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Confirmation Measures

When an argument is inducitvely strong, the premsises are said to confirm the conclusion. For example, suppose that you are interested in whether Ann (whom you do not know very well) is a student at the UMD. If HH is the proposition "Ann is a student at UMD", then your probability of HH is around 0.50.5. Now supppose that you observe, Ann in the Skinner Building carrying a backpack. This evidence confirms HH in the sense that it boosts your probability of HH, even though it does not guarantee that HH is true. On the other hand, observing Ann carrying a backpack in the National Mall would probably not raise your probability of HH, and so does not confirm HH.

As we have noted when discussing the definition of inductively strong arguments, it is important to distinguish between absolute confirmation and incremental confirmation:

  1. XX evidentially supports YY (Pr(Y∣X)>0.5Pr(Y\mid X)>0.5) represents aboslute confirmation where learning that XX is true raises the probability of the hypothesis YY above some fixed threshold (typically the threshold is 0.50.5).

  2. XX is positively relevant to YY (Pr(Y∣X)>Pr(Y)Pr(Y\mid X)>Pr(Y)) represents incremental confirmation. This is the "boost" that the conclusion would receive after learning the premises are true.

So far in this book, we have focused on evaluating a single argument. To evaluate an argument X⇒YX\Rightarrow Y, we first ask:

  1. Is the argument valid? That is, is YY true in all situations that XX is true?

If the answer is no to the above question, then we ask the following:

  1. Should you believe that the conclusion YY is true under the supposition that the premise XX is true?
  2. Is the premise XX relevant to the truth of the conclusion YY?

Of course, it is also important to determine wether the premise XX is true to fully evaluate the above argument.

In this section, we are going to introduce ways to compare two different arguments in terms of both abolute and incremental confirmation. For example, consider the following stochastic truth table:

PPQQRR(Q∧P)(Q\wedge P)(Q∧R)(Q\wedge R)
0.2 T\mathsf{T}T\mathsf{T}T\mathsf{T}T\mathsf{T}T\mathsf{T}
0.2 T\mathsf{T}T\mathsf{T}F\mathsf{F}T\mathsf{T}F\mathsf{F}
0.1 T\mathsf{T}F\mathsf{F}T\mathsf{T}F\mathsf{F}F\mathsf{F}
0.1 T\mathsf{T}F\mathsf{F}F\mathsf{F}F\mathsf{F}F\mathsf{F}
0.15 F\mathsf{F}T\mathsf{T}T\mathsf{T}F\mathsf{F}T\mathsf{T}
0.05 F\mathsf{F}T\mathsf{T}F\mathsf{F}F\mathsf{F}F\mathsf{F}
0.1 F\mathsf{F}F\mathsf{F}T\mathsf{T}F\mathsf{F}F\mathsf{F}
0.1 F\mathsf{F}F\mathsf{F}F\mathsf{F}F\mathsf{F}F\mathsf{F}
Pr(Q∣P) = Pr(Q\mid P)\ =\ Pr(Q∧P)Pr(Q\wedge P)/ / Pr(P)Pr(P)  = \ =\ 0.400.40/ / 0.600.60 = 0.667\ =\ 0.667
Pr(Q∣R) = Pr(Q\mid R)\ =\ Pr(Q∧R)Pr(Q\wedge R)/ / Pr(R)Pr(R)  = \ =\ 0.350.35/ / 0.550.55 = 0.636\ =\ 0.636

In both of the following arguments, the premise evidentially supports the conclusion:

  1. P⇒QP\Rightarrow Q
  2. R⇒QR\Rightarrow Q

But, there is more we can say about these two arguments. Since Pr(Q∣P)>Pr(Q∣R)Pr(Q\mid P)>Pr(Q\mid R), PP provides stronger support for the conclusion QQ than RR does. So, in the absolute sense of confirmation, PP confirms QQ more than RR confirms QQ

Note that Pr(Q)=0.6Pr(Q)=0.6. This means that PP is positively relevant to QQ and RR is positively relevant to QQ. So, both arguments are inductively strong. Is there some way to measure which premise is more positively relevant to QQ? There are different ways to measure how positively relevant a premise is to a conclusion. One natural thought is to measure positive relevance in terms of difference between the conditional and unconditional probabilities.

