Lecture 21. Itō’s formula

Posted on September 19, 2012 by Fabrice Baudoin

Itō’s formula is certainly the most important and useful formula of stochastic calculus. It is the change of variable formula for stochastic integrals. It is a very simple formula whose specificity is the appearance of a quadratic variation term, that reflects the fact that semimartingales have a finite quadratic variation.

Due to its importance, we first provide a heuristic argument on how to derive Itō ‘s formula. Let f : \mathbb{R} \rightarrow \mathbb{R} be a smooth function and x: \mathbb{R} \rightarrow \mathbb{R} be a C^1 path \mathbb{R} \rightarrow \mathbb{R}. We have the following heuristic computation:
f(x_{t+dt})=f(x_t +(x_{t+dt}-x_t))
=f(x_t)+f'(x_t)(x_{t+dt}-x_t)
=f(x_t)+f'(x_t)dx_{t}.
This suggests, by summation, the following correct formula:
f(x_t)=f(x_0)+\int_0^t f'(x_s)dx_s.
Let us now try to consider a Brownian motion (B_t)_{t \ge 0} instead of the smooth path x and let us try to adapt the previous computation to this case. Since Brownian motion has quadratic variation which is not zero, \langle B \rangle_t=t, we need to go at the order 2 in the Taylor expansion of f. This leads to the following heuristic computation:
f(B_{t+dt})=f(B_t +(B_{t+dt}-B_t))
=f(B_t)+f'(B_t)(B_{t+dt}-B_t)+\frac{1}{2} f''(B_t) ((B_{t+dt}-B_t))^2
=f(B_t)+f'(B_t)dB_{t}+\frac{1}{2} f''(B_t)dt.
By summation, we are therefore led to the formula
f(B_t)=f(0)+\int_0^t f'(B_s)dB_{s}+\frac{1}{2}\int_0^t f''(B_s)ds,
which is, as we will see it later perfectly correct.

In what follows, we consider a filtered probability space (\Omega, (\mathcal{F}_t)_{t \ge 0},\mathcal{F},\mathbb{P}) that satisfies the usual conditions. Our starting point to prove Itō’s formula is the following formula which is known as the integration by parts formula for semimartingales:

Proposition. (Integration by parts formula)
Let (X_t)_{t \ge 0} and (Y_t)_{t \ge 0} be two continuous semimartingales, then the process (X_t Y_t )_{t \ge 0} is a continuous semimartingale and we have:
X_t Y_t =X_0 Y_0 +\int_0^t X_s dY_s +\int_0^t Y_s dX_s+\langle X,Y \rangle_t, \quad t \ge 0.

Proof. By bilinearity of the multiplication, we may assume X=Y. Also by considering, if needed, X-X_0 instead of X, we may assume that X_0=0. Let t \ge 0. For every sequence \Delta_n [0,t] such that
\lim_{n \rightarrow +\infty}\mid\Delta_n [0,t]\mid=0, we have
\sum_{k=1}^{n} \left( X_{t^n_k}-X_{t^n_{k-1}}\right)^2=X_t^2-2\sum_{k=1}^{n} X_{t^n_{k-1}} \left(X_{t^n_k} -X_{t^n_{k-1}}\right).
By letting n \to \infty, we therefore obtain the following identity which yields the expected result:
X_t^2=2\int_0^t X_s dX_s +\langle X \rangle_t
\square

We are now in position to prove Itō’s formula in its simpler form.

Theorem. (Itō’s formula I) Let (X_t)_{t \ge 0} be a continuous and adapted semimartingale and let f: \mathbb{R} \rightarrow \mathbb{R} be a function which is twice continuously differentiable. The process (f(X_t))_{t \ge 0} is a semimartingale and the following change of variable formula holds:
f(X_t)=f(X_0)+\int_0^t f'(X_s)dX_s +\frac{1}{2}\int_0^t f''(X_s) d\langle X \rangle_s.

Proof. We assume that the semimartingale (X_t)_{t \ge 0} is bounded. If it is not, we may apply the following arguments to the semimartingale (X_{t \wedge T_n})_{t \ge 0}, where T_n =\inf \{ t \ge 0, X_t \ge n \} and then let n \to \infty. Let \mathcal{A} be the set of two times continuously differentiable functions f for which the formula given in the statement of the theorem holds holds. It is straightforward that \mathcal{A} is a vector space. Let us show that \mathcal{A} is also an algebra, that is also let stable by multiplication. Let f,g \in \mathcal{A}. By using the integration by parts formula with the semimartingales (f(X_t))_{t \ge 0} and (g(X_t))_{t \ge 0}, we obtain
f(X_t)g(X_t)=f(X_0)g(X_0)+\int_0^t f(X_s) dg(X_s) +\int_0^t g(X_s) df(X_s)+\langle f(X),g(X) \rangle_t.
The terms of the previous sum may be separately treated in the following way. Since f,g \in \mathcal{A}, we get:
\int_0^t f(X_s) dg(X_s) =\int_0^t f(X_s) g'(X_s)dX_s +\frac{1}{2}\int_0^t f(X_s)g''(X_s) d \langle X \rangle_s
\int_0^t g(X_s) df(X_s) =\int_0^t g(X_s) f'(X_s)dX_s+\frac{1}{2}\int_0^t g(X_s)f''(X_s) d \langle X \rangle_s
\langle f(X),g(X) \rangle_t= \int_0^t f'(X_s)g'(X_s)d\langle X \rangle_s.
Therefore,
f(X_t)g(X_t) = f(X_0)g(X_0)+\int_0^t f(X_s) g'(X_s)dX_s +\int_0^t g(X_s) f'(X_s)dX_s + \frac{1}{2}\int_0^t f(X_s)g''(X_s) d \langle X \rangle_s+\int_0^t f'(X_s)g'(X_s)d\langle X \rangle_s+\frac{1}{2}\int_0^t g(X_s)f''(X_s) d \langle X \rangle_s
=f(X_0)g(X_0)+\int_0^t (fg)'(X_s)dX_s +\frac{1}{2}\int_0^t (fg)''(X_s) d \langle X \rangle_s.
We deduce that fg \in \mathcal{A}.

