Lecture Five Stochastic

by 5 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 ” August 4-5, 2009 Portland, Oregon

1/37 2/37 The Itˆointegral problem Definition Let W be a 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) of ItˆoIntegral Suppose Y (t), 0 ≤ t ≤ T, is a (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 ∆(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

It remains to show that the cross-terms have zero expectation. For j < k, the increment W (tk+1) − W (tk ) is independent of

∆(tj )∆(tk ) W (tj+1) − W (tj ) , and hence   5 Stochastic Calculus 5.3 Construction for General Integrands E ∆(tj )∆(tk ) W (tj+1) − W (tj ) W (tk+1) − W (tk ) h   i

= E ∆(tj )∆(tk ) W (tj+1) − W (tj ) · E W (tk+1) − W (tk ) h  i  

= E ∆(tj )∆(tk ) W (tj+1) − W (tj ) · 0 h  i = 0.

13/37 14/37 Outline of construction for general integrands Outline of construction (continued)

◮ ◮ P ∞ Given ∆(s), 0 ≤ s ≤ T , satisfying L2(Ω, F, ) is complete, and so the sequence {In(T )}n=1 has T a limit I (T ) in this space. E ∆2(s) ds < ∞, ◮ We define Z0 T ∆(s) dW (s)= I (T ) = lim In(T ). construct an approximating sequence of simple processes →∞ 0 n ∆ (s), 0 ≤ s ≤ T , such that Z n This limit does not depend on the approximating sequence T ∞ 2 {∆n}n=1. lim E ∆(s) − ∆n(s) ds = 0. n→∞ ◮ By choosing approximating sequences that converge rapidly, Z0  we can in fact make the convergence of In(T ) to I (T ) be ◮ T Set In(T )= 0 ∆n(s) dW (s). Itˆo’s isometry implies that almost sure (almost everywhere with respect to P) rather than T in L2. R 2 2 E In(T ) − Im(T )) = E ∆n(s) − ∆m(s) ds. ◮ With additional work, one can choose the approximating 0 Z sequence so that the paths of In(t), 0 ≤ t ≤ T , converge ◮   ∞  Because the sequence {∆n}n=1 converges in uniformly in t ∈ [0, T ] almost surely. This guarantees that L2(Ω × [0, T ], F ⊗ Borel([0, T ]), P × Lebesgue), it is Cauchy there is a limit I (t), 0 ≤ t ≤ T , that is a ∞ P in this space. Therefore, {In(T )}n=1 is Cauchy in L2(Ω, F, ). of t ∈ [0, T ] for P-almost every ω.

15/37 16/37 Theorem E T 2 Under the assumption [ 0 ∆ (s) ds] < ∞, the Itˆointegral tR I (t)= ∆(s) dW (s), 0 ≤ t ≤ T , Z0 is defined and continuous in t ∈ [0, T ]. We have 5 Stochastic Calculus EI (t) = 0, 0 ≤ t ≤ T . 5.4 Example of an ItˆoIntegral

The quadratic variation of the Itˆointegral is

t [I , I ](t)= ∆2(s) ds, 0 ≤ t ≤ T , Z0 and the Itˆointegral satisfies Itˆo’s Isometry

t Var[I (t)] = E I 2(t) = E ∆2(s) ds , 0 ≤ t ≤ T . Z0    17/37 18/37 T W s dW s 0 ( ) ( ) Divide [0, T ] into n equal subintervals. Define By definition, R T jT jT (j + 1)T ∆n(s)= W for ≤ s < . W (s) dW (s) n n n Z0   n−1 jT (j + 1)T jT = lim W W − W . n→∞ n n n j X=0       jT To simplify notation, we denote Wj = W n . Then W0 = W (0) = 0, Wn = W (T ), and  