Confirmation Measure: Difference

In any stochastic truth table, for any formulas XX and YY, let d(X,Y)d(X, Y) be defined as follows:

d(X,Y)=Pr(Y∣X)−Pr(Y).d(X, Y) = Pr(Y\mid X) - Pr(Y).

In the above stochastic truth table, we have:

  • d(P,Q)=Pr(Q∣P)−Pr(Q)=0.667−0.6=0.067d(P,Q)=Pr(Q\mid P)-Pr(Q)=0.667 -0.6 = 0.067; and
  • d(R,Q)=Pr(Q∣R)−Pr(Q)=0.636−0.6=0.036d(R,Q)=Pr(Q\mid R)-Pr(Q)=0.636 -0.6 = 0.036.

For any stochastic truth table, and any formulas XX and YY, the number d(X,Y)d(X, Y) is a measure of the boost in probability that XX receives from YY. There are many other ways to measure how much XX confirms YY. It is beyond the scope of this course to survey all the different confirmation measures. One problem with d(X,Y)d(X, Y) as a measure of incremental confirmation is that if YY is already very probability (i.e., the probability is close to 11), then d(X,Y)d(X, Y) will be very small regardless of how strong the evidence XX is for YY. The next way of measuring positive relevance does not suffer from this problem:

Confirmation Measure: Likelihood Ratios

In any stochastic truth table, for any formulas XX and YY, let â„“(X,Y)\ell(X, Y) be defined as follows:

ℓ(X,Y)=(Pr(X∣Y)−Pr(X∣¬Y))(Pr(X∣Y)+Pr(X∣¬Y))\ell(X, Y) = \frac{(Pr(X\mid Y) - Pr(X\mid \neg Y))}{(Pr(X\mid Y) + Pr(X\mid \neg Y))}

Both d(X,Y)d(X, Y) and ℓ(X,Y)\ell(X, Y) measures the incremental confirmation that XX gives YY. One potential issue that may lead to confusion with the definition of ℓ\ell is that the equation of ℓ(X,Y)\ell(X, Y) uses Pr(X∣Y)Pr(X\mid Y)---the probability of XX conditional on YY---but d(X,Y)d(X, Y) uses Pr(Y∣X)Pr(Y\mid X). If XX is the evidence and YY is the hypothesis, then Pr(X∣Y)Pr(X\mid Y) is the likelihood of observing XX supposing that the hypothesis YY is true and Pr(X∣¬Y)Pr(X\mid\neg Y) is the likelihood of observing XX supposing that the hypothesis YY is false. So, the numerator Pr(X∣Y)−Pr(X∣¬Y)Pr(X\mid Y)-Pr(X\mid\neg Y) is the differece between observing XX assuming YY is true and observing XX assuming that YY is false. To calculate ℓ(P,Q)\ell(P,Q) and ℓ(R,Q)\ell(R,Q) in the above stochastic truth table we need to calculate the following probabilities:

