Ito's Formula

Ito's formula for one Brownian Motion

We want a rule to "differentiate" expression of the form $f(B(t))$, where $f(x)$ is a differentiable function. If $B(t)$ were also differentiable, then the ordinary chain rule would give
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which could be written in differentiable notation as
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However, B(t) is not differentiable, and in particular has nonzero quadratic variation, so the correct formula has an extra term, namely,
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This is Ito's formula in differential form. Integrating this, we obtain Ito's formula in integral form:
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Derivation of Ito's formula

Consider MATH so that
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Let x$_{k},$ $x_{k+1}$ be numbers. Taylor's formula implies
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Fix $\dot{T}>0$ and MATH be a partition of MATH Using Taylor's formula, we write
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We let MATH to obtain
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This is Ito's formula in integral form for the special case
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Homework 6. Page 251, #1, #3. Due 4/4.

Geometric Brownian Motion

Geometric Brownian motion is
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where $\mu $ and $\sigma >0$ are constant. Define
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so
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Then
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According to Ito's formula,
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Thus, Geometric Brownian motion in differential form is
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and Geometric,Brownian motion in integral form is
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Quadratic variation of Geometric Brownian motion

In the integral form of Geometric Brownian motion,
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the Riemann integral
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is differentiable with MATH This term has zero quadratic variation. The Ito integral
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is not differentiable. It has quadratic variation
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Thus the quadratic variation of S is given by the quadratic variation of G. In differentiable notation, we write
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First derivation of the Black-Scholes formula

Wealth of an investor.

An investor begins with nonrandom initial wealth X$_{0}$ and at each time t, holds $\bigtriangleup (t)$ shares of stock. Stock is modelled by a geometric Brownian motion:
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$\bigtriangleup (t)$ can be random, but must be adapted. The investor finances his investing by borrowing or lending at interest rate r. Let X(t) denote the wealth of the investor at time $t$. Then
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Value of an option

Consider an European option which pays $g(S(T))$ at time T. Let MATH denote the value of this option at time t if the stock price is $S(t)=x$. In other words, the value of the option at each time MATH is
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The differential of this value is
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A hedging portfolio starts with some initial wealth X$_{0}$ and invests so that the wealth $X\left( t\right) $ at each time tracks MATH We saw above that
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To ensure that $X(t)=v(t,S(t))$ for all t, we equate coefficients in their differentials. Equating the dB coefficients, we obtain the $\bigtriangleup $-hedging rule:
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Equating the dt coefficients, we obtain
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But we have set MATH and we are seeking to cause $X$ to agree with $v$. Making these substitutions, we obtain
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which simplies to
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In conclusion, we should let $v$ be the solution of the Black-Scholes partial differential equation
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satisfying the terminal condition
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If an investor starts with $X_{0}=v(0,S(0))$ and uses the hedge
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then he will have$X(t)=v(t,S(t))$ for all t, and in particular, MATH
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