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The following question pertains to Wengenroth's textbook "Wahrscheinlichkeitstheorie", de Gruyter 2008 (in German).

The covariance (aka compensator) of the continuous local martingales $X, Y \in \mathcal{CM}^{\text{loc}}(\mathcal{F})$ is defined (in Theorem 9.6 on p. 181) as $$ [X, Y] := XY - X_0Y_0 - X\cdot Y - Y\cdot X $$ where $X\cdot Y$ is the Itô integral of $X$ w.r.t. the integrator $Y$: $(X \cdot Y)_t = \int_0^t X_s dY_s$, and likewise $Y\cdot X$. However, the Itô integral w.r.t. an integrator that is a local martingale has only been defined (p. 179) when the integrand is in $\overline{T}(\mathcal{F})$, i.e. when the integrand can be uniformly approximated by a step-process, where a step-process is defined (p. 176) as a stochastic process that is adapted to the filtration $\mathcal{F}$, and such that each of its paths is a step-function, i.e. a function of the form $\sum_{n = 0}^\infty a_n \mathbb{1}_{[t_n, t_{n + 1})} = \sum_{n = 0}^\infty a_n(\omega) \mathbb{1}_{[t_n(\omega), t_{n + 1}(\omega))}$, for some sequence of real numbers $a(\omega) = (a_1(\omega), a_2(\omega), \dots)$ and some sequence of non-negative real numbers $(0 = t_0(\omega), t_1(\omega), \dots)$ that increases monotonically (possibly weakly) to infinity: $t_n(\omega) \uparrow \infty$.

So the definition of covariance/compensator, given above, begs the question: is every local martingale in $\overline{T}(\mathcal{F})$? If not, then the definition makes no sense, does it? Or am I missing something?

Evan Aad
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    What do you mean by "uniformly approximated"? Uniformly on compact $t$-sets? – saz Jun 13 '15 at 11:26
  • @saz: I mean approximated in the $|||\cdot|||\infty$ half-norm, where $|||X|||\infty := \sup {|X_t(\omega)| : \omega \in \Omega, t \in [0, \infty)}$ [Wengenroth, p. 176]. – Evan Aad Jun 13 '15 at 12:40
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    Okay. The point is actually that $X$ and $Y$ are processes with continuous sample paths and this implies that we can approximate them uniformly with step functions. (It is very important that in your definition of step functions the partition $0=t_0(\omega)< t_1(\omega)<\ldots$ may depend on $\omega$; otherwise we are doomed.) – saz Jun 13 '15 at 12:47
  • @saz: But how can we construct uniformly approximating step-processes that are adapted to the filtration? – Evan Aad Jun 13 '15 at 13:00

1 Answers1

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Fix $\epsilon>0$. Define iteratively a sequence of stopping times by

$$\tau_n := \inf\{t>\tau_{n-1}; |X(t)-X(\tau_{n-1})| \geq \epsilon\}, \qquad n \in \mathbb{N},$$

$\tau_0 := 0$. Then

$$f(t,\omega) := \sum_{n \geq 1} X_{\tau_{n-1}}(\omega) 1_{[\tau_{n-1}(\omega),\tau_n(\omega))}(t)$$

satisfies $\|X-f\|_{\infty} \leq \epsilon$ (since $X$ has continuous sample paths) and $f$ is adapted.

saz
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  • I think your stopping times should be defined as follows $\tau_0 := 0$, $\tau_{n + 1} := (\tau_n + 1) \wedge \inf {t \geq \tau_n \mid: |X_t - X_{\tau_n}| \geq \varepsilon} = (\tau_n + 1) \wedge \inf {t \geq 0 \mid: |X_t - X^{\tau_n}_t| \geq \varepsilon}$ (instead of $1$ I could have used any positive constant). This modification endows the stopping times $\tau_i$ with the following three characteristics, which they lack in your current answer. – Evan Aad Jun 26 '15 at 14:05
  • The $\tau_i$'s are real-valued, so that it makes sense to write $X_{\tau_i}$, 2) $\tau_{i + 1} > \tau_i$, so it makes sense to write $\mathbb{1}{\left[\tau{i - 1}(\omega), \tau_i(\omega)\right)}$. (Also, it makes sure the stopping times increase to infinity, which is a nice, albeit not necessary, touch.),
    • – Evan Aad Jun 26 '15 at 14:06
  • The $\tau_i$'s are stopping times: indeed, according to theorem 7.10.2, "Entrance Times", in Wengenroth's textbook (p. 139), if $Y = (Y_t)_{t \in [0, \infty)}$ is a continuous stochastic process and if $B$ is a closed set of real numbers, then $\rho := \inf {t \geq 0 \mid: Y_t \in B}$ is a stopping time w.r.t. the natural filtration generated by $Y$.
    • – Evan Aad Jun 26 '15 at 14:06
  • I would also like to point out that the continuity of $X$ is not needed in the place where you mentioned it explicitly in your answer; rather it is needed in order to ensure that the $\tau_i$'s are stopping times. Finally, I'd like to note that nowhere in your answer is the fact that $X$ is a local martingale used, so that your answer actually applies to a larger class of processes $X$, namely continuous, real-valued stochastic processes defined over the time interval $[0, \infty)$. It is not even required that $X$ be adapted to any filtration. – Evan Aad Jun 26 '15 at 14:07
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    @EvanAad Yeah, sorry, I wanted to define them inductively, i.e. $\tau_n := \inf{t> \tau_{n-1}; \dots}$ (I fixed it). However, there is no need for $\tau_{n+1}$ to be bounded,if $\tau_{n+1}(\omega) = \infty$, then $$f(t,\omega) = \sum_{j=1}^{n} X_{\tau_{j-1}}(\omega) 1_{[\tau_{j-1}(\omega),\tau_j(\omega))}(t)$$ is well-defined (there is no $X_{\tau_n}(\omega)$ needed). Concerning your last remark: The continuity is needed to ensure that $|X-f|_{\infty} \leq \epsilon$. If $X$ is not continuous (i.e. cadlag), then this estimate does not hold true. – saz Jun 26 '15 at 14:13
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    @EvanAad It doesn't really matter since we can use the completed (right-continuous) filtration instead; but you are right, it makes more sense to use $\geq \epsilon$. Thanks for your comments. – saz Jun 26 '15 at 14:22