Choice-value scale non-identifiability
= Choice-value scale non-identifiability
If $P(a)=\exp(\beta Q(a))/\sum_b\exp(\beta Q(b))$, then for every $c>0$, replacing $Q$ by $cQ$ and $\beta$ by $\beta/c$ leaves each exponent unchanged. The same choices therefore cannot identify both an unrestricted value scale and inverse temperature. Fixing a value scale or adding independent measurements is needed to break this equivalence.