Mark Saroufim - Probability without Measure - It's only a game

Probability without Measure!
Mark Saroufim
University of California San Diego
msaroufi@cs.ucsd.edu
February 18, 2014
Mark Saroufim (UCSD) It’s only a Game! February 18, 2014 1 / 25

Overview
1 History of Probability Theory
Before Kolmogorov
During Kolmogorov
After Kolmogorov
2 Shafer and Vovk
It’s only a game
Winning conditions
Comparison with measure theory
An analogue to variance
3 Efficient Market Hypothesis
Securities Market Protocol

A gambler’s perspective
This was the day before probability theory was even a field in
mathematics, a field without foundations. Pascal and Fermat simply
wanted to win a ton of money betting on horses and wanted to first
see what it meant for a game to be fair.

If I pay a$
You pay a$
The winner gets paid 2a$

If I pay a$
You pay a$
The winner gets paid 2a$
This is what is referred to as inter alia (equal terms)
P[E] = how much money you’re willing to put on a game where you
could win 1$

Looking at the real world
Bernoulli was the first to suggest that probability can be measured from
observation
P{|y/N − p| < } 1 − δ
Now it seems that there could be a more mathematical treatment of
probability..

Kolmogorov’s axioms
The axioms and definitions below relate a set Ω called the sample
space and the set of subsets of Ω, F. Every element in E ∈ F is
called an event
1 If E, F ∈ F then E ∪ F, E ∩ F, E F ∈ F. Or more concisely we say
that F is a field of sets.
2 Ω ⊂ F which with the first axiom means that F is an algebra of sets
3 Every set E ∈ F is assigned a probability which is a non-negative real
value using the function P : E → [0, 1]
4 P[Ω] = 1
5 If E ∩ F = Φ then P[E ∩ F] = P[E] + P[F], more generally we get
what is called the union bound when E and F are not disjoint then
P[E ∩ F] ≤ P[E] + P[F]
6 If ∩∞
n=1En = Φ where En ⊆ En−1 · · · ⊆ E1 we have that
limn→∞ P[En] = 0. This axiom with axiom 2 allows us to call F a
σ-algebra
A random variable x is then understood as a mapping from the size of
elements of F with respect to the probability measure P

Some Set Theory
Kolmogorov’s 6th axiom needs the axiom of choice

Some Set Theory
Definition (ZFC Axiom of Choice)
It is possible to pick one cookie from each jar even if there is an infinite
numbers of jars

Some Set Theory
numbers of jars
Weird things happen: Can divide a ball into two balls of equal volume

Some Set Theory
numbers of jars
Definition (Axiom of Determinacy)
Every 2 player game with perfect information where two players pick
natural numbers at every turn is already determined.

Some Set Theory
numbers of jars
Definition (Axiom of Determinacy)
Every 2 player game with perfect information where two players pick
natural numbers at every turn is already determined.
Weird things also happen: We get that there is no such thing as
non-measurable sets

Sequential Learning
Von Mises was the first to propose that probability could find its
foundations in games
Given a bit string 001111
Predict the odds of a 1 (number shouldn’t change much if we look at
a subsequence called a collective)

Sequential Learning
Von Mises was the first to propose that probability could find its
foundations in games
Given a bit string 001111
Predict the odds of a 1 (number shouldn’t change much if we look at
a subsequence called a collective)
Fortunately we have a method of quantifying how difficult it is to
predict the next bit in a string: Kolmogorov complexity!

Martingales
Originally Martingales are a gambling strategy that can guarantee a win of
1$ given an infinite supply of money
α + 2α + · · · + 2i
α
stop as soon as you win once

Martingales
α + 2α + · · · + 2i
α
Definition (Martingale)
Given a sequence of outcomes x1, . . . , xn we call a capital process L if
E[L(x1, . . . , xn) | x1, . . . , xn−1] = L(x1, . . . , xn)

Martingales
α + 2α + · · · + 2i
α
Definition (Martingale)
Given a sequence of outcomes x1, . . . , xn we call a capital process L if
E[L(x1, . . . , xn) | x1, . . . , xn−1] = L(x1, . . . , xn)
L(E) → ∞ if E has probability 0 (more on this next slide)
Now we define the probability of an event E as
P(E) = inf{L0 | lim
n→∞
Ln ≥ I}

Martingales
Theorem (Doob’s inequality)
P[sup
n
L(x1, . . . , xn) ≥ λ] ≤
1
λ
Look familiar?

Martingales
P[sup
n
L(x1, . . . , xn) ≥ λ] ≤
1
λ
Look familiar?
Markov’s inequality!
P[x ≥ λ] ≤
Ex
λ

Martingales
P[sup
n
L(x1, . . . , xn) ≥ λ] ≤
1
λ
Look familiar?
Markov’s inequality!
P[x ≥ λ] ≤
Ex
λ
Other Chernoeff bounds can be derived in this way as well.

Bounded Fair Coin Game
We’re now ready to setup the first game between what we the skeptic and
nature

nature
Ki is the skeptic’s capital at time i

nature
Mn is the amount of tickets that the skeptic purchases

nature
Mn is the amount of tickets that the skeptic purchases
xn is the value of a ticket (determined by nature)

Theorem
There exists a winning strategy for skeptic but let’s formally define what
we mean by winning

Winning conditions
We claim that the skeptic wins if Kn 0∀n and if one two things happen,
either
lim
n→∞
1
n
n
X
i=1
xi = 0

Winning conditions
either
lim
n→∞
1
n
n
X
i=1
xi = 0
or
lim
n→∞
Kn = ∞

Winning conditions
either
lim
n→∞
1
n
n
X
i=1
xi = 0
or
lim
n→∞
Kn = ∞
The first condition has two different interpretations in the finance
litterature is called a self financing strategy because it means that the
skeptic can remain in the game forever without ever having to borrow
money. Also it means that skeptic wins if nature is forced to play
randomly.

