In Luce (1959) the choice axiom is definied, that for a finite subset $T$ of $U$ such that, for every $S\subset T$, $P_S$ is defined.

  • If $P(x,y)\ne 0,1$ for all $x,y\in T$, then for $R\subset S\subset T$: $P_T(R) = P_S(R)P_T(S)$
  • If $P(x,y)= 0$ for all $x,y\in T$, then for $S\subset T$: $P_T(S) = P_{T-\{x\}}(S - \{x\})$

Now it is written that (Lemma 2, page 7)

If $P(x,y)\ne 0,1$ for all $x,y\in T$, then this axiom is equivalent to $P_S(R) = P_T(R|S)$, for $R\subset S\subset T$.

It is written that this result is obvious except for the condition $P_T(S)>0$.

My (probably naive) question is: why is it obvious?

The conditional measure indcued by $P_T$ is only defined if $P_T(S)>0$ by: $P_T(R|S) = \frac{P_T(R\cap S)}{P_T(S)}$. Again, why is it enough to proof, that $P_T(S)>0$?


1 Answer 1


Supposing that $P(x,y)\ne 0,1$ for all $x,y\in T$, we want to know that the following are equivalent for any $R\subset S \subset T$:

(A) $P_T(R)=P_S(R)P_T(S)$.

(B) $P_S(R)=P_T(R\mid S)$.

The obvious part that you ask about is no more than the second equality below: since $R\subset S$, we have that (A) is equivalent to $$ P_S(R)=\frac{P_T(R)}{P_T(S)}=\frac{P_T(R\cap S)}{P_T(S)}.$$ Thus if $P_T(S)>0$ as well (which is the non-obvious part that Luce demonstrates in Lemma 2 under the stated condition that $P(x,y)\ne 0,1$ for all $x,y\in T$), then the right-hand side above is indeed $P_T(R\mid S)$, and we have the desired equivalence with (B).

  • $\begingroup$ So bottom line, if I dont have that $P_T(S)>0$, the conditional probability $P_T(R|S)$ is not defined? So its not really a proof its just getting why we need such an assumption right? $\endgroup$
    – Druss2k
    Commented Jan 23, 2014 at 12:52

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