Quadratic Variation for $X_t= \int b_s ds$ where $b_s$ is an $F_t$ adapted process.

Question

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?

Answer 1

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.