# Proof of Black Scholes

**Topics:**Normal distribution, Standard deviation, Probability density function

**Pages:**4 (596 words)

**Published:**September 27, 2011

Theorem 1: Suppose Y = ln X is a normal distribution with mean m and variance v, then X has mean exp( m + v /2 ) Proof: The density function of Y= ln X

Therefore the density function of X is given by

Using the change of variable x = exp(y), dx = exp(y) dy, We have

= Note that the integral inside is just the density function of a normal random variable with mean (m-v) and variance v. By definition, the integral evaluates to be 1.

Proof of Black Scholes Formula

Theorem 2: Assume the stock price following the following PDE

Then the option price

for a call option with payoff

is given by

1

Proof: By Ito’s lemma,

If form a portfolio P

Applying Ito’s lemma

Since the portfolio has no risk, by no arbitrage, it must earn the risk free rate,

Therefore we have

Rearranging the terms we have the Black Scholes PDE

With the boundary condition

To solve this PDE, we need the Feynman-Kac theorem: Assume that f is a solution to the boundary value problem:

Then f has the representation:

2

Where S satisfies the following stochastic differential equation

Proof: Suppose that is the solution to the PDE. Let

Applying the Ito’s lemma

Since the last term involves only second order terms only,

Collecting terms we have got

As the first term is simply the PDE, it is zero. Therefore

Integrating from 0 to T

Taking expectation on both side,

Since the integral is a limiting sum of independent Brownian motions increments, i.e. =0 it is zero. Recall that W has independent and stationary increment with a zero mean, i.e. is normally distributed with zero mean. 3

Therefore In other words

End of Proof.

By the Feynman Kac Theorem, the solution to the Black Scholes PDE is given by

Where S follows

Consider Z = ln S, by Ito’s lemma,

Integrate both side from 0 to T, We have

Recall that with mean

has a normal distribution with mean 0, and variance T, and variance...

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