PHIL 109 · Introduction to Formal Reasoning and Decision Making

4 Measures and stochastic truth-tables

4.1 Introduction

Consider the following argument:

• (1) Leona has a heavy, shaggy coat, a long bushy tail, and large tufted ears.
• (2) Most Maine Coons have those characteristics, and relatively few cats which are not Maine Coons do.
• $\therefore$ Leona is a Maine Coon.

That argument is not valid. However, it seems pretty good at first sight. It all depends on what we mean by “most” and “relatively few”. Just as PL offers us a mathematical model of the truth-functional aspects of English, the theory of probability offers us a mathematical model of our ordinary talk in terms of “probable”, “likely”, and so on. Our task in this part is to introduce the theory of probability and give you the tools necessary to analyze arguments that are explicitly probabilistic.

We said that the preceding argument seems pretty good at first sight. But there is a crucial fact missing from the premises: the fact that only about 10% of all cats are Maine Coons.¹ Suppose we add that fact to our argument:

• (1) Leona has a heavy, shaggy coat, a long bushy tail, and large tufted ears.
• (2) Most Maine Coons have those characteristics, and relatively few cats which are not Maine Coons do.
• (3) About 10% of all cats are Maine Coons.
• $\therefore$ Leona is a Maine Coon.

Should we now think that the argument is good? By the end of this part, you will be able to answer questions of this sort. In this chapter, we introduce the mathematical theory of probability as a branch of measure theory.

4.2 Measures

Probability can be defined as a normalized measure over possibility space. That is a mouthful, so we start by talking about measures.

A measure, in math, is a kind of function. We are already familiar with truth-functions: rules that take you from one (or a pair of) truth-value(s) to another truth-value. As we saw, truth-functions can be described in tables. Functions more generally are rules from something (the domain of the function) to something else (the codomain of the function). In the case of a truth-function, the domain and the codomain are both truth-values. Measures are a different kind of function. The domain of a measure is a space (which, as we will see, can be a literal geographical space or a more abstract notion of space as simply a set), and the codomain is the set of non-negative real numbers. Moreover, for a function to be a measure it has to be additive: if a region of the space can be completely decomposed into non-overlapping regions, then the measure of the total region must be the sum of the measures of the non-overlapping regions that compose it.

Let us start with an example: the measure that assigns to each region of New York City its population. We can illustrate part of that measure with a table:

We are representing just part of the function here. For instance, we have not explicitly represented the value for the portion of Long Island that overlaps NYC. Given that that portion is just Brooklyn plus Queens, its population just is the population of Brooklyn plus the population of Queens. So, by assuming that the table above is a measure, we can extract from the table the information about the value of the function applied to that part of Long Island: 4,500,000.

Now contrast that table with the following one, which gives the proportion of the population who are Mets fans:

That function, which takes regions of NYC and returns the proportion of the population in that region which are Mets fans, is not a measure. It satisfies the non-negativity part, but it doesn’t satisfy the additivity part. For consider what result the function should yield for Long Island: not the sum of the values for Brooklyn and Queens (which would yield the absurd 1.35), but rather the population-weighted average of the proportion of Mets fans in Brooklyn and Queens, which is $\frac{2}{3}$.

So, to recap: measures are additive functions from regions of space to non-negative numbers. “Counting” functions, like the population function above, are measures. Proportion functions, like the proportion of Mets function, are not measures because they are not additive. Other measures include the length of portions of a line, the volume of sub-regions of a sphere, etc. (this is where the name for measures comes from). There are also measures where the “space” that is the domain of the function is a space only in a metaphorical sense. Consider, for instance, the function which takes each family member (and each set of members) as input and delivers as output the amount of fruit eaten by them per year:

The “space” in this case is the family in question, and the function is a measure because if we want to know, for instance, how much fruit the parents (Alice and Bruce) consume in a year, the answer is just the sum of what they consume separately. Because the function is also non-negative, this is clearly a measure—the measure of how much fruit any group of members of this family consumes.

4.2.1 Normalized measures

We said we were going to think of probability as a normalized measure over possibility space. We already know what a measure is. But what does it mean to say that a measure is “normalized”? It simply means that we take the number associated with the whole space and then divide the result of the function by that total. For instance, the normalization of the population of NYC gives us the following table:

Notice that the value for the total space for any normalized measure will always be 1. In the case of the table above, the total space just is the total population of NYC. In the case of fruit consumption, the total “space” is the whole family, and its measure is the amount of fruit the whole family eats per year. Normalized measures give us a simple way of representing proportions.

