Definition

An Itô process $X$ is a process such that:

\[X_t = X_0 + \int_0^t a_udu + \int_0^t b_udW_u\]

where $a$ and $b$ are $\mathcal{F}_t$-adapted, continuous at left and $\int_0^t a_u^2 du < \infty$.

$\int_0^t b_udW_u$ is an Itô integral.

Intuitively, an Itô process is a process whose value depends on a trend (drift) and a random parameter (volatility). In other words, $X = X_0 + M + A$ where $X_0$ is the initial value, $M$ is a martingale (includes volatility) and $A$ is the anticipated value (includes drift).

Lemma

The lemma’s main idea is that a smooth function of an Itô process is again an Itô process, with the same Wiener process.

Let $X$ be an Itô process and $f$ a smooth function (i.e. of class $\mathcal{C}^2$), then:

\[df(X_t, t) = \frac{\partial f}{\partial t}(X_t,t)dt + \frac{\partial f}{\partial x}(X_t,t)dX_t + \frac{1}{2} \sigma_t^2 \frac{\partial^2f}{\partial x^2}(X_t,t)(dX_t)^2\]

Using $dX_t = \mu_tdt + \sigma_tdB_t$:

\[df(X_t, t) = \frac{\partial f}{\partial t}(X_t,t)dt + \frac{\partial f}{\partial x}(X_t,t)(\mu_tdt + \sigma_tdB_t) + \frac{1}{2} \sigma_t^2 \frac{\partial^2f}{\partial x^2}(X_t,t)dt\] \[df(X_t, t) = \Big( \frac{\partial f}{\partial t}(X_t,t)+ \frac{\partial f}{\partial x}(X_t,t)\mu_t + \frac{1}{2} \sigma_t^2 \frac{\partial^2f}{\partial x^2}(X_t,t) \Big) dt + \frac{\partial f}{\partial x}(X_t,t)\sigma_tdB_t\]

Notes on the quadratic variation

$(dX_t)^2$ is the quadratic variation of the process $X_t$ (noted $[X_t]$), i.e. the sum of all squared variations:

\[[X_t] = \lim_{||P|| \to 0} \sum_{k=1}^{n}(X_{t_k}-X_{t_{k-1}})^2\]

$P$ is the partition and $||P||$ is the mesh, i.e. the longest of the subinterval.

When $dX_t$ is defined as in $(E)$, we have:

\[(dX_t)^2 = (\mu_tdt + \sigma_tdB_t)^2 = \mu_t^2(dt)^2 + 2\mu_t \sigma_t dt dB_t + \sigma_t^2(dB_t)^2\]

(1) $(dt)^2 = 0$. $dt \sim \frac{1}{N}$ so $(dt)^2 \sim \frac{1}{N^2}$ which tends fast to $0$ when $N \to +\infty$ (continuous time).

(2) $(dB_t)^2 = dt$. Proof (page 11) showing that $(B_t)^2 = t$.

(3) $dB_tdt = 0$. Since $(dB_t)^2 = dt$ as seen previously, we have $dB_tdt = (dt)^{3/2}$. We come to the result using the same reasoning as in (1).