## 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:
$f(x_{t+dt})=f(x_t +(x_{t+dt}-x_t))$
$=f(x_t)+f'(x_t)(x_{t+dt}-x_t)$
$=f(x_t)+f'(x_t)dx_{t}.$
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
$t^{n/2}H_n (\frac{B_t}{\sqrt{t}})=n! \int_0^t \int_0^{t_1} ... \int_0^{t_{n-1}} dB_{s_1}...dB_{s_n}.$

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