4.2.2 Possibility space

A possibility space is the set of things that might happen. Suppose we are wondering when it would be a good day to go to the beach. Monday may have good beach weather—or not! On the other hand, one family member may have a doctor appointment on Monday. As you know, this gives us a space of four possibilities: good beach weather but doctor appointment, good beach weather without doctor appointment, bad beach weather with doctor appointment, bad beach weather without doctor appointment. Those four possibilities represent our possibility space. In general, $n$ atomic propositions will give us a possibility space with $2^{n}$ elements. In other words, truth-tables represent possibility space.

4.2.3 Normalized measures over possibility space and stochastic truth-tables

We are now in a position to fully understand what it means to say that probability is a normalized measure over possibility space. For consider now what happens when we add an additional column to an ordinary truth-table for three variables, and in that column we put numbers, with just the following condition: they have to be non-negative real numbers that add up to 1:

Try it in the Probability Lab Step through this stochastic truth table. The lab names the sentence letters P, Q, ...: here A is P, B is Q, C is R.

Recall that we can think of each row of a truth-table as a possible world. What the table with the $\Prb()$ column added does is assign a non-negative real number to each possible world. If we now assume that the $\Prb()$ column represents an additive function, then we can figure out the value of that function for any proposition whatsoever that can be represented using the atomic propositions $A$, $B$, and $C$. The value of the function applied to a proposition will be the sum of the values of the possible worlds where the proposition is true. Consider, for instance, the proposition $A \to (B \lor C)$. You can verify, by drawing the truth-table for this proposition, that it is true in every row except row 4. Therefore, $\Prb(A \to (B \lor C)) = \frac{1}{5}$. The $\Prb()$ function is a probability function, and because probability functions are additive, we just need the values of the function for each possible world to find out the value of the function applied to any proposition.

We call a truth-table with a column for a probability function added a stochastic truth-table. Notice some interesting consequences of stochastic truth-tables. First, the probability of any tautology will be 1. That is because any tautology is true in every possible world, and we know that the sum of the probabilities for each possible world adds up to 1. On the other hand, given that contradictions are true in no possible world, the probability of any contradiction will be 0. Next, consider a disjunction $(\varphi \lor \psi)$ where $\varphi$ and $\psi$ are mutually exclusive:

So $(\varphi \lor \psi)$ will be true in those worlds where $\varphi$ is true and in those worlds where $\psi$ is true. If $\varphi$ and $\psi$ were not mutually exclusive, simply adding up those worlds could lead to double-counting. But because they are mutually exclusive, $\Prb(\varphi \lor \psi) = \Prb(\varphi) + \Prb(\psi)$.

4.3 Conditional probability

So far we have been talking about the probability of a single proposition, however complex. But we are also interested in the conditional probability of one proposition given, or conditional on, another proposition. Suppose that we have a bag with 60 marbles. Each marble is either large or small, and each marble is either red or blue. We are told exactly how many marbles there are of each combination: 30 are large and red, 10 are large and blue, 15 are small and red, and 5 are small and blue. Suppose that we will draw a marble from the bag, and that each marble has the same probability of being drawn. We can ask questions about unconditional probability. For instance, the probability of drawing a large marble is $\frac{2}{3}$, because 40 of the 60 marbles are large; and the probability of drawing a red marble is $\frac{3}{4}$, because 45 of the 60 marbles are red. But suppose we want to know the probability that we draw a large marble, conditional on us drawing a red marble. How would we go about thinking about that? We are not asking what is the probability of any single proposition, not even of the proposition that if we draw a red marble then we draw a large one (more on that later). Rather, we are asking how the probabilities of two different propositions relate to each other. The intuitive idea behind conditional probability (applied to this case) is that we are zooming in on that region of the space of possibilities where we draw a red marble, and ask what proportion of that space is occupied by large marbles.

There is a definition of conditional probability that answers to this intuitive idea. We write the probability of $\varphi$ given $\psi$ like this: $\Prb(\varphi \given \psi)$, and we define it as a ratio of unconditional probabilities:

That is to say, the probability of $\varphi$ given $\psi$ is the ratio of the probability of the conjunction of $\varphi$ and $\psi$ to the unconditional probability of $\psi$. A crucial proviso is that conditional probabilities are undefined if the proposition we are conditioning on has a probability of 0—because division by 0 is itself undefined. The idea of conditional probability is that we are zooming in on that region of possibility space where $\psi$ is true, and we are asking in what proportion of that space $\varphi$ is true as well.

