Defining the stochastic integral as $$\int_0^t\phi_sdB_s$$if I can prove that $\phi_s \in \lambda^2$ (namely that $\phi_s$ is progressive and E[$\int_0^\infty \phi_s^2ds]< \infty$), which are the additional information that it gives me compared to know that $\phi_s \in \lambda^2(T)$ ( $\phi_s$ progressive and E[$\int_0^T \phi_s^2ds]< \infty$)?

And why these conditions imply the existence of the stochastic integral?

  • $\begingroup$ Hopefully you mean $L^2$ instead of $\lambda^2.$ Unless $\lambda$ is another measure... $\endgroup$ Commented Dec 24, 2019 at 3:07
  • $\begingroup$ $\lambda ^2=L^2([0,+\infty] \times \Omega, \mathcal P_r, ds \otimes \mathbb P)$ $\endgroup$
    – Buddy_
    Commented Dec 24, 2019 at 10:03
  • $\begingroup$ I think you have typos in your question and you mean $E\int_0^T \phi^2_sds < \infty$ instead of $E\int_0^T \phi^2_s B_s < \infty.$ $\endgroup$
    – UBM
    Commented Dec 24, 2019 at 10:04
  • $\begingroup$ @UBM yes sorry, now I correct $\endgroup$
    – Buddy_
    Commented Dec 24, 2019 at 10:05

1 Answer 1


In the construction process of the stochastic integral $\int_0^t f(s) dB(s)$, the integral is first defined for integrands that are simple processes and then is extended to general integrands $f$. These general integrands $f$ have to satisfy the condition $E\int_0^T |f(s)|^2 ds < \infty \quad \text{(*)}.$ We need this condition to guarantee that $f$ can be approximated by a sequence of simple processes $\{ g_n \}$. Since we already have defined the integral for simple integrands $g_n$, the stochastic integral for a general integrand $f$ is defined as the limit in $L^2$ of the sequence of random variables $\{ \int_0^T g_n(s) dB(s) \}$ where $\{ g_n \}$ converges in $L^2$ to $f.$

More specifically, assume we have defined the Ito integral (wrt to the BM) for integrands that are simple process and we have also proved the following two properties for that ("simple") integral:

  • Linearity of the integral.

  • Ito isometry.

We use condition (*) to define the space of integrands for which the integral can be defined. Let $\mathscr{M}^{2}([a,b];\mathbb{R})$ be the space of all real-valued measurable {$\mathcal{F}_{t}\}$-adapted process $f=\{f(t)\}_{a\leq t\leq b}$ such that $E\int_{a}^{b}|f(t)|^{2}dt<\infty.$

Lemma 1. For any $f\in\mathscr{M}^{2}([a,b];\mathbb{R}),$ there exists a sequence $\{g_{n}\}$ of simple processes such that $$\lim_{n\rightarrow\infty}E\int_{a}^{b}|f(t)-g_{n}(t)|^{2}dt=0.$$

(This lemma is proved using condition (*). For a proof of the Lemma check Oksendal- pag 27)

Now, in the next theorem, we use Lemma 1 to prove the existence of the integral.

Theorem Let $f \in \mathscr{M}^{2}([a,b];\mathbb{R})$ and let $\{g_{n}\}$ be a sequence of simple processes such that \begin{equation} \lim_{n\rightarrow\infty}E\int_{a}^{b}|f(t)-g_{n}(t)|^{2}dt=0. \end{equation} Then there exists a random variable $X \in L^{2}(\Omega,\mathbb{R})$ such that \begin{equation} \lim_{n\rightarrow\infty}E \Big|\int_{a}^{b}g_{n}(t)dB(t)-X\Big|^{2}=0. \end{equation}

