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 $a=t_0<\cdots<t_n=t$ 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*}


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

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.

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  • $\begingroup$ 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 $\endgroup$ – alpastor May 1 '19 at 14:07
  • $\begingroup$ @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.) $\endgroup$ – 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.

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

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