Lecture Five Stochastic Calculus
by 5 Stochastic Calculus Steven E. Shreve 5.1 ItˆoIntegral for a Simple Integrand Department of Mathematical Sciences 5.2 Properties for Simple Integrands Carnegie Mellon University 5.3 Construction for General Integrands 5.4 Example of an ItˆoIntegral for the 5.5 Itˆo’s Formula for One Process MAA Short Course 5.6 Solution to Exercise “Financial Mathematics” August 4-5, 2009 Portland, Oregon
1/37 2/37 The Itˆointegral problem Definition Let W be a Brownian motion defined on a probability space (Ω, F, P). A process ∆(s, ω), a function of s ≥ 0 and ω ∈ Ω, is adapted if the dependence of ∆(s, ω) on ω is as a function of the 5 Stochastic Calculus initial path fragment W (u, ω), 0 ≤ u ≤ s. In particular, ∆(s) is independent of W (t) − W (s) whenever 0 ≤ s ≤ t. 5.1 ItˆoIntegral for a Simple Integrand We want to make sense of t ∆(s) dW (s), 0 ≤ t ≤ T . Z0 Remark If g(s) is a differentiable function, then we can define
t t ∆(s) dg(s)= ∆(u)g ′(s) ds. Z0 Z0 This won’t work for Brownian motion, however, because the paths of Brownian motion are not differentiable. 3/37 4/37 Simple Integrand Interpretation of Simple Integrand
Let Π = {t0, t1,..., tn} be a partition of [0, T ], i.e., ◮ Think of W (s) as the price per share of an asset at time s. ◮ Think of t0, t1,..., tn−1 as the trading dates in the asset. 0= t0 ≤ t1 ≤···≤ tn = T . ◮ Think of ∆(t0), ∆(t1),..., ∆(tn−1) as the number of shares Assume that ∆(s) is constant in s on each subinterval [tk , tk+1). of the asset acquired at each trading date and held to the We call such a ∆ a simple process. next trading date. Gain from trading.
I (t) = ∆(t0)[W (t) − W (t0)] = ∆(t0)W (t), 0 ≤ t ≤ t1,
I (t) = ∆(t0)[W (t1) − W (t0)] + ∆(t1)[W (t) − W (t1)],
t1 ≤ t ≤ t2,
t1 t2 t3 t4 s I (t) = ∆(t0)[W (t1) − W (t0)] + ∆(t1)[W (t2) − W (t1)] +∆(t2)[W (t) − W (t2)], t2 ≤ t ≤ t3. One path of ∆ The process I is the Itˆointegral of the simple process ∆, i.e., Example t I (t)= ∆(s) dW (s), 0 ≤ t ≤ T . ∆(s)= W (tk ), tk ≤ s < tk+1 Z0 5/37 6/37 Expectation of Itˆointegral
Theorem The Itˆointegral of a simple process has expectation zero. Proof: 5 Stochastic Calculus By definition n−1 5.2 Properties for Simple Integrands I (T )= ∆(tj ) W (tj+1) − W (tj ) . j=0 X Compute expectation term by term. Because ∆(tj ) is independent of W (tj+1) − W (tj ), we have
E ∆(tj ) W (tj+1) − W (tj ) = E∆(tj ) · E W (tj+1) − W (tj ) E = ∆(tj ) · 0 = 0.
7/37 8/37 Exercise (5.1) Quadratic Variation of ItˆoIntegral Suppose Y (t), 0 ≤ t ≤ T, is a stochastic process (a function of t and ω) such that if 0 ≤ s ≤ t, then the increment Y (t) − Y (s) is Theorem The simple Itˆointegral independent of the path of Y up to time s and has expectation zero. Let {∆(s)}0≤s≤T be a simple process adapted to Y, i.e., t there is a partition Π= {t0, t1,..., tn} of [0, T ] such that ∆(s) is I (t)= ∆(u) dW (u) 0 constant in s in each subinterval [tj , tj+1), and for each s ∈ [0, T ], Z the random variable ∆(s) depends on ω only through the path of has quadratic variation Y up to time s, and hence ∆(s) is independent of Y (t) − Y (s) for T all t ∈ [s, T ]. Define the Itˆointegral [I , I ](T )= ∆2(u) du (QV ) 0 n−1 Z I (T )= ∆(t ) Y (t ) − Y (t ) . and variance j j+1 j T j 2 2 X=0 E I (T ) = E ∆ (u) du. (VAR) E (i) Show that I (T ) = 0. Z0 (ii) A simple arbitrage is a simple process ∆ such that I (T ) ≥ 0 Remark almost surely and P{I (T ) > 0} > 0. Show that there is no Both sides of (QV) are random, but the expressions in (VAR) are simple arbitrage under the assumptions of this exercise. not. (VAR) is called Itˆo’s Isometry. 9/37 10/37 Proof of (QV) Proof of (VAR) n−1 For s ∈ [tj , tj+1], we have ∆(s) = ∆(tj ) and I (T )= ∆(tj ) W (tj+1) − W (tj ) . I (s) = I (t ) + ∆(t ) W (s) − W (t ) j=0 j j j X h i Squaring and taking expectations, we obtain = I (tj ) − ∆(tj )W (tj ) + ∆(tj )W (s). n−1 h i 2 2 2 On this subinterval, quadratic variation of I comes from the E I (T ) = E ∆ (tj ) W (tj+1) − W (tj ) quadratic variation of W , which is scaled by ∆(tj ). Therefore k=j X h i 2 E . [I , I ](tj+1) − [I , I ](tj ) = ∆ (tj ) [W , W ](tj+1) − [W , W ](tj ) +2 ∆(tj )∆(tk ) W (tj+1) − W (tj ) W (tk+1) − W (tk ) j