Lewis's triviality result

In the mathematical theory of probability, David Lewis's triviality result is a theorem about the impossibility of systematically equating the conditional probability ${\displaystyle P(B\mid A)}$ with the probability of a so-called conditional event, ${\displaystyle A\rightarrow B}$.

The statement "The probability that if ${\displaystyle A}$, then ${\displaystyle B}$, is 20%" means (put intuitively) that event ${\displaystyle B}$ may be expected to occur in 20% of the outcomes where event ${\displaystyle A}$ occurs. The standard formal expression of this is ${\displaystyle P(B\mid A)=0.20}$, where the conditional probability ${\displaystyle P(B\mid A)}$ equals, by definition, ${\displaystyle P(A\cap B)/P(A)}$.

Beginning in the 1960s, several philosophical logicians—most notably Ernest Adams and Robert Stalnaker—floated the idea that one might also write ${\displaystyle P(A\rightarrow B)=0.20}$, where ${\displaystyle A\rightarrow B}$ is the conditional event "If ${\displaystyle A}$, then ${\displaystyle B}$".^[1] That is, given events ${\displaystyle A}$ and ${\displaystyle B}$, one might suppose there is an event, ${\displaystyle A\rightarrow B}$, such that ${\displaystyle P(A\rightarrow B)}$ could be counted on to equal ${\displaystyle P(B\mid A)}$, so long as ${\displaystyle P(A)>0}$.

Part of the appeal of this move would be the possibility of embedding conditional expressions within more complex constructions. One could write, say, ${\displaystyle P(A\cup (B\rightarrow C))=0.75}$, to express someone's high subjective degree of confidence ("75% sure") that either ${\displaystyle A}$, or else if ${\displaystyle B}$, then ${\displaystyle C}$. Compound constructions containing conditional expressions might also be useful in the programming of automated decision-making systems.^[2]

How might such a convention be combined with standard probability theory? The most direct extension of the standard theory would be to treat ${\displaystyle A\rightarrow B}$ as an event like any other, i.e., as a set of outcomes. Adding ${\displaystyle A\rightarrow B}$ to the familiar Venn- or Euler diagram of ${\displaystyle A}$ and ${\displaystyle B}$ would then result in something like Fig. 1, where ${\displaystyle s,t,\ldots ,z}$ are probabilities allocated to the eight respective regions, such that ${\displaystyle s+t+\cdots +z=1}$.

が に等しいためには、 、つまり、領域内の確率が、領域内の確率の領域内の割合に等しいことが必要です。一般に、この等式は当然真ではないため、これを確実に真にするには、確率関数に対する新しい制約が必要です。つまり、コルモゴロフの確率公理 を満たすことに加えて、 となる任意のイベントおよびに対して という新しい制約も満たす必要があります。 ${\displaystyle P(A\rightarrow B)}$${\displaystyle P(B\mid A)}$${\displaystyle t+v+w+y=(s+t)/(s+t+x+y)}$${\displaystyle A\rightarrow B}$${\displaystyle A\cap B}$${\displaystyle A}$${\displaystyle P(A\rightarrow B)=P(B\mid A)}$${\displaystyle A}$${\displaystyle B}$${\displaystyle P(A)>0}$

Lewis (1976) は、上記の提案に関して一見致命的な問題を指摘しました。それは、一連のイベントが自明でないと仮定すると、新しい制限されたクラスの- 関数は、イベントの発生を前提として確率関数を新しい関数 に変換する操作である条件付けの下で閉じないということです。つまり、 の場合、 である限り、 は一般には真ではありません。これは、合理性が適切に動作する確率関数を必要とする場合、完全に合理的な人 (またはコンピューティング システム) は、任意のイベントが発生したことを知っただけで非合理的になることを意味します。Bas van Fraassen はこの結果を「まさに爆弾」と呼びました (1976、p. 273)。 ${\displaystyle P}$${\displaystyle P}$${\displaystyle P_{C}(\cdot )=P(\cdot \mid C)}$${\displaystyle C}$${\displaystyle P(A\rightarrow B)=P(B\mid A)}$${\displaystyle P_{C}(A\rightarrow B)=P_{C}(B\mid A)}$${\displaystyle P(C)>0}$${\displaystyle C}$

