NOTE: if Pr(Q|P) = Pr(Q), then Pr(P|Q) = Pr(P); that is, if Q is independent of P, then P is independent of Q.

Proof:

Pr(P&Q) = Pr(P) x Pr(Q|P). Hence, Pr(Q|P) = [Pr(P&Q)/Pr(P)]. Suppose now that Q is independent of P, that is, Pr(Q|P) = Pr(Q). Then, [Pr(P&Q)/Pr(P)] = Pr(Q). Multiplying both sides by Pr(P)/Pr(Q), one gets [Pr(P&Q)/Pr(Q)] = Pr(P). But [Pr(P&Q)/Pr(Q)] = Pr(P|Q). Hence, Pr(P|Q) = Pr(P).

B.
**Proof of Bayes' Theorem**

We know that Pr(P&Q) = Pr(P) x Pr(Q|P), and Pr(Q&P) = Pr(Q)
x Pr(P|Q). But (P&Q) and (Q&P) are logically equivalent,
an consequently their probability is the same. So,

Pr(P) x Pr(Q|P) = Pr(Q) x Pr(P|Q). Dividing both sides by Pr(P),
one gets Pr(Q|P) = Pr(P|Q) x [Pr(Q)/Pr(P)], which is Bayes'
Theorem.

C.
**A more extended version of Bayes' Theorem**

Since P is logically equivalent to [(P&Q) v (P&-Q)] (check
it out!), Pr(P) = Pr{[(P&Q) v (P&-Q)]}. But since (P&Q)
and (P&-Q) are mutually exclusive, Pr{[(P&Q) v (P&-Q)]} = Pr(P&Q)
+ Pr (P&-Q).

So, Pr(P) = Pr(P&Q) + Pr (P&-Q). But Pr(P&Q)
= Pr(Q) x Pr(P|Q), and Pr (P&-Q) = Pr(-Q) x Pr(P|-Q). So,
Pr(P) = Pr(Q) x Pr(P|Q) + Pr(-Q) x Pr(P|-Q). Hence, by
substituting the last equality in Bayes' Theorem, one gets:

Pr(P|Q) x Pr(Q)

Pr(Q|P) = -------------------------------------------

Pr(Q) x Pr(Q|P) + Pr(-Q) x Pr(P|-Q)

D.
**Pr(A&B) is less or equal to Pr(A)**

Proof:

The proof is a __reductio__. Suppose that Pr(A&B)
> Pr(A). Then, Pr(A&B) > Pr[(A&B) v (A&-B)]. But
since (A&B) and (A&-B) are mutually exclusive,

Pr(A&B) > Pr(A&B) + Pr(A&-B). Simplifying, one gets
0 > Pr(A&-B), which is imposible because no probability is less
than 0.

E.
**Pr(AvB) is greater or equal to Pr(A)**

Proof:

The proof is a __reductio__. Suppose that Pr(A) > Pr(AvB).
Then, Pr(A) > Pr(A) + Pr(B) - Pr(A&B). Hence, Pr(A&B) > Pr(B),
which is impossible.

F.
**The Logical Consequence Principle**: If A entails B (i.e., if
"A; hence B" is valid), then Pr(B) is not less than Pr(A)

Proof:

Suupose that A entails B. Then -A v B is a tautology. (WHY?)
Hence, Pr(-A v B) = 1. So, Pr(-A) + Pr(B) - Pr(-A&B) = 1.
Consequently, 1-Pr(A) + PrB) - Pr(-A&B) = 1.

So, Pr(B) - Pr(A) = Pr(-A&B). But Pr(-A&B) is not less
than 0. Hence, Pr(B) is not less than Pr(A).

G.

A bet on P is fair **just in case** W/L = [1-Pr(P)]/Pr(P).

Proof:

A bet on P is fair just in case the betting quotient equal Pr(P), that
is Pr(P) = L/(L+W); that is, L Pr(P) + W Pr(P) = L; that is, W Pr(P) =
L - L Pr(P); that is, WPr(P) = L(1-Pr(P)); so,

W/L = [1-Pr(P)]/Pr(P) .