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Let

  • $(\Omega,\mathcal{A},\operatorname{P})$ be a probability space
  • $\mathbb{F}=(\mathcal{F}_t)_{t\ge 0}$ be a filtration on $(\Omega,\mathcal{A})$
  • $H=(H_t)_{t\ge 0}$ be a real-valued stochastic process and
    • $H$ be product measurable, i.e. $$(\omega,t)\mapsto H_t(\omega)$$ measurable with respect to $\mathcal{A}\otimes\mathcal{B}([0,\infty))$-$\mathcal{B}(\mathbb{R})$
    • $H$ be adapted to $\mathbb{F}$ and $$\left\|H\right\|^2:=\operatorname{E}\left[\int_0^\infty H_t^2\;dt\right]<\infty$$

Now, let $\mathcal{E}_0$ be the set of all such $H$ and $\mathcal{E}$ be the set of all $H$ of the form $$H_t(\omega)=\sum_{i=1}^nh_{i-1}(\omega)1_{(t_{i-1},t_i]}$$ where $n\in\mathbb{N}$, $0=t_0<\ldots<t_n$ and $h_{i-1}$ is bounded and $\mathcal{F}_{t_{i-1}}$-measurable.


I want to show: If $H$ is $\mathbb{F}$-progressively measurable, i.e. for all $t\ge 0$ $$\Omega\times[0,t]\to\mathbb{R}\;,\;\;\;(\omega,s)\mapsto X_s(\omega)$$ is measurable with respect to $\mathcal{F}_t\otimes\mathcal{B}([0,t])$-$\mathcal{B}(\mathbb{R})$, and $\left\|H\right\|^2<\infty$, then $H\in\overline{E}$, where $\overline{E}$ denotes the closure of $\mathcal{E}$ with respect to $\mathcal{E}_0$:

Please note that $\mathbb{F}$-progressive measurability implies product measurability and $\mathbb{F}$-adaptedness.


So, I need to show that for all such $H$, there exists a sequence $(H^n)_{n\in\mathbb{N}}\subseteq\mathcal{E}$ with $$\left\|H^n-H\right\|^2\stackrel{n\to\infty}{\to}0$$

Probably, it's simple. But I really don't understand why it suffices to show that for all $T>0$ there exists such a sequence $(H^n)_{n\in\mathbb{N}}$ with $$\operatorname{E}\left[\int_0^T\left(H_t^n-H_t\right)^2\;dt\right]\stackrel{n\to\infty}{\to}0$$

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Fix $\epsilon>0$. Since $\|H\|^2 < \infty$, we can choose $R>0$ sufficiently large such that

$$\mathbb{E} \left( \int_R^{\infty} H_t^2 \, dt \right) < \epsilon.$$

By assumption, there exists $H^n \in \mathcal{E}$ such that

$$\mathbb{E} \left( \int_0^R |H_t^n-H_t|^2 \, dt \right) \leq \epsilon.$$

Without loss of generality, we may assume $H^n_t = 0$ for all $t>R$ (otherwise, consider $H^n \cdot 1_{[0,R]}$). Consequently, we obtain

$$\|H^n-H\|^2 = \mathbb{E} \left( \int_0^R |H^n_t-H_t)|^2 \, dt \right) + \mathbb{E} \left( \int_R^{\infty} |H_t|^2 \, dt \right) \leq 2 \epsilon.$$

This shows that we construct a sequence $(H^n) \subseteq \mathcal{E}$ such that $\|H^n-H\|^2 \to 0$.

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  • $\begingroup$ Let $$H^n_t(\omega):=\sum_{i=0}^{n-1}H_{\frac inT}(\omega)1_{\left(\frac inT,\frac {i+1}nT\right]}(t)\;\;\;\text{for }(\omega,t)\in\Omega\times[0,\infty)$$ If $H$ is left-continuous, then $$H^n_t(\omega)\stackrel{n\to\infty}{\to}H_t(\omega)\;\;\;\text{for all }(\omega,t)\in\Omega\times(0,\infty)\tag{1}$$ (Please correct me, but $(1)$ should be false for $t=0$, since $H_0^n\equiv 0\not\equiv H_0$ in general). What I want to do now, is use $(1)$ to show $$\operatorname{E}\left[\int_0^T(H^n_t-H_t)^2\;dt\right]\stackrel {n \to \infty}{\to}0$$ and I think we can use Dominated convergence. $\endgroup$ – 0xbadf00d May 12 '15 at 11:32
  • $\begingroup$ So, if we assume that $H$ (viewed as a random variable $\Omega\times [0,\infty)\to\mathbb{R}$) is bounded, we should be fine, right? $\endgroup$ – 0xbadf00d May 12 '15 at 11:40
  • $\begingroup$ @0xbadf00d Just add $H_0 1_{\{0\}}(t)$ to $H^n_t$, then $H_t^n \to H_t$ holds also true for $t=0$. And yes, if we assume boundedness, then we can use dominated convergence, but in general $H$ may not be bounded. $\endgroup$ – saz May 12 '15 at 15:09
  • $\begingroup$ math.stackexchange.com/questions/1279298/… $\endgroup$ – 0xbadf00d May 12 '15 at 19:56

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