PPQQRR(P∧Q)(P\wedge Q)(P∧¬Q)(P\wedge \neg Q)¬Q\neg Q(R∧Q)(R\wedge Q)(R∧¬Q)(R\wedge \neg Q)¬Q\neg Q
0.2 T\mathsf{T}T\mathsf{T}T\mathsf{T}T\mathsf{T}F\mathsf{F}F\mathsf{F}T\mathsf{T}F\mathsf{F}F\mathsf{F}
0.2 T\mathsf{T}T\mathsf{T}F\mathsf{F}T\mathsf{T}F\mathsf{F}F\mathsf{F}F\mathsf{F}F\mathsf{F}F\mathsf{F}
0.1 T\mathsf{T}F\mathsf{F}T\mathsf{T}F\mathsf{F}T\mathsf{T}T\mathsf{T}F\mathsf{F}T\mathsf{T}T\mathsf{T}
0.1 T\mathsf{T}F\mathsf{F}F\mathsf{F}F\mathsf{F}T\mathsf{T}T\mathsf{T}F\mathsf{F}F\mathsf{F}T\mathsf{T}
0.15 F\mathsf{F}T\mathsf{T}T\mathsf{T}F\mathsf{F}F\mathsf{F}F\mathsf{F}T\mathsf{T}F\mathsf{F}F\mathsf{F}
0.05 F\mathsf{F}T\mathsf{T}F\mathsf{F}F\mathsf{F}F\mathsf{F}F\mathsf{F}F\mathsf{F}F\mathsf{F}F\mathsf{F}
0.1 F\mathsf{F}F\mathsf{F}T\mathsf{T}F\mathsf{F}F\mathsf{F}T\mathsf{T}F\mathsf{F}T\mathsf{T}T\mathsf{T}
0.1 F\mathsf{F}F\mathsf{F}F\mathsf{F}F\mathsf{F}F\mathsf{F}T\mathsf{T}F\mathsf{F}F\mathsf{F}T\mathsf{T}
Pr(P∣Q) = Pr(P\mid Q)\ =\ Pr(P∧Q)Pr(P\wedge Q)/ / Pr(Q)Pr(Q)  = \ =\ 0.400.40/ / 0.600.60 = 0.667\ =\ 0.667
Pr(P∣¬Q) = Pr(P\mid \neg Q)\ =\ Pr(P∧¬Q)Pr(P\wedge \neg Q)/ / Pr(¬Q)Pr(\neg Q)  = \ =\ 0.200.20/ / 0.400.40 = 0.500\ =\ 0.500
Pr(R∣Q) = Pr(R\mid Q)\ =\ Pr(R∧Q)Pr(R\wedge Q)/ / Pr(Q)Pr(Q)  = \ =\ 0.350.35/ / 0.600.60 = 0.583\ =\ 0.583
Pr(R∣¬Q) = Pr(R\mid \neg Q)\ =\ Pr(R∧¬Q)Pr(R\wedge \neg Q)/ / Pr(¬Q)Pr(\neg Q)  = \ =\ 0.200.20/ / 0.400.40 = 0.500\ =\ 0.500

In the above stochastic truth table, we then have:

  • â„“(P,Q)=Pr(Q∣P)−Pr(Q∣¬P)Pr(Q∣P)+Pr(Q∣¬P)=0.667−0.50.667+0.5≈0.143\ell(P,Q)=\frac{Pr(Q\mid P)-Pr(Q\mid \neg P)}{Pr(Q\mid P)+Pr(Q\mid \neg P)}=\frac{0.667 - 0.5}{0.667 + 0.5}\approx 0.143; and

  • â„“(R,Q)=Pr(R∣P)−Pr(R∣¬P)Pr(R∣P)+Pr(R∣¬P)=0.583−0.50.583+0.5≈0.076\ell(R,Q)=\frac{Pr(R\mid P)-Pr(R\mid \neg P)}{Pr(R\mid P)+Pr(R\mid \neg P)}=\frac{0.583 - 0.5}{0.583 + 0.5}\approx 0.076.

Although d(X,Y)d(X, Y) and â„“(X,Y)\ell(X, Y) are often different numbers (as in the above example), they are related in the following way:

Observation

In any stochastic truth table, for all formulas XX and YY:

  1. XX is positively relevant to YY if and only if d(X,Y)>0d(X, Y) > 0 if and only if â„“(X,Y)>0\ell(X, Y)>0
  2. XX and YY are independent if and only if d(X,Y)=0d(X, Y) = 0 if and only if â„“(X,Y)=0\ell(X, Y) = 0
  3. XX is negatively relevant to YY if and only if d(X,Y)<0d(X, Y) < 0 if and only if â„“(X,Y)<0\ell(X, Y) < 0

There are many ways to compare and contrast the above two different ways to measure incremental confirmation. One reason to prefer ℓ\ell is illustrated by computing d(P,P∨Q)d(P, P\vee Q) and ℓ(P,P∨Q)\ell(P,P\vee Q) in the following stochastic truth table:

