Lecture 17. Some properties of the Itō integral

In this lecture, we give study properties of the Itō integral that was defined in the previous lecture. The following proposition is easy to prove and its proof is left to the reader as an exercise.

Proposition: Let $u,v \in L^2 (\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P})$,
show that
$\mathbb{E} \left( \int_0^{+\infty} u_s dB_s \right)=0$
and
$\mathbb{E} \left( \int_0^{+\infty} u_s dB_s \int_0^{+\infty} v_s dB_s \right)=\mathbb{E} \left( \int_0^{+\infty} u_s v_s ds \right).$

Associated with Itō’s integral, we can construct an integral process, its fundamental property is that it is continuous martingale.

Proposition: Let $u \in L^2 (\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P})$. The process
$\left( \int_0^t u_s dB_s \right)_{t \ge 0}=\left( \int_0^{+\infty} u_s 1_{[0,t]}(s)dB_s \right)_{t \ge 0}$
is a martingale with respect to the filtration $(\mathcal{F}_t)_{t \ge 0}$ that admits a continuous modification.

Proof:

We first prove the martingale property. If
$u_t=\sum_{i=0}^{n-1} F_i 1_{(t_i,t_{i+1}]} (t)$
is in $\mathcal{E}$, then for every $t \ge s$,
$\mathbb{E} \left( \int_0^t u_v dB_v \mid \mathcal{F}_s \right)$
$=\mathbb{E} \left( \sum_{i=0}^{n-1} F_i (B_{t_{i+1}\wedge t }-B_{t_i \wedge t}) \mid \mathcal{F}_s \right)$
$=\sum_{i=0}^{n-1} F_i (B_{t_{i+1}\wedge s}-B_{t_i \wedge s})$
$=\int_0^s u_v dB_v.$

Thus if $u \in \mathcal{E}$, the process
$\left( \int_0^t u_s dB_s \right)_{t \ge 0}=\left( \int_0^{+\infty} u_s 1_{[0,t]}(s)dB_s \right)_{t \ge 0}$
is a martingale with respect to the filtration $(\mathcal{F}_t)_{t \ge 0}$. Since $\mathcal{E}$ is dense in $L^2 (\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P})$, and since it is easily checked that a limit in $L^2 (\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P})$ of martingales is still a martingale, we deduce the expected result.

We now prove the existence of a continuous version.

If $u \in \mathcal{E}$, the continuity of the integral process easily stems from the continuity of the Brownian paths. Let $u \in L^2 (\Omega,(\mathcal{F}_t)_{t \ge 0},\mathbb{P})$ and let $u^n$ be a sequence in $\mathcal{E}$ that converges to $u$. From Doob’s inequality, we have for $m,n \ge 0$ and $\varepsilon > 0$,
$\mathbb{P} \left( \sup_{ t \ge 0} \left| \int_0^t (u^n_s-u^m_s)dB_s\right| \ge \varepsilon \right)$
$\le \frac{\mathbb{E} \left( \mid\int_0^{+\infty} (u^n_s-u^m_s)dB_s \mid^2\right)}{\varepsilon^2}$
$\le \frac{\mathbb{E} \left( \int_0^{+\infty}(u^n_s-u^m_s)^2 ds \right)}{\varepsilon^2}$.
There exists thus a sequence $(n_k)_{k \ge 0}$ such that
$\mathbb{P} \left( \sup_{ t \ge 0} \left| \int_0^t (u^{n_{k+1}}_s-u^{n_k}_s)dB_s \right| \ge \frac{1}{2^k} \right) \le \frac{1}{2^k}.$
From Borel-Cantelli lemma, the sequence of processes $\left( \int_0^t u^{n_k}_sdB_s\right)_{t \ge 0}$ converges then almost surely uniformly to the process $\left( \int_0^t u_sdB_s\right)_{t \ge 0}$ which is therefore continuous $\square$

As a straightforward consequence of the previous proposition and Doob’s inequalities, we obtain

Proposition Let $u \in L^2 (\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P})$.