Winning conditions
either
lim
n→∞
1
n
n
X
i=1
xi = 0
or
lim
n→∞
Kn = ∞
The first condition has two different interpretations in the finance
litterature is called a self financing strategy because it means that the
skeptic can remain in the game forever without ever having to borrow
money. Also it means that skeptic wins if nature is forced to play
randomly.
The second makes sense, you win if you become infinitely rich but
since this is unlikely this condition embodies the infinitary hypothesis
which says that there is no strategy that avoids bankruptcy that
guarantees that skeptic become infinitely rich.

Back to the Fair Coin Game
Theorem
There exists a winning strategy for skeptic

Back to the Fair Coin Game
Theorem
Law of Large Numbers.
Skeptic bets on heads, this forces nature not to play heads often or else
skeptic will become infinitely rich. So nature will start playing tails, when
that happens skeptic puts an on tails.

What if xn ∈ [−1, 1] instead of {−1, 1}?
Theorem

Proof of Bounded Fair Coin Game
We will need some terminology to tackle this problem we define a real
valued function on Ω called P which is a strategy that takes situations
s = x1, x2, . . . , xn and decides the number of tickets to buy P(s).

KP
(x1x2 . . . xn) = KP
(x1x2 . . . xn−1) + P(x1x2 . . . xn)xn

KP
(x1x2 . . . xn) = KP
(x1x2 . . . xn−1) + P(x1x2 . . . xn)xn
Definition
Skeptic forces an event E if KP(s) = ∞∀s ∈ Ec

Lemma
The skeptic can force
lim sup
n→∞
1
n
n
X
i=1
xi ≤
and
lim sup
n→∞
1
n
n
X
i=1
xi ≥

Proof.
take 1 as the starting capital
1 + KP
(x1x2 . . . xn) = (1 + KP
(x1 . . . xn−1))(1 + xn) =
n
Y
i=1
(1 + xi ) C
where C is a constant so take the log on both sides
n
X
i=1
ln(1 + xi ) ≤ D
Now use ln(1 + t) ≥ t − t2 when t ≥ −1/2
1
n
n
X
i=1
xi ≤
D
n
+
and we get the top part of the lemma. Replace by − to get the second

Bounded Forecast games
Somebody has got to be setting the prices, let a forecaster announce price
of ticket at iteration n as mn
Theorem
There exists a winning strategy for skeptic by reduction to the bounded
fair coin game.

Bounded Forecast games
Somebody has got to be setting the prices, let a forecaster announce price
of ticket at iteration n as mn
Theorem
There exists a winning strategy for skeptic by reduction to the bounded
fair coin game.
Proof.
First divide all prices by C to normalize prices to [−1, 1] then set mn = 0
and we recover the previous game. Note we also need to change the first
condition to limn→∞
1
n
Pn
i=1(xi − mi ) = 0

Measure theoretic law of large numbers
Assuming Xi are i.i.d random variables with mean µ and variance σ2 we
define An = X1+X2+···+Xn
n then E[An] = nµ
n = µ and similarly
Var[An] = nσ2
n2 = σ2/n. By Chebyshev’s inequality we get the weak law of
large numbers
P( |An − µ| ≥ ) ≤ Var[An]/2
=
σ2
n2

n then E[An] = nµ
Var[An] = nσ2
large numbers
P( |An − µ| ≥ ) ≤ Var[An]/2
=
σ2
n2
To prove Chebyshev’s we define A = {w ∈ Ω | |X(w)| ≥ α}.
X(w) ≥ αIA(w). Take the expectation on both sides to get
E(|X|) ≥ αE(IA) = αP(A)

n then E[An] = nµ
Var[An] = nσ2
large numbers
P( |An − µ| ≥ ) ≤ Var[An]/2
=
σ2
n2
E(|X|) ≥ αE(IA) = αP(A)
In game theoretic proof we don’t need i.i.d assumption

n then E[An] = nµ
Var[An] = nσ2
large numbers
P( |An − µ| ≥ ) ≤ Var[An]/2
=
σ2
n2
E(|X|) ≥ αE(IA) = αP(A)
In game theoretic proof we don’t need i.i.d assumption we don’t even to
assume a distribution exists!

Unbounded game

Unbounded game
Theorem
If
P∞
n=1
vn
n2 ∞ then the skeptic has a winning strategy

Unbounded game
Theorem
If
P∞
n=1
vn
n2 ∞ then the skeptic has a winning strategy
Proof.
Similar in nature to proof of the bounded fair coin game. Main idea is that
the skeptic’s capital is a supermartingale (a sequence that decreases in
expectation)

What about an application
Suppose you’re a clever young guy/gal who wants to make money off of
these ideas

What about an application
Suppose you’re a clever young guy/gal who wants to make money off of
these ideas
A natural next step is to make an infinitely large amount of money off the
stock market

Efficient Market Hypothesis
Unfortunately it seems that its difficult to have consistently better returns
than the market and we will prove this. We make two assumptions that
transaction costs are neglible (not as controversial as it sounds) and that
the capital of a specific investor isn’t too big relative to the market.

Proof.
Maybe next time, Finance theory might need its own talk :)

References
Shafer and Vovk (2001)
Probability and Finance It’s only a Game!
Ramon Van Handel
Stochastic Calculus
Peter Clark
All I ever needed to know from Set Theory

Let’s think about how this could change machine learning, talk to me and
let’s write a paper about it!

Mark Saroufim - Probability without Measure - It's only a game

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