5
$\begingroup$

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 $P_T(S)>0$?

$\endgroup$

1 Answer 1

1
$\begingroup$

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).

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.