Example
PPQQ
0.2 T\mathsf{T}T\mathsf{T}
0.1 T\mathsf{T}F\mathsf{F}
0.3 F\mathsf{F}T\mathsf{T}
0.4 F\mathsf{F}F\mathsf{F}
The evidence is PP
The hypothesis is (P∨Q)(P\vee Q)
Pr((P∨Q)∣P) = Pr((P\vee Q)\mid P)\ = \ 1.0001.000
Pr((P∨Q)) = Pr((P\vee Q))\ = \ 0.6000.600
d(P,(P∨Q)) = d(P, (P\vee Q))\ = \ Pr((P∨Q)∣P)−Pr((P∨Q)) = Pr((P\vee Q)\mid P) - Pr((P\vee Q))\ = \ 0.4000.400
Pr(P∣(P∨Q)) = Pr(P\mid (P\vee Q))\ = \ 0.5000.500
Pr(P∣¬(P∨Q)) = Pr(P\mid \neg (P\vee Q))\ = \ 0.0000.000
ℓ(P,(P∨Q)) = \ell(P, (P\vee Q))\ = \ (Pr(P∣(P∨Q))−Pr(P∣¬(P∨Q)))/(Pr(P∣(P∨Q))+Pr(P∣¬(P∨Q))) = (Pr(P\mid (P\vee Q)) - Pr(P\mid \neg (P\vee Q)))/(Pr(P\mid (P\vee Q)) + Pr(P\mid \neg (P\vee Q)))\ = \ 1.0001.000

Intuitively, since P⊨P∨QP\models P\vee Q non-trivially (in the sense that the premise is not a contradiction and the conclusion is not a tautology), the premise PP should give the greatest confirmation to the conclusion P∨QP\vee Q. While ℓ(P,P∨Q)\ell(P, P\vee Q) is assigned the maximum value of 1, the measure d(P,P∨Q)d(P, P\vee Q) is only 0.4. In fact we have the following key observation:

Observation

In any stochastic truth table, For any formulas XX and YY, in any stochastic truth table:

d(X,Y)={Pr(¬Y) if P⊨Y,Pr(X)>0,0<Pr(Y)<1−Pr(Y) if P⊨¬Y,Pr(X)>0,0<Pr(Y)<1d(X, Y)=\begin{cases} Pr(\neg Y) & \text{ if } P\models Y, Pr(X)>0, 0< Pr(Y) < 1 \\-Pr(Y) & \text{ if } P\models \neg Y, Pr(X)>0, 0< Pr(Y) < 1\\\end{cases}
ℓ(X,Y)={1 if P⊨Y,Pr(X)>0,0<Pr(Y)<1−1 if P⊨¬Y,Pr(X)>0,0<Pr(Y)<1\ell(X, Y)=\begin{cases} 1 & \text{ if } P\models Y, Pr(X)>0, 0< Pr(Y) < 1 \\-1 & \text{ if } P\models \neg Y, Pr(X)>0, 0< Pr(Y) < 1\\\end{cases}

Practice Questions#

  1. Find d(E,H)d(E,H) and â„“(E,H)\ell(E,H) in the following stochastic truth table.

    EEHH
    0.25 T\mathsf{T}T\mathsf{T}
    0.25 T\mathsf{T}F\mathsf{F}
    0.25 F\mathsf{F}T\mathsf{T}
    0.25 F\mathsf{F}F\mathsf{F}
  2. Find d(E,H)d(E,H) and â„“(E,H)\ell(E,H) in the following stochastic truth table.

    EEHH
    0 T\mathsf{T}T\mathsf{T}
    0.5 T\mathsf{T}F\mathsf{F}
    0.5 F\mathsf{F}T\mathsf{T}
    0 F\mathsf{F}F\mathsf{F}
  3. Find d(P,Q)d(P,Q), d(P,R)d(P,R), â„“(P,Q)\ell(P,Q) and â„“(P,R)\ell(P,R) in the following stochastic truth table.

    PPQQRR
    0.1 T\mathsf{T}T\mathsf{T}T\mathsf{T}
    0.2 T\mathsf{T}T\mathsf{T}F\mathsf{F}
    0.1 T\mathsf{T}F\mathsf{F}T\mathsf{T}
    0.2 T\mathsf{T}F\mathsf{F}F\mathsf{F}
    0.2 F\mathsf{F}T\mathsf{T}T\mathsf{T}
    0.1 F\mathsf{F}T\mathsf{T}F\mathsf{F}
    0.05 F\mathsf{F}F\mathsf{F}T\mathsf{T}
    0.05 F\mathsf{F}F\mathsf{F}F\mathsf{F}
  4. Find d(P,Q)d(P,Q), d(P,R)d(P,R), â„“(P,Q)\ell(P,Q) and â„“(P,R)\ell(P,R) in the following stochastic truth table.

