Consider the Itô-Process $$X(t) = X_{0} + \int_{0}^{t}\mu(s,X(s))\,ds + \int_{0}^{t}\sigma(s,X(s))\,dW(s)$$ where you can safely assume that the drift $\mu$ and the volatility $\sigma$ satisfy the necessary regularity conditions for this SDE to have a strong solution.

I am interested in any tools that would help me to make sense of / investigate an integral of the form $$``\;Y(t)=\int_{0}^{t}\max[\,dX(s),0\,]\,ds\;``$$ i.e. the time integral of the ''positive innovations'' of $X$. In other words, $Y$ is the process that grows when $X$ grows (by the same magnitude), but is idle when $X$ shrinks.

Any comments, ideas and suggestions are highly appreciated. Thank you!

  • $\begingroup$ It doesn't seem like this is sensible on its face. If you replace max by min you should get something with the same distribution up to a sign, and if you subtract them you appear to have the total variation of $X(s)$ up to time $t$, which is typically infinite for all $t$ almost surely. So $Y(t)$ would just be infinity, unless you renormalize somehow. $\endgroup$ – Nate Eldredge Nov 7 '16 at 23:01
  • $\begingroup$ Can you say more about why you want to do this? $\endgroup$ – Nate Eldredge Nov 7 '16 at 23:09

One way to formalize the concept is to approximate $X(t)$ by a Markov jump process, then define the concept for this spatial approximation (where it makes perfect sense), and then take the limit. To construct this approximation, one approximates the infinitesimal generator $L_t$ of the SDE by a suitable spatial difference approximation (see Chapter 2 of this reference for more detail). Let $X^h(t)$ be the resulting Markov jump process where $h$ is a jump size parameter. Note that this process only moves by jumps, and if the approximation is stable, the number of jumps in a finite time interval is a.s. bounded. Define the positive innovation of $X(t)$ as: $$ Y(t) = \lim_{h \to 0} \sum_{s \le t} \max( \Delta X^h(s), 0 ) $$ where $\Delta X^h(s)$ denotes the jump of $X^h$ at $s$ defined as $\Delta X^h(s) = X^h(s+) - X^h(s-)$. This definition leads to a nontrivial $Y(t)$. Indeed, since $\max(a,0)=(a+|a|)/2 \ge a/2$, we have that $$ \sum_{s \le t} |\Delta X^h(s)| \ge \sum_{s \le t} \max( \Delta X^h(s), 0 ) \ge \sum_{s \le t} \frac{1}{2} \Delta X^h(s) \tag{$*$} $$ almost surely. Note that the expected values of the random variables appearing in the upper and lower bounds are well-defined in the limit as $h \to 0$. For example, the expected value of the lower bound converges to $\mathbb{E}_x \int_0^t \mu(s,X(s))/2 ds$, I think.


First, let us confirm numerically that $Y(t)$ is an integrable random variable. For this purpose, suppose $X(t)$ is an OU process with unit drift/noise coefficients; and let $\overline{Y}^h(t)$, $Y^h(t)$ and $\underline{Y}^h(t)$ denote the upper, middle, and lower bounds appearing in ($*$). The figures below plot these quantities for $t=1$ with initial condition $X(0)=1$.

enter image description here

Why are these random variables integrable? The action of the infinitesimal generator of this particular approximation is given by: $$ L^h f(x) = \frac{e^{-h x}}{2 h^2} \left( f(x+h) - f(x) \right) + \frac{e^{h x}}{2 h^2} \left( f(x-h) - f(x) \right) $$ If $f \in C^4_b(\mathbb{R})$, a Taylor expansion about $h=0$ shows that $ L^h f(x) = L f(x) + O(h^2) $ where $L f(x) = - x f'(x) + f''(x)/2$. Moreover, for any $h>0$, this Markov jump process approximation is right continuous with left limits and is a process of finite variation with zero continuous part. Thus, Ito's formula for this process reduces to a telescoping sum: $$ f(X^h(t)) - f(X^h(0)) = \sum_{s \le t} \left( f(X^h(s)) - f(X^h(s-)) \right) $$ Add to both sides of the equation $- \int_0^t L^h f(X^h(s)) ds$ to obtain $$ f(X^h(t)) - f(X^h(0)) - \int_0^t L^h f(X^h(s)) ds = \sum_{s \le t} \left( f(X^h(s)) - f(X^h(s-)) \right) - \int_0^t L^h f(X^h(s)) ds $$ where the LHS is a local martingale (and for suitable functions a true martingale). Thus, we see that $- \int_0^t L^h f(X^h(s)) ds$ is the compensator for $\sum_{s \le t} \left( f(X^h(s)) - f(X^h(s-)) \right)$. Now choose $f(x) = x^2$ and $f(x)=x$ to obtain upper and lower bounds on $Y^h(t)$.

  • $\begingroup$ What do you mean when you say $Y(t)$ is "nontrivial"? I agree it won't be 0, but it seems to me it will typically be $\infty$. I would expect your upper bound to blow up as $h \to 0$. $\endgroup$ – Nate Eldredge Nov 7 '16 at 23:03
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    $\begingroup$ @NateEldredge Trivial means that the pre-limit of $Y(t)$ tends to zero a.s. as $h \to 0$. I'll add more detail to address your concern shortly. $\endgroup$ – Nawaf Bou-Rabee Nov 7 '16 at 23:28
  • $\begingroup$ @NateEldredge I updated my answer based on your feedback. $\endgroup$ – Nawaf Bou-Rabee Nov 9 '16 at 18:19

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