T n−1 W (s) dW (s) = lim Wj (Wj+1 − Wj ). n→∞ t1 t2 t3 t = T s 0 j 4 Z X=0

One path of W (s) and ∆4(s) 19/37 20/37 n−1 n−1 n−1 n−1 From the previous page, we have 1 2 1 2 1 2 (Wj+1 − Wj ) = Wj+1 − Wj Wj+1 + Wj 2 2 2 n−1 n−1 j=0 j=0 j=0 j=0 1 2 1 2 X X X X Wj (Wj+1 − Wj )= Wn − (Wj+1 − Wj ) . n n−1 n−1 2 2 1 1 j=0 j=0 = W 2 − W W + W 2 X X 2 k j j+1 2 j k=1 j=0 j=0 Letting n → ∞, we get X X X n−1 n−1 n−1 T 1 2 1 2 1 2 1 2 1 1 2 1 = W + W − Wj Wj + W W (s) dW (s)= W (T ) − [W , W ](T ) = W (T ) − T . 2 n 2 k +1 2 j 2 2 2 2 k j j 0 X=0 X=0 X=0 Z n−1 n−1 1 Remark = W 2 + W 2 − W W 2 n j j j+1 If g is a differentiable function with g(0) = 0, then j j X=0 X=0 T T T n−1 ′ 1 2 1 2 1 2 g(s) dg(s)= g(s)g (s) ds = g (s) = g (T ). = W + Wj (Wj − Wj+1). 2 0 2 2 n 0 0 j=0 Z Z X T n−1 n−1 1 The extra term 2 T in 0 W (s)dW (s) comes from the nonzero 1 2 1 2 Wj (Wj+1 − Wj )= W − (Wj+1 − Wj ) . quadratic variation of Brownian motion. 2 n 2 R j=0 j=0 X X 21/37 22/37

Exercise (5.2) Show that 5 Stochastic Calculus n−1 (j + 1)T (j + 1)T jT lim W W − W 5.5 Itˆo’s Formula for One Process n→∞ n n n j X=0       1 1 = W 2(T ) + T . 2 2

23/37 24/37 Along the path of a Brownian motion, we want to “differentiate” Remark f (W (t)), where f (x) is a differentiable function. If the path of the The mathematically meaningful form of Itˆo’s formula is Itˆo’s Brownian motion W (t) could be differentiated with respect to t, formula in form: then the ordinary would give t T ′ 1 ′′ d ′ ′ f (W (T )) − f (W (0)) = f (W (t)) dW (t)+ f (W (t)) dt. f (W (t)) = f (W (t))W (t), 0 2 0 dt Z Z This is because we have definitions for both appearing on which could be written in differential notation as the right-hand side. The first, ′ ′ ′ df (W (t)) = f (W (t)) W (t) dt = f (W (t)) dW (t). T f ′(W (t)) dW (t) Because W has nonzero quadratic variation, the correct formula Z0 has an extra term, namely, is an Itˆointegral. The second 1 df (W (t)) = f ′(W (t)) dW (t)+ f ′′(W (t)) dt . T 2 f ′′(W (t)) dt dW (t)dW (t) Z0 This is Itˆo’s formula in differential form. |{z} is a with respect to time, computed path by path.

25/37 26/37 Application of Itˆo’s Formula Derivation of Itˆo’s Formula 1 2 Consider f (x)= 2 x , so that Consider f (x)= 1 x2, so that f ′(x)= x, f ′′(x) = 1. 2 f ′(x)= x, f ′′(x) = 1. Itˆo’s formula in integral form

Let xj+1 and xj be numbers. Taylor’s formula implies T 1 T f W (T ) −f W (0) = f ′ W (s) dW (s)+ f ′′ W (s) ds 2 1 Z0 Z0 f (x ) − f (x ) = (x − x )f ′(x )+ (x − x )2f ′′(x ).     j+1 j j+1 j j 2 j+1 j j becomes In this case, Taylor’s formula to second order is exact because f is 1 T 1 T T 1 W 2(T )= W (s) dW (s)+ 1 ds = W (s) dW (s)+ T , a quadratic function. 2 2 2 Z0 Z0 Z0 In the general case, the above equation is only approximate, and the error is of the order of (x − x )3. The total error will or equivalently, k+1 k have limit zero in the last step of the following argument (see T 1 1 Exercise 4.6(iii) of Lecture 4). W (u) dW (u)= W 2(T ) − T . 2 2 Z0