ルイスの証明は以下のとおりです。一連のイベントが非自明であるとは、2 つの可能なイベントとが相互に排他的であるが、すべての可能性を網羅するものではない、つまり、、、 であるときです。このような 2 つのイベントが存在するということは、イベント も存在することを意味します。また、条件付きイベントが認められる場合は、イベント も存在します。この証明は、このような最小限に非自明な一連のイベントが存在するという仮定から矛盾を導き出します。 ${\displaystyle A}$${\displaystyle B}$${\displaystyle P(A)>0}$${\displaystyle P(B)>0}$${\displaystyle P(A\cap B)=0}$${\displaystyle P(A\cup B)<1}$${\displaystyle A\cup B}$${\displaystyle (A\cup B)\rightarrow A}$

1. 条件付けの後、まず 上での確率を​​考え、次にその補集合 上での確率を​​考えます。 ${\displaystyle (A\cup B)\rightarrow A}$${\displaystyle A}$${\displaystyle A'}$
   • を条件とすると となります。しかし、 -関数に対する新しい制約により となります。したがって、 となります。${\displaystyle A}$${\displaystyle P_{A}((A\cup B)\rightarrow A)=P(((A\cup B)\rightarrow A)\cap A)/P(A)}$${\displaystyle P}$${\displaystyle P_{A}((A\cup B)\rightarrow A)=P_{A}((A\cup B)\cap A)/P_{A}(A\cup B)=}$ ${\displaystyle P((A\cup B)\cap A\mid A)/P(A\cup B\mid A)=1/1=1}$${\displaystyle P(((A\cup B)\rightarrow A)\cap A)=P(A)}$
   • を条件付けると が得られます。しかし、 も得られます。(との相互排他性によりが保証されます。) したがって、 です。${\displaystyle A'}$${\displaystyle P_{A'}((A\cup B)\rightarrow A)=P(((A\cup B)\rightarrow A)\cap A')/P(A')}$${\displaystyle P_{A'}((A\cup B)\rightarrow A)=P_{A'}((A\cup B)\cap A)/P_{A'}(A\cup B)=}$ ${\displaystyle P((A\cup B)\cap A\mid A')/P(A\cup B\mid A')=0/P(A\cup B\mid A')=0}$${\displaystyle B}$${\displaystyle A}$${\displaystyle P(A\cup B\mid A')\neq 0}$${\displaystyle P(((A\cup B)\rightarrow A)\cap A')=0}$
2. 恒等式を として具体化します。ステップ 1 の結果により、左辺は に簡約されますが、右辺は、-関数に関する新しい制約により に等しくなります。したがって、となり、 となり、 という規定に矛盾します。これで証明は完了です。${\displaystyle P(X\cap Y)+P(X\cap Y')=P(X)}$${\displaystyle P(((A\cup B)\rightarrow A)\cap A)+P(((A\cup B)\rightarrow A)\cap A')=}$ ${\displaystyle P((A\cup B)\rightarrow A)}$${\displaystyle P(A)}$${\displaystyle P}$${\displaystyle P((A\cup B)\cap A)/P(A\cup B)=P(A)/P(A\cup B)}$${\displaystyle P(A)=P(A)/P(A\cup B)}$${\displaystyle P(A\cup B)=1}$${\displaystyle P(A\cup B)<1}$

A graphical version of the proof starts with Fig. 2, where the ${\displaystyle A}$ and ${\displaystyle B}$ from Fig. 1 are now disjoint and ${\displaystyle A\rightarrow B}$ has been replaced by ${\displaystyle (A\cup B)\rightarrow A}$.^[3] By the assumption that ${\displaystyle A}$ and ${\displaystyle B}$ are possible, ${\displaystyle x+y>0}$ and ${\displaystyle u+v>0}$. By the assumption that together ${\displaystyle A}$ and ${\displaystyle B}$ do not together exhaust all possibilities, ${\displaystyle u+v+x+y<1}$. And by the new constraint on probability functions, ${\displaystyle P((A\cup B)\rightarrow A)=P(A\mid A\cup B)=}$ ${\displaystyle P(A\cap (A\cup B))/P(A\cup B)=P(A)/P(A\cup B)}$, which means that

(1) ${\displaystyle y+v+w={\frac {x+y}{x+y+u+v}},}$

Conditioning on an event involves zeroing out the probabilities outside the event's region and increasing the probabilities inside the region by a common scale factor. Here, conditioning on ${\displaystyle A}$ will zero out ${\displaystyle u,v}$ and ${\displaystyle w}$ and scale up ${\displaystyle x}$ and ${\displaystyle y}$, to ${\displaystyle x_{A}}$ and ${\displaystyle y_{A}}$, respectively, and so

