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The vNM theorem suggests that weak-ordering, continuity, and independence is equivalent to the existence of expected utility, unique up to an affine transformation.

In Savage's axioms of expected utility, the P5 is the non-nullity assumption: there exists two outcomes $x,y$ such that $x\succ y$. Also, in Debreu's theorem of cardinal utility, there is an essentiality assumption serving a similar role.

My question: does the vNM axioms imply P5?

If not, let's consider a preference $p\succsim q$ for all $p,q$, this preference clearly does not have a cardinal utility representation.

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A preference such that $p\succeq q$ for all lotteries $p$ and $q$ clearly does (!) have a cardinal utility representation: The utility function over outcomes must be constant, and this is clearly also necessary.

In Savage's theory, the axiom plays a different role. If a decision-maker is indifferent between all outcomes, a subjective expected utility representation still exists. For example, a constant function on outcomes together with an arbitrary probability distribution over states. In that case, there is clearly no unique probability distribution (with at least two states), and even the Bernoulli function need not be unique up to positive affine transformations.

For Debreu's theorem, a similar assumption is needed to ensure that there are really three factors in a substantive way. His result does not work with only two factors, and a third factor that does not matter is like having only two factors.

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  • $\begingroup$ First paragraph: your constant utility function seems not unique up to a positive affine transformation. For example, both $U=-1$ and $U=1$ works. Usually, a "cardinal utility" function is something unique up to positive linear or positive affine transformation. If it is only unique up to a positive monotonic transformation, then it is called an "ordinal utility". $\endgroup$
    – dodo
    Commented Jun 26 at 10:29
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    $\begingroup$ The two functions you gave are related by a positive affine transformation: $1=-1+2$ $\endgroup$ Commented Jun 26 at 11:49

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