1
$\begingroup$

I quote Øksendal (2003).

Let $\mathcal{V}=\mathcal{V}(S,T)$ be the class of functions $f(t,\omega):[0,\infty)\times\Omega\to\mathbb{R}$ such that $(t,\omega)\to f(t,\omega)$ is $\mathcal{B}\times\mathcal{F}$-measurable (where $\mathcal{B}$ denotes the Borel $\sigma$-algebra on $[0,\infty)$), $f(t,\omega)$ is $\mathcal{F}_t$-adapted and $\mathbb{E}\bigg[\int_{S}^T f(t,\omega)^2 dt\bigg]<\infty$. [...]

Starting from a probability space $\left(\Omega,\mathbb{P},\mathcal{E}\right)$ and a Brownian motion $\left(B_t\right)_{t\ge0}$, if $\phi(t,\omega)$ is bounded and elementary, then $$\mathbb{E}\left[\left(\int_S^T\phi(t,\omega)dB_t(\omega)\right)^2\right]=\mathbb{E}\left[\int_S^T\phi(t,\omega)^2 dt\right]\tag{1}$$ [...]
If $f\in\mathcal{V}$ one can show that it is possible to choose elementary functions $\phi_n\in\mathcal{V}$ such that: $$\mathbb{E}\left[\int_S^T|f-\phi_n|^2 dt\right]\to0\tag{2}$$ Then, define $$\mathcal{I}\left[f\right](\omega)=\int_S^T f(t,\omega)dB_t(\omega)=\lim_{n\to\infty}\int_S^T\phi_n(t,\omega)dB_t(\omega)\tag{3}$$ The limit exists as an element of $L^2(\mathbb{P})$, since $\left\{\int_S^T\phi_n(t,\omega)dB_t(\omega)\right\}$ forms a Cauchy sequence in $L^2(\mathbb{P})$, by $(1)$.



What I cannot understand is the statement in bold above. Why is that true?

$\endgroup$
2
$\begingroup$

Note that (2) implies: $$\mathbb{E}\left[\int_S^T|\phi_n(t)-\phi_m(t)|^2 dt\right]\to0. \tag{4}$$ Thus \begin{align} E \left|\int_S^T \phi_n(t)dB_t-\int_S^T \phi_m(t)dB_t \right|^2 &=E \left|\int_S^T (\phi_n(t)-\phi_m(t))dB_t \right|^2 \\ &=E \int_S^T |\phi_n(t)-\phi_m(t)|^2 dt \rightarrow 0. \end{align} The last equality follows from the Ito isometry and the convergence to zero follows follows from (4). This proves that $\{\int_S^T \phi_n(t)dB_t \}$ is a Cauchy sequence in $L^2$.

Edit: proof that (2) implies (4): \begin{align} E\left[\int_S^T|\phi_n(t)-\phi_m(t)|^2 dt\right]&=E\left[\int_S^T|\phi_n(t)-f(t)+f(t)-\phi_m(t)|^2 dt\right] \\ &\leq E\left[\int_S^T 2|\phi_n(t)-f(t)|^2+2|f(t)-\phi_m(t)|^2 dt\right] \\ &= 2E\left[\int_S^T |\phi_n(t)-f(t)|^2dt\right]+2E\left[\int_S^T|f(t)-\phi_m(t)|^2 dt\right] \rightarrow 0. \end{align} The '$\leq$' follows from the inequality $(a+b)^2 \leq 2a^2 + 2b^2$ and the convergence to zero from (2).

$\endgroup$
4
  • $\begingroup$ Why does $(2)$ imply $\mathbb{E}\left[\int_S^T|\phi_n(t)-\phi_m(t)|^2 dt\right]\to0$? $\endgroup$ – Strictly_increasing Nov 7 '20 at 18:18
  • $\begingroup$ @Strictly_increasing: I edited my previous answer, hope it helps. $\endgroup$ – UBM Nov 7 '20 at 18:34
  • $\begingroup$ But $(2)$ establishes that $\mathbb{E}\left[\int_S^T|f-\phi_n|^2 dt\right]\to0$ for $f\in\mathcal{V}$, and not that $\mathbb{E}\left[\int_S^T|\phi-\phi_n|^2 dt\right]\to0$, doesn't it? $\endgroup$ – Strictly_increasing Nov 7 '20 at 18:36
  • 1
    $\begingroup$ Sorry, I was accidentally calling $\phi$ what it was suppose to be $f.$ I fixed it in the answer. $\endgroup$ – UBM Nov 7 '20 at 18:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.