(2) ${\displaystyle y_{A}+0+0={\frac {x_{A}+y_{A}}{x_{A}+y_{A}+0+0}},}$ which simplifies to ${\displaystyle y_{A}=1.}$

Conditioning instead on ${\displaystyle A'}$ will zero out ${\displaystyle x}$ and ${\displaystyle y}$ and scale up ${\displaystyle u,v}$ and ${\displaystyle w}$, and so

(3) ${\displaystyle 0+v_{A'}+w_{A'}={\frac {0+0}{0+0+u_{A}+v_{A}}},}$ which simplifies to ${\displaystyle v_{A'}+w_{A'}=0.}$

From (2), it follows that ${\displaystyle x_{A}=0}$, and since ${\displaystyle x_{A}}$ is the scaled-up value of ${\displaystyle x}$, it must also be that ${\displaystyle x=0}$. Similarly, from (3), ${\displaystyle v=w=0}$. But then (1) reduces to ${\displaystyle y=y/(y+u)}$, which implies that ${\displaystyle y+u=1}$, which contradicts the stipulation that ${\displaystyle u+v+x+y<1}$.

In a follow-up article, Lewis (1986) noted that the triviality proof can proceed by conditioning not on ${\displaystyle A}$ and ${\displaystyle A'}$ but instead, by turns, on each of a finite set of mutually exclusive and jointly exhaustive events ${\displaystyle A,C,D,E,\ldots \,.}$ He also gave a variant of the proof that involved not total conditioning, in which the probability of either ${\displaystyle A}$ or ${\displaystyle A'}$ is set to 1, but partial conditioning (i.e., Jeffrey conditioning), by which probability is incrementally shifted from ${\displaystyle A'}$ to ${\displaystyle A}$.

Separately, Hájek (1989) pointed out that even without conditioning, if the number of outcomes is large but finite, then in general ${\displaystyle P(B\mid A)=P(A\cap B)/P(A)}$, being a ratio of two outputs of the ${\displaystyle P}$-function, will take on more values than any single output of the function can. So, for instance, if in Fig. 1 ${\displaystyle s,t,\ldots }$ are all multiples of 0.01 (as would be the case if there were exactly 100 equiprobable outcomes), then ${\displaystyle P(A\rightarrow B)}$ must be a multiple of 0.01, as well, but ${\displaystyle P(A\cap B)/P(A)}$ need not be. That being the case, ${\displaystyle P(A\rightarrow B)}$ cannot reliably be made to equal ${\displaystyle P(B\mid A)}$.

Hájek (1994) also argued that the condition ${\displaystyle P(A\rightarrow B)=P(B\mid A)}$ caused acceptable ${\displaystyle P}$-functions to be implausibly sparse and isolated from one another. One way to put the point: standardly, any weighted average of two probability function is itself a probability function, so that between any two ${\displaystyle P}$-functions there will be a continuum of weighted-average ${\displaystyle P}$-functions along which one of the original ${\displaystyle P}$-functions gradually transforms into the other. But these continua disappear if the added ${\displaystyle P(A\rightarrow B)=P(B\mid A)}$ condition is imposed. Now an average of two acceptable ${\displaystyle P}$-functions will in general not be an acceptable ${\displaystyle P}$-function.

Assuming that ${\displaystyle P(A\rightarrow B)=P(B\mid A)}$ holds for a minimally nontrivial set of events and for any ${\displaystyle P}$-function leads to a contradiction. Thus ${\displaystyle P(A\rightarrow B)=P(B\mid A)}$ can hold for any ${\displaystyle P}$-function only for trivial sets of events—that is the triviality result. However, the proof relies on background assumptions that may be challenged. It may be proposed, for instance, that the referent event of an expression like “${\displaystyle A\rightarrow B}$” is not fixed for a given ${\displaystyle A}$ and ${\displaystyle B}$, but instead changes as the probability function changes. Or it may be proposed that conditioning on ${\displaystyle C}$ should follow a rule other than ${\displaystyle P_{C}(\cdot )=P(\cdot \mid C)}$.

But the most common response, among proponents of the ${\displaystyle P(A\rightarrow B)=P(B\mid A)}$ condition, has been to explore ways to model conditional events as something other than subsets of a universe set of outcomes. Even before Lewis published his result, Schay (1968) had modeled conditional events as ordered pairs of sets of outcomes. With that approach and others in the same spirit, conditional events and their associated combination and complementation operations do not constitute the usual algebra of sets of standard probability theory, but rather a more exotic type of structure, known as a conditional event algebra.