27/37 28/37 From the previous page, we have Fix T > 0 and let Π= {t0, t1,..., tn} be a partition of [0, T ]. f (W (T )) − f (W (0)) f (W (T )) − f (W (0)) n−1 n−1 2 1 = W (t ) W (t ) − W (t ) + W (t ) − W (t ) . j j+1 j 2 j+1 j n−1 j " # k " # X=0 X=0 = f (W (tj+1)) − f (W (tj )) j " # We let kΠk → 0, to obtain X=0 n−1 f (W (T )) − f (W (0)) ′ = W (tj+1) − W (tj ) f (W (tj )) T 1 j=0 " # = W (s) dW (s)+ [W , W ](T ) X 0 2 n−1 2 Z T 1 + W (t ) − W (t ) f ′′(W (t )) T 1 T 2 j+1 j j = f ′(W (s)) dW (s)+| {z f ′′}(W (s)) ds. j " # X=0 0 2 0 Z Z 1 n−1 n−1 2 1 = W (t ) W (t ) − W (t ) + W (t ) − W (t ) . This is Itˆo’s formula in integral form for the special| {z case} j j+1 j 2 j+1 j j=0 " # j=0 " # 1 X X f (x)= x2. 2

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y = f (x) Exercise (5.3) Let u ∈ R be constant and define f (x)= eux . Use Itˆo’s formula ′ f (W (tj ))(W (tj+1) − W (tj )) applied to f (W (t)) to obtain the moment-generating function formula uW (T ) 1 u2T Ee = e 2 . (Compare with Exercise 4.4 of Lecture 4.) W (tj ) W (tj+1) ′ f W (tj+1) − f W (tj ) = f W (tj ) W (tj+1) − W (tj ) Exercise (5.4) Let f (x)= x4. Use Itˆo’s formula applied to f (W (t)) to obtain the   + Small Error  ′ fourth-moment formula f W (tj+1) − f W (tj ) = f W (tj ) W (tj+1) − W (tj ) 1 ′′ 2 E 4 2   + f W(t ) W (t ) − W(t ) W (T ) = 3T . 2 j j+1 j + Smaller Error  (Compare with Exercise 4.5 of Lecture 4.) We need the higher accuracy before summing. Otherwise, the accumulated small errors do not vanish as the step-size goes to zero. 31/37 32/37 Solution to Exercise 5.1

(i) As in the proof of the theorem preceding the exercise, we use 5 Stochastic Calculus independence to compute EI (T ) term by term: 5.7 Solution to Exercise E ∆(tj ) Y (tj+1) − Y (tj ) = E∆(tj ) · E Y (tj+1) − Y (tj ) E   = ∆(tj ) · 0 = 0.  (ii) If I (T ) ≥ 0 almost surely and P{I (T ) > 0} > 0, then EI (T ) > 0. This contradicts part (i).

33/37 34/37 Solution to Exercise 5.2 Solution to Exercise 5.3 ′ ′′ 2 jT We have f (x)= uf (x) and f (x)= u f (x). Therefore, Itˆo’s Let tj = . The quadratic variation result for Brownian motion is n formula becomes n−1 T T 2 uW (T ) uW (0) uW (t) 1 2 uW (t) lim W (tj+1) − W (tj ) = T . e = e + u e dW (t)+ u e dt. n→∞ 2 j=0 Z0 Z0 X  Taking expectations and using the fact that the expectation of the The Example shows that Itˆointegral is zero, we obtain

n−1 T 1 2 1 uW (T ) 1 2 uW (t) lim W (tj )(W (tj+1) − W (tj ) = W (T ) − T . Ee =1+ u Ee dt. n→∞ 2 2 2 j=0 0 X  Z We differentiate both sides with respect to T to obtain Adding these two equations, we obtain d uW (T ) 1 2 uW (T ) n−1 Ee = u Ee . 1 2 1 dT 2 lim W (tj+1) W (tj+1) − W (tj ) = W (T )+ T . n→∞ 2 2 j The unique solution to this ordinary differential equation satisfying X=0  EeuW (0) =1 is uW (T ) 1 u2T This is the desired result. Ee = e 2 .

35/37 36/37 Exercise 5.4

With f (x)= x4, we have f ′(x) = 4x3 and f ′′(x) = 12x2. Therefore, Itˆo’s formula becomes

T T W 4(T ) = 4 W 3(t) dW (t) + 6 W 2(t) dt. Z0 Z0 Taking expectations of both sides and using the fact that the Itˆo integral has expectation zero, we obtain

T EW 4(T ) = 6 EW 2(t) dt 0 Z T = 6 t dt Z0 = 3T 2.

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