# Find the quadratic variation process of $\int f(s) \, dB_s$

Let $$f \in L^2[a,b]$$ and let $$\displaystyle M(t)=\int_a^tf(s)dB(s)$$.
Find the quadratic variation process, $$[M]_t$$ , of $$M(t)$$.

Here the quadratic variation process is the limit in probability of $$\sum\limits_{i=1}^n(M(t_i)-M(t_{i-1}))^2$$ where $$a=t_0<\cdots is a partition of $$[a,t]$$ and the limit is taken as $$\Vert\Delta_n\Vert=\max\limits_{1\le i \le n}(t_i-t_{i-1}) \to 0$$.

Also above, $$B(t)$$ is the standard Brownian Motion.

Im guess that $$[M]_t=\int_a^tf(s)^2ds$$ but I am having trouble showing this. Here is what I have tried.

\begin{align} & \phantom{ {}={} } P\left( \left\vert \sum\limits_{i=1}^n\left(M(t_i)-M(t_{i-1})\right)^2 - \int_a^tf(s)^2ds \right\vert > \epsilon \right) \\ &= P\left( \left\vert \sum\limits_{i=1}^n\left(\int_{t_{i-1}}^{t_i}f(s)dB(s))\right)^2 - \int_a^tf(s)^2ds \right\vert > \epsilon \right) \\ &\le\frac{ \mathrm{Var}\sum\limits_{i=1}^n \left(\int_{t_{i-1}}^{t_i}f(s)dB(s)\right)^2}{\epsilon^2} \\ &=\dfrac{\sum\limits_{i=1}^n2\left(\int_{t_{i-1}}^{t_i}f(s)^2ds\right)^2}{\epsilon^2} \end{align}

Where above the inequality comes from Chebychev since $$E\left(\int_{t_{i-1}}^{t_i}f(s)dB(s) \right)^2=E\left(\int_{t_{i-1}}^{t_i}f(s)^2ds\right)$$ and $$\left(\int_{t_{i-1}}^{t_i}f(s)dB(s) \right)$$ are independent because of the independent increments of a Brownian Motion and lastly since $$\left(\int_{t_{i-1}}^{t_i}f(s)dB(s) \right)^2$$ follows a $$\mathrm{Gamma}\left(\frac12,2\int_{t_{i-1}}^{t_i}f(s)^2ds \right)$$ density. I am stuck at this point though.

The assertion follows if we can show that

$$\lim_{\delta \to 0} \sup_{\|\Delta\| \leq \delta} \sum_{i=1}^n \left( \int_{t_{i-1}}^{t_i} f(s)^2 \, ds \right)^2 = 0. \tag{1}$$

Recall the following result (see e.g. here or here)

Let $$u \in L^1([a,b])$$ be an integrable function. Then $$u$$ is uniformly integrable, i.e. for any $$k \in \mathbb{N}$$ there exists a constant $$r>0$$ such that $$\int_A |u(s)| \, ds \leq \frac{1}{k}$$ for all measurable sets $$A \subseteq [a,b]$$ with Lebesgue meausre $$\leq r$$.

Fix $$k \in \mathbb{N}$$. Since $$u := f^2$$ is integrable, we can choose $$r>0$$ such that $$\int_A |f(s)|^2 \, ds \leq 1/k$$ for any measurable set $$A$$ with Lebesgue measure $$\leq r$$. If $$\Delta_n$$ is a partition of $$[a,t]$$ with $$\|\Delta_n\| \leq r$$ we get

\begin{align*} \sum_{i=1}^n \left( \int_{t_{i-1}}^{t_i} f(s)^2 \, ds \right)^2&\leq \frac{1}{k} \sum_{i=1}^n \left( \int_{t_{i-1}}^{t_i} f(s)^2 \, ds \right) \\ &= \frac{1}{k} \int_a^t f(s)^2 \, ds. \end{align*}

Hence,

$$\limsup_{\delta \to 0} \sup_{\|\Delta\| \leq \delta} \sum_{i=1}^n \left( \int_{t_{i-1}}^{t_i} f(s)^2 \, ds \right)^2 \leq \frac{1}{k},$$

and since $$k \in \mathbb{N}$$ is arbitrary this proves the assertion.

A final remark regarding your reasoning: To get the last equality in your computations I would rather use that $$\int_u^v f(s) \, dB_s$$ is Gaussian with mean zero and variance $$\int_u^v f(s)^2 \, ds$$ (.. note that this allows you to compute all moments of $$\int_u^v f(s) \, dB_s$$). There is no need to know the distribution of the squared integral.

• Thank you, when I was working on it I was saying $\sum_{i=1}^n \left( \int_{t_{i-1}}^{t_i} f(s)^2 \, ds \right)^2 < \sum_{i=1}^n \beta^2 = n \beta^2$ instead of doing what you did by saying $\sum_{i=1}^n \left( \int_{t_{i-1}}^{t_i} f(s)^2 \, ds \right)^2 < \sum_{i=1}^n \beta \left( \int_{t_{i-1}}^{t_i} f(s)^2 \, ds \right) = \beta \left( \int_{a}^{b} f(s)^2 \, ds \right)$ for some arbitrary $\beta >0$ when the partition is sufficiently fine by the uniform continuity of the integral – alpastor May 1 '19 at 14:07
• @alpastor Actually I'm using nothing but the uniform continuity of the integral (... and I realize now that uniform continuity is somewhat easier to prove than the uniform integrability; the mapping $g(t) := \int_a^t f(s)^2 \, ds$ is bounded, continuous and monotone; hence uniformly continuous.) – saz May 1 '19 at 15:21

You may be overthinking this. Because $$f$$ is square integrable, the function $$g(u):=\int_a^u f(s)^2\,ds$$ is continuous. Consequently, \eqalign{ \sum_{i=1}^n\left(\int_{t_{i-1}}^{t_i} f(s)^2\,ds\right)^2 &\le\max_{1\le i\le n}[g(t_i)-g(t_{i-1}]\cdot \sum_{i=1}^n\int_{t_{i-1}}^{t_i} f(s)^2\,ds\cr &=\max_{1\le i\le n}[g(t_i)-g(t_{i-1}]\cdot \int_{a}^{t} f(s)^2\,ds\cr } and the max above tends to $$0$$ as $$n\to\infty$$ because $$g$$ is uniformly continuous on $$[a,t]$$. This is all that's needed to finish your Chebyshev estimate argument.

• Yep, I wrote in my comment responding to saz above where I was getting confused. I see how to do it now though. Thanks! – alpastor May 1 '19 at 18:08