Lecture 21. Itō’s formula

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))
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

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.
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 Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s