    PPQQRR
    0.2 T\mathsf{T}T\mathsf{T}T\mathsf{T}
    0 T\mathsf{T}T\mathsf{T}F\mathsf{F}
    0.1 T\mathsf{T}F\mathsf{F}T\mathsf{T}
    0.3 T\mathsf{T}F\mathsf{F}F\mathsf{F}
    0.2 F\mathsf{F}T\mathsf{T}T\mathsf{T}
    0.1 F\mathsf{F}T\mathsf{T}F\mathsf{F}
    0 F\mathsf{F}F\mathsf{F}T\mathsf{T}
    0.1 F\mathsf{F}F\mathsf{F}F\mathsf{F}
  5. Find d(P,(Q∨R))d(P,(Q\vee R)) and ℓ(P,(Q∨R))\ell(P,(Q\vee R)) in the following stochastic truth table.

    PPQQRR
    0.2 T\mathsf{T}T\mathsf{T}T\mathsf{T}
    0 T\mathsf{T}T\mathsf{T}F\mathsf{F}
    0.1 T\mathsf{T}F\mathsf{F}T\mathsf{T}
    0.3 T\mathsf{T}F\mathsf{F}F\mathsf{F}
    0.2 F\mathsf{F}T\mathsf{T}T\mathsf{T}
    0.1 F\mathsf{F}T\mathsf{T}F\mathsf{F}
    0 F\mathsf{F}F\mathsf{F}T\mathsf{T}
    0.1 F\mathsf{F}F\mathsf{F}F\mathsf{F}
  6. Find d((Q∨R),Q)d((Q\vee R), Q) and ℓ((Q∨R),Q)\ell((Q\vee R), Q) in the following stochastic truth table.

    QQRR
    0.3 T\mathsf{T}T\mathsf{T}
    0.1 T\mathsf{T}F\mathsf{F}
    0.3 F\mathsf{F}T\mathsf{T}
    0.3 F\mathsf{F}F\mathsf{F}
  7. Find d((Q∨R),P)d((Q\vee R), P) and ℓ((Q∨R),P)\ell((Q\vee R), P) in the following stochastic truth table.

    PPQQRR
    0.2 T\mathsf{T}T\mathsf{T}T\mathsf{T}
    0 T\mathsf{T}T\mathsf{T}F\mathsf{F}
    0.1 T\mathsf{T}F\mathsf{F}T\mathsf{T}
    0.3 T\mathsf{T}F\mathsf{F}F\mathsf{F}
    0.2 F\mathsf{F}T\mathsf{T}T\mathsf{T}
    0.1 F\mathsf{F}T\mathsf{T}F\mathsf{F}
    0 F\mathsf{F}F\mathsf{F}T\mathsf{T}
    0.1 F\mathsf{F}F\mathsf{F}F\mathsf{F}
  8. Find d(P→Q,Q)d(P \rightarrow Q, Q) and ℓ(P→Q,Q)\ell(P \rightarrow Q, Q) in the following stochastic truth table.

    PPQQ
    0.1 T\mathsf{T}T\mathsf{T}
    0.2 T\mathsf{T}F\mathsf{F}
    0.4 F\mathsf{F}T\mathsf{T}
    0.3 F\mathsf{F}F\mathsf{F}
  9. Find d(Q,P→Q)d(Q, P \rightarrow Q) and ℓ(Q,P→Q)\ell( Q, P \rightarrow Q) in the following stochastic truth table.

    PPQQ
    0.1 T\mathsf{T}T\mathsf{T}
    0.2 T\mathsf{T}F\mathsf{F}
    0.4 F\mathsf{F}T\mathsf{T}
    0.3 F\mathsf{F}F\mathsf{F}
  10. Find d(P,P→Q)d(P, P \rightarrow Q) and ℓ(P,P→Q)\ell( P, P \rightarrow Q) in the following stochastic truth table.

    PPQQ
    0.1 T\mathsf{T}T\mathsf{T}
    0.2 T\mathsf{T}F\mathsf{F}
    0.4 F\mathsf{F}T\mathsf{T}
    0.3 F\mathsf{F}F\mathsf{F}