Going back to our marbles example, we were wondering what is the probability of drawing a large marble, conditional on drawing a red one. Given our definition of conditional probability, we can figure out the answer just by applying the formula. Let $L$ be we draw a large marble, and $R$ be we draw a red marble. We are looking, then, for the solution to the following equation: $\Prb(L \given R) = \frac{\Prb(L \land R)}{\Prb(R)}$. We are given all the information we need to solve that equation in the description of the case: $\Prb(R) = \frac{3}{4}$ and $\Prb(L \land R) = \frac{1}{2}$, so $\Prb(L \given R) = \frac{\sfrac{1}{2}}{\sfrac{3}{4}} = \frac{2}{3}$.

We can also figure out conditional probabilities by using stochastic truth-tables. The description of the case gives us all the unconditional probabilities we need:

How can we represent conditional probabilities with stochastic truth-tables? One thing to consider is that conditional probabilities are probabilities, and so when we add a column for conditional probabilities it will have to be filled with non-negative numbers that add up to 1. Now notice, first, that, for any proposition $\varphi, \Prb(\varphi \given \varphi) = 1$.³ So, if we are considering adding a column for the probabilities conditional on $R$, for instance, the rows where $R$ is true will have to add up to 1, which means that the rows where $R$ is false will have to be given 0:⁴

Now, how do we fill-in the two remaining rows? By preserving the ratios that they had before. Notice that the probability of row 1 was exactly twice that of row 3. In order to make the rows sum up to 1 again (a process called re-normalization), we therefore have to assign $\frac{2}{3}$ to row 1 and $\frac{1}{3}$ to row 3:

Try it in the Probability Lab Step through this stochastic truth table, with its conditional column. The lab names the sentence letters P, Q, ...: here L is P, R is Q.

We can now read conditional probabilities off the table in just the same way as we read off unconditional ones: the probability of a proposition conditional on $R$ is the sum of the values of the $\Prb(\cdot \given R)$ column in the rows where that proposition is true. For instance, $L$ is true in rows 1 and 2, so $\Prb(L \given R) = \frac{2}{3} + 0 = \frac{2}{3}$, which is exactly the answer we obtained from the definition of conditional probability. Similarly, $\neg L$ is true in rows 3 and 4, so $\Prb(\neg L \given R) = \frac{1}{3} + 0 = \frac{1}{3}$, and the two conditional probabilities add up to 1, as they should.

Remember that we said before that we should not confuse conditional probabilities with unconditional ones, not even when the proposition whose unconditional probability we are talking about is a conditional proposition. Notice, for instance, that whereas $\Prb(L \given R) = \frac{2}{3}$, $\Prb(R \to L) = \frac{3}{4}$.

Exercises

Consider the following stochastic truth-table:

Practice in the Probability Lab The lab names the sentence letters P, Q, ...: here A is P, B is Q, C is R.

Calculate the probabilities for the following formulas:

1. $(A \leftrightarrow B)$

   Show solution

   $\Prb(A\leftrightarrow B)=2/5$.

2. $(C \to B)$

   Show solution

   $\Prb(C\to B)=1$.

3. $(A \land C)$

   Show solution

   $\Prb(A\land C)=1/10$.

4. $(\neg A \land C)$

   Show solution

   $\Prb(\lnot A\land C)=1/10$.

5. Add a column for $\Prb(\cdot \given (B \leftrightarrow C))$ to the above table.

   Show solution

   The column $\Prb(\cdot\given B\leftrightarrow C)$ gives, for rows TTT, TTF, TFT, TFF, FTT, FTF, FFT, FFF: $1/6,\ 0,\ 0,\ 1/2,\ 1/6,\ 0,\ 0,\ 1/6$ (each old value divided by $\Prb(B\leftrightarrow C)=3/5$, and $0$ outside the $B\leftrightarrow C$ rows).

6. Is there a function $\Prb()$ that meets all the following constraints?

   • $\Prb(A \to (B \leftrightarrow C)) = 1$

   • $\Prb(B) = \Prb(\neg B)$

   • $\Prb(C) = 2 \times \Prb(C \land A)$

   • $\Prb(B \land C \land \neg A) = \frac{1}{5}$

   If not, prove that it’s impossible. If so, provide a probability table and demonstrate that the resulting distribution satisfies each of the four constraints. (Hint: Start by building a probability table; then figure out what each of the constraints says about the probability values in the table; then figure out if it’s possible to meet all of the constraints at once.)

   Show solution

   Yes, it is possible. One distribution that works, for rows TTT, …, FFF in the same order: $1/5,\ 0,\ 0,\ 1/4,\ 1/5,\ 1/10,\ 0,\ 1/4$. It sums to $1$ and satisfies all four constraints.