#### Supplement to Epistemic Utility Arguments for Probabilism

## Proof of Theorem 5

We wish to prove the following theorem (Leitgeb and Pettigrew 2010a):

Theorem 5.Local and Global ComparabilityandAgreement on Urgencyentail

L(A,v,r) = λ|v(A) −r|²G(b,v) = λΣ|v(v) −_{i}b(v)|², where again_{i}v_{1}, …,v_{n}are the atoms ofF.

Suppose **Local and Global Comparability** and
**Agreement on Urgency**. By **Local and Global
Comparability**, there is a strictly increasing function
*f* : ℜ → ℜ such that *f*(0) = 0 and

L(A,v,r) =f(|v(A) −r|) andG(b,v) =f(||−b||)v

It follows that there is a strictly increasing function *g* :
ℜ → ℜ such that *g*(0) = 0 and

L(A,v,r) =g(|v(A) −r|²) andG(b,v) =g(||−b||²)v

To prove the theorem, it suffices to show that there is λ
> 0 such that *g*(*x*) = λ*x*, for all
*x* ≥ 0. For, if this is true, then

*L*(*A*,*v*,*r*) = λ|*v*(*A*) −*r*|²*G*(*b*,*v*) = λ||−*b*||².*v*

as required.

To prove this, it suffices to show that *g*′ is
constant and positive on the non-negative reals. We prove this in two
stages:

- First, we show that
*g*′(1 +*x*) =*g*′(1), for*x*≥ 0. - Second, we show that
*g*′(*x*) =*g*′(*x*+ 1), for*x*≥ 0.

From this, we infer that *g*′(*x*) =
*g*′(1), for all *x* ≥ 0. That is,
*g*′ is constant. Since *g* is strictly increasing,
we know that *g*′ must be a positive constant, as
required.

To prove that *g*′(1 + *x*) =
*g*′(1), for *x* ≥ 0, we take *a* ≥ 0
and let *b*′(*v*_{2}) = √*a*,
and *b*′(*v*_{3}) = … =
*b*′(*v*_{n}) = 0.

Then we have

LUrg _{L, W}(x,v_{1}|b)= d/dxLExp_{L, W}(x,v_{1}|b)= d/dxΣb(v)L(v_{1},v,x)= d/dx[b(v_{1})g((1 −x)²) +b(v_{2})g(x²) + … +b(v_{n})g(x²)]= −2(1 − x)b(v_{1})g′((1 −x)²) + 2xb(v_{2})g′(x²) + … + 2xb(v_{n})g′(x²)

Thus, LUrg_{L, W}(0,
*v*_{1} |
*b*) =
−2*b*(*v*_{1})*g*′(1).

And we also have

GUrg _{G, W}(b′,x,v_{1}|b)= d/dxΣb(v)G(v,b′(x/v_{1}))= d/dxΣb(v)g(||−b′(x/v_{1})ˆ||²)v= d/dx[b(v_{1})g((1 −x)² +a) +b(v_{2})g(x² + (1 − √a)²) +b(v_{3})g(x² +a+ 1) + … +b(v_{n})g(x² +a+ 1)]= −2(1− x)b(v_{1})g′((1 −x)² +a) + 2xb(v_{2})g′(x² + (1 − √a)²) + 2xb(v_{3})g(x² +a+ 1) + … + 2xb(v_{n})g′(x² +a+ 1)

Thus, GUrg_{G, W}(*b*′, 0,
*v*_{1} |
*b*) =
−2*b*(*v*_{1})*g*′(1 + *a*).
Thus, by **Agreement on urgency**, we have
*g*′(1 + *a*) = *g*′(1), for all
*a* ≥ 0, as required.

To prove that *g*′(*x*) =
*g*′(*x* + 1), for *x* ≥ 0, we take
*a* ≥ 0, and let *b*(*v*_{1}) =
*b*(*v*_{3}) = … =
*b*(*v*_{n}) = 0 and
*b*(*v*_{2}) = 1 and
*b*′(*v*_{2}) =
*b*′(*v*_{3}) = … =
*b*′(*v*_{n}) = 0.

Then we have

LUrg _{L, W}(x,v_{1}|b)= d/dxg((v_{2}(v_{1}) −x)²)= d/dxg(x²)= 2 xg′(x²)

and

GUrg _{G, W}(b′,x,v_{1}|b)= d/dxg(||v_{2}ˆ −b′(x/v_{1})ˆ||²)= d/dxg(x² + 1)= 2 xg′(x² + 1).

Thus, by **Agreement on urgency**, we have
*g*′(*x*²) =
*g*′(*x*² + 1), as required.

This completes our proof.