As a conclusion, \mathcal{A} is an algebra of functions. Since \mathcal{A} contains the function x \rightarrow x, we deduce that \mathcal{A} actually contains every polynomial function. Now in order to show that every function f which is twice continuously differentiable is actually in \mathcal{A}, we first observe that since X is assumed to be bounded, it take its values in a compact set. It is then possible to find a sequence of polynomials P_n such that, on this compact set, P_n uniformly converges toward f, P'_n uniformly converges toward f' and P''_n uniformly converges toward f'' \square

As a particular case of the previous formula, if we apply this formula with X as a Brownian motion, we get the formula that was already pointed out at the beginning of the section: If f: \mathbb{R} \rightarrow \mathbb{R} is twice continuously differentiable function, then
f(B_t)=f(0)+\int_0^t f'(B_s)dB_{s}+\frac{1}{2}\int_0^t f''(B_s)ds.

It is easy to derive the following variations of Itō’s formula:

Theorem: (Itō’s formula II) Let (X_t)_{t \ge 0} be a continuous and adapted semimartingale, and let (A_t)_{t \ge 0} be an adapted bounded variation process. If f: \mathbb{R}\times \mathbb{R} \rightarrow \mathbb{R} is a function that is once continuously differentiable with respect to its first variable and that is twice continuously differentiable with respect to its second variable, then for t \ge 0:
f(A_t, X_t)=f(A_0, X_0)+\int_0^t \frac{\partial f}{\partial t} (A_s, X_s)dA_s+\int_0^t \frac{\partial f}{\partial x} (A_s, X_s)dX_s +\frac{1}{2}\int_0^t \frac{\partial^2 f}{\partial x^2}(A_s, X_s) d \langle X \rangle_s.

Theorem. (Itō’s formula III) Let (X^1_t)_{t \ge 0},…,(X^n_t)_{t \ge 0} be n adapted and continuous semimartingales and let f: \mathbb{R}^n \rightarrow \mathbb{R} be a twice continuously differentiable function. We have:
f(X^1_t,...,X^n_t)= f(X^1_0,...,X^n_0)+\sum_{i=1}^n\int_0^t \frac{\partial f}{\partial x_i} (X^1_s,...,X^n_s)dX^i_s +\frac{1}{2} \sum_{i,j=1}^n \int_0^t\frac{\partial^2 f}{\partial x_i \partial x_j} (X^1_s,...,X^n_s) d\langle X^i,X^j \rangle_s.

Exercise. Let f: \mathbb{R}_{\ge 0} \times \mathbb{R} \rightarrow \mathbb{C} be a function that is once continuously differentiable with respect to its first variable and twice continuously differentiable with respect to its second variable that satisfies
\frac{1}{2} \frac{\partial^2 f}{\partial x^2}+\frac{\partial f}{\partial t}=0. Show that if (M_t)_{t \ge 0} is a continuous local martingale, then (f(\langle M \rangle_t,M_t))_{t \ge 0} is a continuous local martingale. Deduce that for \lambda \in \mathbb{C}, the process \left( \exp (\lambda M_t -\frac{1}{2} \lambda^2 \langle M \rangle_t )\right)_{t \ge 0} is a local martingale.

Exercise. The Hermite polynomial of order n is defined as
H_n (x)=(-1)^n e^{\frac{x^2}{2}} \frac{d^n}{dx^n} e^{-\frac{x^2}{2}}.

  • Compute H_0, H_1,H_2,H_3.
  • Show that if (B_t)_{t \ge 0} is a Brownian motion, then the process \left(t^{n/2}H_n (\frac{B_t}{\sqrt{t}})\right)_{t \ge 0} is a martingale.
  • Show that
    t^{n/2}H_n (\frac{B_t}{\sqrt{t}})=n! \int_0^t \int_0^{t_1} ... \int_0^{t_{n-1}} dB_{s_1}...dB_{s_n}.

This entry was posted in Stochastic Calculus lectures. Bookmark the permalink.

Leave a comment