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Assume we have an Itô process of the form :

$$X_t=X_a+\int_a^t f(s)dB(s)+\int g(s)ds$$

(or $dX_t=f(t)dB(t)+g(t)dt$).

I would like to calculate the quadratic variation of the process using the definition:

$$\sum_i (X_{t_{i}}-X_{t_{i-1}})^2=\sum_i \bigg(X_a+\int_a^{t_{i}}f(s)dB(s)+\int_a^{t_{i}}g(s)ds-X_a-\int_a^{t_{i-1}}f(s)dB(s)-\int_a^{t_{i-1}}g(s)ds\bigg)^2$$

I guess I could the write this as

$$=\sum_i\bigg(\int_{t_{i-1}}^{t_{i}}f(s)dB(s)+\int_{t_{i-1}}^{t_{i}}g(s)ds\bigg)^2$$

If I assume $f$ is a simple stochastic process I write

$$=\sum_i\bigg(f(s_{i-1})(B_{s_i}-B_{s_{i-1}})+g(s_{i-1})(s_i-s_{i-1})\bigg)^2$$ $$=\sum |f(s_{i-1})|^2(B_{s_i}-B_{s_{i-1}})^2+2\sum f(s_{i-1})g(s_{i-1})(B_{s_i}-B_{s_{i-1}})(s_i-s_{i-1})+\sum g(s_{i-1})^2(s_i-s_{i-1})^2$$

I suppose that the second and third summation tend to zero when the mesh of our partitions goes to zero (although I don't know how to formalize it here). The third should go to zero since the quadratic variation of a function of finite variation should be zero, the second summation I guess by the continuity of the brownian motion. (?)

And the first one should converge to the Riemann integral $$\int_a^t f(s)^2 ds$$

Now my doubts are:

  • I may have performed some barely-legal-steps in my calculations, could you point them out?
  • How to show that indeed the first summation converges (in which sense?) to the Riemann integral?
  • What if $f$ is not simple, how can I extend this to that case?

Thanks in advance.

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    $\begingroup$ What are your assumptions on $f$, $g$? (Random or deterministic? Bounded or just square integrable? Continuous or just measurable?) $\endgroup$
    – saz
    Jan 25, 2020 at 6:44
  • $\begingroup$ I should have clarified that aspect, $f$ is adapted and with trajectories in $L^2[a,b]$ almost sure, $g$ instead is adapted with trajectories in $L^1[a,b]$ almost sure. $\endgroup$
    – Chaos
    Jan 25, 2020 at 8:39
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    $\begingroup$ Could you please add this information to the body of your question? Makes it easier for other readers to understand what exactly you are interested in $\endgroup$
    – saz
    Jan 25, 2020 at 10:29
  • $\begingroup$ I am editing the body of the question. Thanks for pointing that out $\endgroup$
    – Chaos
    Jan 25, 2020 at 11:51

1 Answer 1

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Since you only impose mild assumptions on $f$, $g$, the proof is somewhat technical, e.g. we cannot work with Riemann sums because $f^2$ might not be Riemann integrable.

Without loss of generality, I will assume that $a=0$ and $X_0=0$. Write $X_t = M_t+A_t$ where $$M_t := \int_0^t f(s) \, dB_s \qquad A_t := \int_0^t g(s) \, ds.$$ If we denote by $\langle \cdot,\cdot \rangle$ the quadratic (co)variation, then $$\langle X,X \rangle_t = \langle M+A,M+A \rangle_t = \langle M,M \rangle_t + 2 \langle M,A \rangle_t + \langle A,A \rangle_t. \tag{1}$$ This follows by a straight-forward computation similar to that in your question. We are going to show that \begin{align*} \langle M,M \rangle_t &= \int_0^t f(s)^2 \, ds \tag{2} \\ \langle M,A \rangle_t &= 0 \tag{3} \\ \langle A,A \rangle_t &= 0. \tag{4} \end{align*}

Proof of $(4)$:

Let $g=g(t,\omega)$ be a measurable function such that $g(\cdot,\omega) \in L^1([0,T])$ for any $T>0$. Then $t \mapsto A_t(\omega) = \int_0^t g(s,\omega) \, ds$ is a continuous function for each $\omega$ and so $A_{\bullet}(\omega)$ is uniformly continuous on $[0,T]$. If $\Pi=\{0=t_0<\ldots<t_n=T\}$ is a partition of $[0,T]$ with mesh size $|\Pi|$, then \begin{align*} \sum_i |A_{t_{i+1}}-A_{t_i}|^2 &\leq \sup_{|s-t| \leq |\Pi|, s,t \in [0,T]} |A_{s}-A_t| \sum_{i=1}^n |A_{t_{i+1}}-A_{t_i}| \\ &\leq \sup_{|s-t| \leq |\Pi|, s,t \in [0,T]} |A_{s}-A_t| \int_0^T |g(s)| \, ds. \end{align*} Because of the uniform continuity on $[0,T]$, the right-hand side converges a.s. to $0$ as the mesh size $|\Pi|$ tends to zero. This proves $\langle A,A \rangle_T=0$.

