# [Stochastic Process] Black-Scholes Formula

## 1. Introduction

Motivation: Find the cost of buying options to prevent arbitrage opportunity.
Definition: Let be the price of a stock at time s, considered on a time horizon . The  following actions are available:
• At any time s, , you can buy or sell shares of stock for
• At time , there are N options available. The cost of option is per option which allows you to purchase share at time for price .
Objective: the goal is to determine so that no arbitrage opportunity exists.
Approach: By the arbitrage theorem, this requires finding a probability measure so that each bet has zero expected pay off.

We try the following probability measure on : suppose that is geometric Brownian motion. That is, , where .

## 2. Analyze Bet 0: Purchase 1 share of stock at time s

Now, we consider a discount factor . Then, the expected present value of the stock at time t is

Choose such that

Then .

Thus the expected present value of the stock at t equals the present value of the stock when it is purchased at time s. That is, the discount rate exactly equals to the expected rate of the stock returns.

## 3. Analyze Bet i: Purchase 1 share of option i (at time s =0)

First, we drop the subscript i to simplify notation.

Expected present value of the return on this bet is

=

Setting this equal to zero implies that

=

has value 0 when , i.e., when

Thus the integral becomes

=

Now, apply a change of variables:
, i.e.,

## 4. Summary

No arbitrage exists if we find costs and a probability distribution such that expected outcome of every bet is 0.
• If we suppose that the stock price follow geometric Brownian motion and we choose the option costs according to the above formula, then the expected outcome of every bet is 0.
• Note: the stock price does not actually need to follow geometric Brownian motion. We are saying that if stock price follow Brownian motion, then the expected outcome of very bet would be 0, so no arbitrage exists.
A few sanity checks
• When , cost of option is
• When , cost of option is
• As t increases, c increases
• As k increases, c decreases
• As increases, c increases
• As increase, c increase (assuming )
• As , c decreases

# [Stochastic Process] Arbitrage

1. Introduction

Definition
the simultaneous buying and selling of securities, currency, or commodities in different markets or in derivative forms in order to take advantage of differing prices for the same asset.

2. Option Pricing

Question:

How to choose x and y:

How to maximize profit:

Key Assumption: There is no limit to buying or selling of options. In practice,

you may only be able to buy, but no sell, for example.

3. Aibitrage Theorem

Definition:
Consider n possible wagers on m possible outcomes: .
Let to be the outcome of wagers i if outcome j occurs.
If is bet on wager , then is earned if outcome j occurs.

Arbitrage Theorem:

• such that ,
• such that
Intuitively,
• First theorem: there is a probability vector such that the expected outcome of every bet is 0, or
• There existing a betting scheme that leads to a sure win.

## TODO: explain more about these two theorem

# [Stochastic Process] Geometric Brownian Motion

1. Motivation

Definition: Let be Brownian motion with drift coefficient and variance parameter . Let . Then is geometric Brownian motion.

Motivation: Let be the price of a stock at time (where n is distance); Let be the fractional increase/decrease in the price of the stock from time n-1 to time n.
We suppose that are i.i.d. Then

The process looks like a random walk.

2. Property of Geometric Brownian Motion

Proof:
=
=
• E[Y(t) | Y(s) = y_s] = y_s e^{mu(t-s) + sigma^2(t-s)/2}
Proof: If W ~ , then is lognormal with mean
Thus, {X(t) – X(s)} ~ , since is Brownian motion with drift

• Note: if , then . Thus is increasing even though the jump process with is symmetric
3. Example
Question: You invest 1000 dollars in the stock market. Suppose that the stock market can be modeled using geometric Brownian motion with an average daily return of 0.03% and a standard deviation of 1.02%. what is the probability that your money increases after 1 year (260 business days?) 10 years? 30 years?

Answer: Let , where is standard Brownian Motion.
Let where is standard Brownian motion, is Brownian motion with drift, and is geometric Brownian motion.

=
=
=
=
=

# [Stochastic Process] Brownian Motion

Course notes for “Stochastic Process”
2014 Fall
1. Motivation
Brownian motion can be thought of a symmetric random walk where the jumps sizes are very small and where jumps occur very frequently.

• Each jump size are
• The time before two jumps are .

### 1.1 What’s the mean and variance?

Let denote whether the i-th jump is to the right(+1) or to the left(-1), we have with probability and respectively.
Thus, and .

Let denote the state of  Markov Chain after n jumps, then

The denote the continuous Markov Chain after n jumps, then

Then we have
and .

Let ,  then
.

## 2. Properties of Brownian Motion

• (1)
• (2)
• (3)  X(t) has independent increments
i.e., is independent of assuming the intervals of and are disjoint.
• (4)  X(t) has stationary increments
i.e., has the same distribution as if .
Example: What’s the distribution of ?
Answer: by the stationary property, we have ~ .

3. Standard Brownian Motion (SBM)

• SBM ~
• Let , then .
4. Brownian Motion with Drift

Definition 1: Let be the standard Brownian motion. Let , then is Brownian motion with drift .

Definition 2: . is Brownian Motion with drift and variance parameter if

• has stationary and independent increments
• ~

Example: Let be Brownian motion with and drift . What is Pr{X(30) >0 | X(10) = -3}Pr{X(30) – X(20) >3 | X(10) =3 }Pr{X(30) – X(10) >3  }Pr{X(20) – X(0) >3  }Pr{X(20)>3  }Pr{N(2,80) > 3} = Pr{X(0,1) > frac{3-2}{sqrt{80}}1-Phi(frac{1}{4sqrt{5}})X(s) = x | X(t) = Bfrac{s}{t} cdot Bfrac{s}{t} cdot (t-s)s = t/2X(t)N(0, sigma^2 t)10 after 6 hours, what is the probability that the stack was above its starting value after 3 hours?

This is a Brownian bridge process where ~ = .
Thus

Example 2: If a bicycle race between two competitors, Let denote the amount of time (in seconds) by which the racer that started in the insider position is ahead when 100t percent of the race has been completed, and suppose that , can be effectively modeled as a Brownian Motion process with variance parameter .

6. First Passage Time

Let denote the first time that standard Brownian motion hits level a (starting at X(0) = 0), assuming a >0, then we have

• : you know that at some time before t, the process hits a. From that point forward, you are just as likely to be above a as below a.
• : cannot be above a, because the first passage time to a is after t.
Thus,
.

Change variables: , we have .
By symmetry, we get .