I agree with your calculus. In case EY was only talking about the binary case or made a mistake, I checked Wikipedia, which agrees* with your definition of the quadratic scoring rule and claims that it is strictly proper. It also suggests the spherical scoring rule, where the return for an outcome that you weighted p is p/sqrt(sum of p_i^2), which does seem to work analytically.
However, I don’t get the same numbers as you. I get the expected return of .5,.2,.3 as 0.6 and the return of .51,.19,.3 as 0.60173. (but the sign is the same, so it makes little difference)
* Edit: no, actually it does not agree. That’s the problem.
Thanks. I realize now I calculated those numbers I cited while leaving out the payoff from the .3 option since it wasn’t changing, then forgot to add it back in. Strange what Wikipedia says when there was this counterexample. If I have some time later I’ll check through the sources linked in the article.
I agree with your calculus. In case EY was only talking about the binary case or made a mistake, I checked Wikipedia, which agrees* with your definition of the quadratic scoring rule and claims that it is strictly proper. It also suggests the spherical scoring rule, where the return for an outcome that you weighted p is p/sqrt(sum of p_i^2), which does seem to work analytically.
However, I don’t get the same numbers as you. I get the expected return of .5,.2,.3 as 0.6 and the return of .51,.19,.3 as 0.60173. (but the sign is the same, so it makes little difference)
* Edit: no, actually it does not agree. That’s the problem.
Thanks. I realize now I calculated those numbers I cited while leaving out the payoff from the .3 option since it wasn’t changing, then forgot to add it back in. Strange what Wikipedia says when there was this counterexample. If I have some time later I’ll check through the sources linked in the article.