Proof of $(3)$:

This is quite similar to the previous proof. Take measurable $f,g$ such that $f(\cdot,\omega) \in L^2([0,T])$ and $g(\cdot,\omega) \in L^1([0,T])$ for $T>0$. The stochastic integral $M_t = \int_0^t f(s) \, dB_s$ has continuous sample paths with probability $1$. Exactly as in the previous part, we get $$\sum_i |M_{t_{i+1}}-M_{t_i}| \, |A_{t_{i+1}}-A_{t_i}| \leq \sup_{|s-t| \leq |\Pi|, s,t \in [0,T]} |M_s-M_t| \int_0^T |g(s)| \,ds.$$ Because of the uniform continuity on compact time intervals, the right-hand side converges to $0$ as $|\Pi| \to 0$. Hence, $\langle M,A \rangle_T=0$ for all $T>0$.

Proof of $(2)$:

For simple functions this is a straight-forward calculation, see this question. To extend $(2)$ to a larger class of functions, we need to use approximation techniques. For brevity of notation set $$S_{\Pi}(Y,Z) := \sum_{i} (Y_{t_{i+1}}-Y_{t_i})(Z_{t_{i+1}}-Z_{t_i})$$ and $S_{\Pi}(Y) =: S_{\Pi}(Y,Y)$.

Case 1: $f$ satisfies $\mathbb{E}\int_0^T f(s)^2 \, ds < \infty$ for each $T>0$.

Since $f$ is progressively measurable and satisfies the above integrability condition, there exists a sequence of simple functions $(f_n)_{n \in \mathbb{N}}$ such that $$\mathbb{E}\int_0^T |f(s)-f_n(s)|^2 \, ds \to 0, \qquad T>0 \tag{5}$$ and $$\mathbb{E} \left| \int_0^T f_n(s) \, dB_s - \int_0^T f(s) \, dB_s \right|^2 \to 0, \qquad T>0. \tag{6}$$ Set $M_n(t):=\int_0^t f_n(s) \, dB_s$ and fix $T>0$. We have $$\langle M,M \rangle_T = \langle M-M_n,M-M_n \rangle_T + 2 \langle M-M_n,M_n \rangle_T + \langle M_n,M_n \rangle_T. \tag{7}$$ Let $\Pi$ be a partition of $[0,T]$ . Taking expectation and applying Itô's isometry, we find \begin{align*} \mathbb{E}(S_{\Pi}(M-M_n)) &= \sum_i \mathbb{E}\int_{t_i}^{t_{i+1}} (f_n(s)-f(s))^2 \, ds \\ &= \mathbb{E}\int_0^T (f_n(s)-f(s))^2 \, ds \end{align*} Letting $|\Pi| \to 0$ using Fatou's lemma, we get

$$\mathbb{E}(\langle M-M_n \rangle_T) \leq \mathbb{E}\int_0^T (f_n(s)-f(s))^2 \, ds \xrightarrow[n \to \infty]{(5)} 0, $$ which shows that $\langle M-M_n\rangle_T \to 0$ in $L^1$. Similarily, an application of Itô's isometry (combined with the polarization identity, see here) shows that $$\mathbb{E}(S_{\Pi}(M-M_n,M_n)) = \mathbb{E}\int_0^T (f_n(s)-f(s)) f_n(s) \, ds.$$ Applying the Cauchy-Schwarz inequality and using $(5)$, it follows that the right-hand side converges to $0$ as $n \to \infty$ (uniformly in $\Pi$), and so $\langle M-M_n,M_n \rangle_T \to 0$ in $L^1$. Finally we already know that $\langle M_n,M_n \rangle_T=\int_0^T f_n(s)^2$ and so $$\lim_{n \to \infty} \langle M_n,M_n \rangle_T = \int_0^T f(s)^2 \, ds.$$ Letting $n \to \infty$ in $(7)$ proves the assertion.

Case 2: $\int_0^t f(s)^2 \, ds < \infty$ with probability $1$.

In order to extend $(2)$ such function we need to truncate $f$, e.g. consider $f_n := (-n) \vee f \wedge n$. Each $f_n$ satisfies the integrability assumption from Case 1, and so we know the quadratic variation of $\int_0^t f_n(s) \, dB_s$. Now, similar as in the previous part, we can use this knowledge to compute the quadratic varion of $\int_0^T f(s) \, dB_s$. Let me know in case that you really want to see all the details.

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  • $\begingroup$ +1 Wow such an amazing answer, I'm accepting it since it solves my original question, I'll read all the details and eventually I'll let you know if something is not clear! Thanks again $\endgroup$
    – Chaos
    Jan 25, 2020 at 11:48
  • $\begingroup$ @RScrlli Sure, let me know if something is not clear. $\endgroup$
    – saz
    Jan 25, 2020 at 11:48
  • $\begingroup$ +1 Fantastic answer $\endgroup$
    – Mdoc
    Jan 25, 2020 at 21:20
  • $\begingroup$ Hey @saz sorry to bother you, I don't quite get the result after the formula $(7)$. I don't see how $\mathbb E(S(M-M_n))\rightarrow 0$ as $n\rightarrow \infty$ implies that the quadratic variation of $(M-M_n)$ equals $0$. As far as I understand the q.v. is calculated by letting the mesh of the partition go to $0$. But in this case we are letting $n$ go to $\infty$. Thanks in advance! $\endgroup$
    – Chaos
    Feb 7, 2020 at 8:54
  • $\begingroup$ Does it has to do with the Law of Large Numbers? $\endgroup$
    – Chaos
    Feb 7, 2020 at 9:05

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