#### Supplement to Epistemic Utility Arguments for Probabilism

## Proof of Theorems 8 and 9

We wish to prove the following theorems (Leitgeb and Pettigrew 2010b):

Theorem 8.SupposeL(A,v,r) = λ|v(A) −r|². Supposeb∈BandE∈F. Then the following are equivalent:

- For all
AinFandrin ℜ,LExp_{L, E}(b(A),A|b) ≤ LExp_{L, E}(r,A|b)bis a probability function onFandb(E) = 1.

That is, only probabilistic credence functions that assign credence
1 to *E* expect themselves to be at least as good as they expect
any other credence function to be in the light of evidence
*E*.

Theorem 9.SupposeL(A,v,r) = λ|v(A) −r|². Supposeb,b′ inBare probability functions,EinF, andb(E) > 0. Then the following are equivalent:

- For all
AinFandrin ℜ,LExp_{L, E}(b′(A),A|b) ≤ LExp_{L, E}(r,A|b)b′(•) =b( • |E).

That is, an initial probabilistic credence function *b*
expects conditionalization to give the best credence function in the
presence of evidence *E*.

We prove these two theorems together because they both depend on the
following lemma. We use Σ_{A} to denote the sum
over those *v* in **V** that make proposition
*A* true. And we use Σ to denote the sum over all
*v* in **V**.

Lemma 1.SupposeL(A,v,r) = λ|v(A) −r|²,b∈Bandb(v) > 0 for somevsuch thatv(E) = 1. Then the following two propositions are equivalent:

d/dxLExp_{L, E}(x,A|b) = 0x= [Σ_{A&E}b(v)] / [Σ_{E}b(v)]

The proof is straightforward calculus:

d/dxLExp_{L, E}(x,Ab)= d/dx[Σ_{A&E}b(v)(1 −x)² + Σ_{¬A&E}b(v)x²]= 2 xΣ_{E}b(v) − 2Σ_{A&E}b(v)= 0

if, and only if,

as required.x= [Σ_{A&E}b(v)] / Σ_{E}b(v)

From this lemma, Theorems 8 and 9 follow.

By Lemma 1, Theorem 8(1) holds if, and only if,

b(A) = [Σ_{A&E}b(v)] / [Σ_{E}b(v)]

which can easily be verified to hold if, and only if, *b* is
a probability function on **F** with
*b*(*E*) = 1. That is, if, and only if, Theorem 8(2)
holds.

Similarly, by Lemma 1, Theorem 9(1) holds if, and only if,

b’(A) = [Σ_{A&E}b(v)] / [Σ_{E}b(v)]

for all *A* in **F**, which holds if, and only
if, *b*’(*A*) = *b*(*A* |
*E*) for
all *A* in **F**. That is, if, and only if, Theorem
9(2) holds.