0

Let $b_S$ be an $F_t$ adapted process, Borel measurable in $t$ st $\int |b_s|^2ds < \infty$ (a.s). Setting $X_t=\int^t_0 b_sds$ and partitioning the interval $[0,t]$ i.e. $0=t^n_0<t^n_1... $ such that $d_n=\max_i |t^n_{i+1}-t^n_i| \rightarrow 0$ as $n \rightarrow \infty$, prove that

$\sum_i |X_{t^n_{i+1}} - X_{t^n_{i}} |^2 \rightarrow 0$ (a.s)

Heres my answer, I was wondering if you guys agreed with it:

  • $|X_{t_{i+1}} - X_{t_{i}} |^2 = |\int^{t_{i+1}}_0 b_s ds - \int^{t_{i}}_0 b_s ds|^2 = |\int^{t_{i+1}}_{t_{i}} b_s ds|^2 \leq \int^{t_{i+1}}_{t_{i}} |b_s|^2 ds$
  • $E|X_{t^n_{i+1}} - X_{t^n_{i}} |^2$ via Fatou is $\leq \int^{t^n_{i+1}}_{t^n_{i}} E|b_s|^2 ds = E|b_s|^2(t^n_{i+1}-t^n_i) \leq E|b_s|^2d_n$

Therefore as $n \rightarrow \infty$, $E|X_{t^n_{i+1}} - X_{t^n_{i}} |^2 \rightarrow 0$ hence by summing this over all $i$, $E\sum_i |X_{t^n_{i+1}} - X_{t^n_{i}} |^2 \rightarrow 0$ hence $\sum_i |X_{t^n_{i+1}} - X_{t^n_{i}} |^2 \rightarrow 0$ (a.s) - yes?

saz
  • 123,507
user127159
  • 93

1 Answers1

2

First of all, note that by Jensen's inequality

$$|X_{t_{i+1}}-X_{t_i}|^2 = \left| \int_{t_i}^{t_{i+1}} b_s ds \right|^2 \leq (t_{i+1}-t_i) \int_{t_i}^{t_{i+1}} b_s^2 \, ds. \tag{1}$$

Hence,

$$\sum_i |X_{t_{i+1}}-X_{t_i}|^2 \leq \sup_i |t_{i+1}-t_i|\cdot \sum_i \int_{t_i}^{t_{i+1}} b_s^2 \, ds = \sup_i |t_{i+1}-t_i|\cdot \int_0^t b_s^2 \, ds.$$

Letting the mesh size $\sup_i |t_{i+1}-t_i| \to 0$ finishes the proof.

saz
  • 123,507
  • I don't see how Jensen's Inequality applies in Equation 1... – user127159 Mar 26 '14 at 15:50
  • 1
    Well, Jensen's inequality states that $$V \left( \int b(s) , d\mu(s) \right) \leq \int V(b(s)) , d\mu(s)$$ where $V$ is a convex function and $\mu$ a probability measure. Apply this inequality for $V(x)=x^2$ and $\mu(s) = \frac{ds}{t_{i+1}-t_i}$. – saz Mar 26 '14 at 16:03
  • Alrite - Btw I still don't understand where I went wrong with my solution... – user127159 Mar 26 '14 at 16:15
  • 1
    @user127159 First of all, you made a mistake when applying the Jensen inequality. Moreover, there are some integrals missing [$\mathbb{E}(b_s^2)$ should read $\mathbb{E}\int b_s^2 , ds$] and since you only assume $\int b_s^2 ds < \infty$ a.s. this does in general not imply that this expectation is finite. Finally, $L^1$-convergence does not imply pointwise convergence. – saz Mar 26 '14 at 16:20