• For every $\lambda > 0$,
$\mathbb{P} \left( \sup_{t \ge 0} \left| \int_0^t u_s dB_s \right| \ge \lambda \right) \le \frac{\mathbb{E} \left( \int_0^{+\infty} u_s^2 ds \right)}{\lambda^2};$
• $\mathbb{E}\left( \left( \sup_{t \ge 0} \left| \int_0^t u_s dB_s \right|\right)^2 \right) \le 4 \mathbb{E} \left( \int_0^{+\infty} u_s^2 ds \right).$

For $u \in L^2 (\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P})$, the Riemann sums $\sum_{k=0}^{n-1} u_{\frac{kt}{n}} \left( (B_{\frac{(k+1)t}{n}}-B_{\frac{kt}{n}} \right)$,
need not to almost surely converge to $\int_0^t u_s dB_s$. However the following proposition shows that under weak regularity assumptions we have a convergence in probability.

Proposition: Let $u \in L^2 (\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P})$ be a left continuous process. Let $t \ge 0$. For every sequence of subdivisions $\Delta_n [0,t]$ such that $\lim_{n \rightarrow +\infty}\mid\Delta_n [0,t]\mid=0$, the following convergence holds in probability:
$\lim_{n \rightarrow +\infty} \sum_{k=0}^{n-1}u_{t^n_{k}} \left( B_{t^n_{k+1}}-B_{t^n_{k}}\right)=\int_0^t u_s dB_s.$

Proof.

Let us first assume that $u$ is bounded almost surely. We have
$\sum_{k=0}^{n-1}u_{t^n_{k}} \left( B_{t^n_{k+1}} -B_{t^n_{k}}\right)=\int_0^t u^n_s dB_s,$
where $u^n_s =\sum_{k=0}^{n-1}u_{t^n_{k}} 1_{(t^n_{k}, t^n_{k+1}]}(s)$. The Itō’s isometry and the Lebesgue dominated convergence theorem shows then that $\int_0^t u^n_s dB_s$ converges to $\int_0^t u_s dB_s$ in $L^2$ and therefore in probability. For general $u$‘s we can use a localization procedure. For $N \ge 0$, consider the random time $T_N=\inf \{ t \ge 0, | u_t | \ge N \}.$

We have for every $\varepsilon >0$,
$\mathbb{P} \left( \left| \sum_{k=0}^{n-1}u_{t^n_{k}} \left( B_{t^n_{k+1}} -B_{t^n_{k}}\right) - \int_0^t u_s dB_s \right| \ge \varepsilon \right)$
$\le \mathbb{P}(T_N \le t)+\mathbb{P} \left( \left| \sum_{k=0}^{n-1}u_{t^n_{k}} \left( B_{t^n_{k+1}} -B_{t^n_{k}}\right) - \int_0^t u_s dB_s \right| \ge \varepsilon , T_N \ge t\right)$
$\le \mathbb{P}(T_N \le t)+\mathbb{P} \left( \left| \sum_{k=0}^{n-1}u_{t^n_{k}} 1_{|u_{t^n_{k}} |\le M } \left( B_{t^n_{k+1}} -B_{t^n_{k}}\right) - \int_0^t u_s 1_{ |u_{s} |\le M }dB_s \right| \ge \varepsilon \right).$
This easily implies the convergence in probability.

$\square$

Exercise:

• Show that for $t \ge 0$,
$\int_0^t B_s dB_s =\frac{1}{2} \left( B_t^2 -t \right).$
What is surprising in this formula ?
• Show that when $n \rightarrow +\infty$, the sequence
$\sum_{k=0}^{n-1} B_{\frac{(k+1)t}{n}} \left( B_{\frac{(k+1)t}{n}}-B_{\frac{kt}{n}} \right),$
converges to a random variable that shall be computed.

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