Proof. We know that $X_{n}:=\int_{a}^{b}g_{n}(t)dB(t) \in L^{2}(\Omega,\mathbb{R})$ for all $n \in \mathbb{N}$. We also know that $L^{2}(\Omega,\mathbb{R})$ is a Hilbert space with the norm $||X||^{2}:=E|X|^{2}=\int_{\Omega}|X(\omega)|^{2}dP(\omega)$, thus $L^{2}(\Omega,\mathbb{R})$ is a complete space. Thus, we only have to prove that $\{X_{n}\}$ is a Cauchy sequence in $L^{2}(\Omega,\mathbb{R})$. Let $\epsilon>0$. By Lemma 1, there exists an $n_{0} \in \mathbb{N}$ such that $n \geq n_{0}$ implies $E\int_{a}^{b}|f(t)-g_{n}(t)|^{2}dt < \frac{\epsilon}{4} $. Thus, $ \forall m,n \geq n_{0}$, we have \begin{align*} ||X_{m}-X_{n}||^{2} & = E|X_{m} - X_{n}|^{2}\\ & = E\Big|\int_{a}^{b}g_{m}(t)dB(t) - \int_{a}^{b}g_{n}(t)dB(t) \Big|^{2}\\ & = E\Big|\int_{a}^{b}[g_{m}(t) - g_{n}(t)]dB(t) \Big|^{2} && \tag*{(linearity int)} \\ & = E\int_{a}^{b}|g_{m}(t) - g_{n}(t)|^{2}dt && \tag*{(Ito Isometry)} \\ & = E\int_{a}^{b}|(g_{m}(t) - f(t)) + (f(t) - g_{n}(t))|^{2}dt \\ & \leq 2E\int_{a}^{b}(g_{m}(t) - f(t))^{2}dt + 2E\int_{a}^{b}(g_{n}(t) - f(t))^{2}dt \\ & < \epsilon. \tag*{ (Lemma 1)} \end{align*} Therefore, $\{X_{n}\}$ is a Cauchy sequence $L^{2}(\Omega,\mathbb{R})$ and the limit X exists.

The $L^{2}(\Omega,\mathbb{R})$ random variable X is called the Ito integral of $f$ with respect to the BM and is denoted by $$\int_{a}^{b}f(t)dB(t).$$

If we use $T$ in $E\int_0^T |f(s)|^2 ds < \infty$, we are only guaranteeing the existence of the stochastic integral integral on [0,T].

Note 1: we can define the integral for a more general class of integrands by imposing the weaker condition $\int_0^T |f(s)|^2 ds < \infty \text{ (**) }.$ The class of integrands that satisfy condition $\text{(**)}$ is "bigger" (i.e. it contains the class $\mathscr{M}^{2}([a,b];\mathbb{R}))$ but you loose the martingale property. The process $ I_t = \{ \int_0^t f(s) dB(s) < \infty;$ $0 \leq t \leq T \}$ for integrands $f \in \mathscr{M}^{2}([a,b];\mathbb{R})),$ is an square-integrable martingale. But if $f$ satifies "only" (**), then the process $\{ I_t \}$ is "only" a local martingale.

Note 2: You said in your question that the integrand $f$ must be progessively measurable, but when the integrator is the Brownian motion, it is possible to define the integral for integrands that are adapted (to the B.M. filtration) which is a wider class than the progessively measurable class.

  • $\begingroup$ Could you better explain why it guarantees the existence of the stochastic integral, namely why it imply the existence of the $L^2$ limit? $\endgroup$
    – Buddy_
    Commented Dec 24, 2019 at 11:33
  • $\begingroup$ I edited the answer and give a proof above. The summary is: (condition (*) $\implies$ Lemma 1) and (Lemma 1 $\implies$ existence of the integral). $\endgroup$
    – UBM
    Commented Dec 24, 2019 at 15:41
  • $\begingroup$ very grateful, amazing answer. $\endgroup$
    – Buddy_
    Commented Dec 25, 2019 at 10:51
  • $\begingroup$ I have a question: if I have a stochastic integral define as: $\int_0^{\infty}X_sdB_s$ and I want to verify if this integral is well defined. Then I have to check that $X_s \in \lambda^2_loc$ namely I have to verify that, for all t $\int_0^tX_s^2ds < \infty$ is it right? instead If I wanna verify that $\int_0^tX_sdB_s$ is well defined is sufficient that I check that $X_s \in \lambda^2_{loc}(T)$ that is checking that $\int_0^TX_s^2ds < \infty$ (no more for all T). is it right? and the same reasoning if I wanna verify when an integral is well defined for a $X_s$ in $\lambda^2$ or $\lambda^2(T)$ $\endgroup$
    – Buddy_
    Commented Dec 30, 2019 at 18:34
  • $\begingroup$ in other word, if I want to verify the existence of a stochastic integral, I have to check if the integrands is in $\lambda$ or $\lambda(T)$ depending on the interval of the ingral. If it goes until infinity then I have to check for the integrand in $lambda$ otherwise in $\lambda(T)$ $\endgroup$
    – Buddy_
    Commented Dec 30, 2019 at 18:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .