 $P,P\rightarrow Q\Rightarrow Q$.

Consider an argument $X\Rightarrow Y$. We have introduced three ways to evaluate how $X$ supports $Y$:

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

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

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

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 $X, Y$, and $Z$, if $X\models Y$, then $X,Z\models Y$.

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 $X$, $Y$ and $Z$ and a stochastic truth table such that $X$ evidentially supports $Y$, but conditional on $Z$, $X$ does not evidentially supports $Y$.

To illustrate this phenomenon, find a stochastic truth table for $P$, $Q$ and $R$ such that:

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

## 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 $B$ mean that "Linda is a bank teller" and $F$ mean "Linda is active in the feminist movement". Then, anyone who reports that 2 is more probable than 1, is saying that $Pr(B\wedge F) > Pr(B)$. However, this contradicts the laws of probability: Since $(B\wedge F)\rightarrow B$ is a tautology, in any stochastic truth table $Pr(B)\ge Pr(B\wedge F)$. 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 $X$, $Y$ and $Z$ and all stochastic truth tables, if $Z\rightarrow Y$ is a tautology, then

$Pr(Z\mid X)\le Pr(Y\mid X)$

Let $E$ 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, $Pr(B\wedge F\mid E)\le Pr(B\mid E)$. So if $E$ evidentially supports $B\wedge F$, then $E$ must evidentially support $B$. More generally, we have the following:

Observation

For all formulas $X$, $Y$ and $Z$ and stochastic truth tables, if $Z\rightarrow Y$ is a tautology and $X$ evidentially supports $Z$, then $X$ evidentially supports $Y$.

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 $B$ and $B\wedge F$. Unlike evidential support, positive relevance is not preserved under tautological implications.

Observation

There are formulas $X$, $Y$ and $Z$ and a stochastic truth table such that $Z\rightarrow Y$ is a tautology, $X$ is positively relevant to $Z$, but $X$ is not positively relevant to $Y$.

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

1. $Pr(B\wedge F\mid E) > Pr(B\wedge F)$ ($E$ is positively relevant to $B\wedge F$)
2. $Pr(B \mid E) < Pr(B)$ ($E$ is negatively relevant to $B$)

## 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 $X$, $Y$ and $Z$, if $X\models Z$ and $Y\models Z$, then $X\veeY\models Z$.

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

1. The truth value assignment makes $X$ true. Then, since $X\models Z$, the truth value assignment makes $Z$ true

2. The truth value assignment makes $X$ false. Then, since $\neg X\models Z$, the truth value assignment makes $Z$ true.

Since $Z$ is true in every truth value assignment, $Z$ 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 $X$ evidentially supports $Y$ and $\neg X$ evidentially supports $Y$, then $Y$ is probable: For all formulas $X$ and $Y$ and all stochastic truth tables, if $Pr(Y\mid X)>0.5$ and $Pr(Y\mid \neg X)>0.5$, then $Pr(Y)>0.5$.

More generally, we can show that $X$ evidentially supports $Y$ by demonstrating that both of the following are true:

1. $X$ evidentially supports $Y$ conditional on $Z$; and
2. $X$ evidentially supports $Y$ conditional on $\neg Z$.

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

Observation

For all formulas $X$, $Y$ and $Z$ and all stochastic truth tables,
if $Pr(X\mid Y\wedge Z)>0.5$ and $Pr(X\mid Y \wedge \neg Z)>0.5$,
then $Pr(X\mid Y)>0.5$.

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

1. $Y$ is positively relevant to $X$ conditional on $Z$ and
2. $Y$ is positively relevant to $X$ conditional on $\neg Z$,

then $Y$ is positively relevant to $X$.

The first complication is that "$Y$ is positively relevant to $X$ conditional on $Z$" can be interpreted in two ways:

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

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 $X$, $Y$ and $Z$ and all stochastic truth tables,
if $Pr(X\mid Y\wedge Z)>Pr(X)$ and $Pr(X\mid Y \wedge \neg Z)>Pr(X)$,
then $Pr(X\mid Y)>Pr(X)$.

Surprisingly, there is no analogous result when "$Y$ is positively relevant to $X$ conditional on $Z$" uses the first interpretation:

Failure of the Unconditional Sure-Thing Principle

There are formulas $X$, $Y$ and $Z$ and a stochastic truth table, such that

1. $Pr(X\mid Y\wedge Z)>Pr(X\mid Z)$,
2. $Pr(X\mid Y \wedge \neg Z)>Pr(X\mid \neg Z)$, but
3. $Pr(X\mid Y)\le Pr(X)$.

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:

$\frac{1}{5} < \frac{2}{8}$
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:

$\frac{6}{8} < \frac{4}{5}$

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:

$\frac{7}{13} > \frac{6}{13}$

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 $P\Rightarrow Q\vee \neg Q$ valid? Does $P$ evidentially supports $(Q\vee\neg Q)$? Is $P$ positively relevant to $Q$?
1. Is $P\wedge\neg P\Rightarrow Q$ valid? Does $P\wedge\neg P$ evidentially supports $Q$? Is $P\wedge\neg P$ positively relevant to $Q$?
1. Is $P\wedge Q\Rightarrow P$ valid? Does $P\wedge Q$ evidentially supports $P$? Is $P\wedge Q$ positively relevant to $Q$?
1. Can you find a stochastic truth table for $P$, $Q$ and $R$ such that $P$ evidentially supports $Q$, but $P$ does not evidentially support $Q\vee R$?
1. Can you find a stochastic truth table for $P$, $Q$ and $R$ such that $P$ is positively relevant to $Q$, $P$ is positively relevant to $R$, but $P$ is not positively relevant to $Q\vee R$?