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A quick summary of the things regarding my question is this:

You have a probability space $(\Omega, \mathcal{F},P)$ and a filtration $\{\mathcal{F}_t\}$.

You have a $f(t,\omega): [0,\infty)\times\Omega \rightarrow \mathbb{R}$ satisfying.

  1. It is $\mathcal{B}\times\mathcal{F}$-measurable.

  2. $f(t,\omega)$ is $\mathcal{F}$-adapted.

  3. For two real numbers S,T $E[\int_S^Tf(t,\omega)^2dt]<\infty$.

A function $\phi$ is elementary if it satisfies the above points and is of the form :

$$\phi(t,\omega)=\Sigma_j e_j(\omega)\mathcal{X}_{[t_j,t_{j+1})}$$

Øksendal then shows that there is a sequence of elemntary functions $\phi_n$ such that $E[\int_S^T|f-\phi_n|^2dt]\rightarrow 0$.

Then he defines the integral as the $L^2$ limit of $\int_S^T\phi_n(t,\omega)dB_t(\omega)$. He says that the limit exists in $L^2$ since this is a Cauchy sequence. He refers to the Itô-isometry which he has proved for elementary functions, using this I get myself trying to show that the limit is Cauchy in $L^2$ :

$$E[(\int_S^T\phi_n(t,\omega)dB_t(\omega)-\int_S^T\phi_m(t,\omega)dB_t(\omega))^2]=E[(\int_S^T(\phi_n-\phi_m)(t,\omega)dB_t(\omega))^2]=E[\int_S^T(\phi_n-\phi_m)^2dt]$$

But I still can't see how this can get arbitrary small as $n,m$ get big enough. I tried writing $(\phi_n-\phi_m)^2=(\phi_n-f+f-\phi_m)$ and then multiplaying out using the elementary square rules, and then trying to use Hölder's-inequality. But I wasn't able to finish it.

Can you please help me?

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If $x,y$ are non-negative real numbers then $2xy\leq x^2+y^2$, hence $$ |\phi_m-\phi_n|^2\leq (|\phi_m-f|+|f-\phi_n|)^2=|\phi_m-f|^2+2|\phi_m-f||\phi_n-f|+||\phi_n-f|^2$$ $$\leq 2(|\phi_m-f|^2+|\phi_n-f|^2)$$

Now you can integrate from $S$ to $T$ and take expectation.

The big picture of what's going on is that every convergent sequence in a metric space is Cauchy.

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