MATH 56A SPRING 2008 STOCHASTIC PROCESSES 197

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1 MATH 56A SPRING 8 STOCHASTIC PROCESSES Itô s formula. First I stated the theorem. Then I did a simple example to make sure we understand what it says. Then I proved it. The key point is Lévy s theorem on quadratic variation statement of the theorem. Theorem 9.19 (Itô). Suppose that f(x) is a C (twice continuously differentiable) function and W t is standard Brownian motion. Then f(w t ) f(w ) = f (W s )dw s + 1 f (W s )ds where f (W s )dw s is the stochastic integral that we defined last time. The key point is the unexpected ds in the formula. Example 9.. Take f(x) = ax + bx + c. Then f(w ) = f() = c, So, the LHS of Itô s formula is: f (x) = ax + b f (x) = a f(w t ) f(w ) = aw t + bw t The RHS is (aw s + b) dw s + 1 a ds = = a aw s dw s + If we cancel the bw s terms we have: aw t = a b dw s + at W s dw s + bw t + at W s dw s + at. The infinitesimal version of this is (after dividing by a): dw s = W t dw t + dt

2 198 STOCHASTIC INTEGRATION proof of Itô s formula. I proved the infinitesimal version of Itô s formula which says: df(w t ) = f (W t )dw t + 1 f (W t )dt This is the limit as δt of the Taylor formula which we saw earlier in the derivation of the heat equation: δf(w t ) = f(w t+δt ) f(w t ) = Taylor f (W t )δw t + 1 f (W t )(δw t ) + O(δW 3 ) In usual calculus we ignore the second term. But in stochastic calculus we keep the second term and ignore the third term since O(δW 3 ) = o(δt). As δt goes to, δf(w t ) df(w t ) So, what we need to prove is that δw t dw t (δw t ) dt with probability one. This will follow from Lévy s theorem: Theorem 9.1 (Lévy). The quadratic variation of Brownian motion is = t almost surely. The first point is: Why does this complete the proof of Itô s formula? To see this we need to write the infinitesimal version of Lévy s theorem: d = dt Now, recall the definition of quadratic variation: := lim δt (δw ) So, δ = (δw ) by definition. Or: d = (dw t ) Therefore, the (δw t ) in Taylor s formula gives the dt in Itô s formula as δt.

3 MATH 56A SPRING 8 STOCHASTIC PROCESSES 199 Proof of Lévy s Theorem. We know that δw t = W t+δt W t N(, δt) This implies by an easy calculation that E((δW t ) ) = δt E((δW t ) 4 ) = 3(δt) So, the variance of (δw t ) is Var((δW t ) ) = E((δW t ) 4 ) E((δW t ) ) = 3(δt) (δt) = (δt). So, the standard deviation of (δw t ) is δt. In other words, (δw t ) = δt ± δt }{{} error The error term is bigger than the term itself! Lévy s Theorem is saying that the error term ± δt is negligible when compared to the main term δt!!! Now go back to the original statement. = lim (δw ) δt where the number of terms in the sum is N = t/δt. Since the expected value of each term is E((δW ) ) = δt we know that E( ) = δt = t δt δt = t This means that is equal to t on average. (So, the distribution of possible values of forms a bell shaped curve centered at t. The width of the curve at t is the standard deviation which is the square root of the variance.) To prove Lévy s theorem we need to prove that the variance of is zero. Since Var(X +Y ) = Var(X)+Var(Y ) for independent X, Y we have: ( Var (δw ) ) = Var ( (δw ) ) = (δt) But we have t/δt terms in this sum so the sum is = t δt (δt) = tδt which converges to zero as δt. This proves: Var( ) =.

4 STOCHASTIC INTEGRATION So, is equal to its expected value t with probability one. probability distribution is a Dirac delta function at t.) Exercise 9.. Calculate E(X n ) for X N(, σ ). By definition: Integrate by parts: E(X n ) = u = x n 1 x n e x /σ πσ dx du = (n 1)x n dx (Its dv = x e x /σ πσ dx v = σ e x /σ πσ The product uv vanished at both tails. So, So, E(X n ) = (n 1)σ x n /σ e x dx πσ E(X n ) = (n 1)σ E(X n ) Since E(X) = µ =, this formula shows that E(X n 1 ) = for all n. But, E(X ) = 1. So, and E(X ) = σ E(X 4 ) = 3σ E(X ) = 3σ 4 Exercise 9.3. Show that if X t is continuous with bounded variation then X t = Definition 9.4. The variation of X t is the total distance travelled by X from time to time t: variation X t := lim δt δxt For a C 1 (continuously differentiable) function f(t) this is variation f = f (s) ds

5 MATH 56A SPRING 8 STOCHASTIC PROCESSES 1 X t = lim (δx) δt Since X t is continuous, δx as δt. So, this is the same as lim δx (δx) = lim δx =. }{{}}{{} bounded δx δx This uses the Lebesgue style idea of cutting up the image instead of the domain: Instead of cutting equal time intervals, we cut up equal space intervals. Then we can factor out the δx. (The Lebegue integral is given by taking the limit as δy = y i+1 y i : g dµ = lim y i µ ({x g(x) (y i, y i+1 ]}) Ω δy where µ is the measure on subsets of the domain. If µ = P, this is = y f(y)dy = E(Y ). where f(y) is the density function of Y = g.)