# A dilemma about the definition of the stochastic integral $\int_a^b\Phi\:{\rm d}W$

Let

• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $T>0$
• $(\mathcal F_t)_{t\in[0,\:T]}$ be a filtration of $\mathcal A$
• $W$ be an $\mathcal F$-Brownian motion on $(\Omega,\mathcal A,\operatorname P)$

$\Phi:\Omega\times[0,T]\to\mathbb R$ is called elementary $\mathcal F$-predictable $:\Leftrightarrow$ $$\Phi=\sum_{i=1}^k1_{(t_{i-1},\:t_i]}^{[0,\:T]}\eta_i\tag1$$ for some $k\in\mathbb N$, $0\le t_0<\cdots<t_k\le T$ and $\mathcal F_{t_{i-1}}$-measurable $\eta_i:\Omega\to\mathbb R$ with $|\eta_i(\Omega)|\in\mathbb N$ for $i\in\left\{1,\ldots,k\right\}$. Let $$\mathcal E:=\left\{\Phi:\Omega\times[0,T]\to\mathbb R\mid\Phi\text{ is elementary }\mathcal F\text{-predictable}\right\}$$ be equipped with the norm inherited from $L^2\left(\operatorname P\otimes\left.\lambda^1\right|_{[0,\:T]}\right)$ and $$\mathcal I^2:=\left\{\Phi\in\mathcal L^2\left(\operatorname P\otimes\left.\lambda^1\right|_{[0,\:T]}\right):\Phi\text{ is }\mathcal F\text{-predictable}\right\}\;.$$ If $\Phi\in\mathcal E$, then $$\int\Phi\:{\rm d}W:=\sum_{i=1}^k\eta_i\left(W_{t_i}-W_{t_{i-1}}\right)$$ and $$\int_a^b\Phi\:{\rm d}W:=\int 1_{(a,\:b]}^{[0,\:T]}\Phi\:{\rm d}W\tag2$$ for $0\le a\le b\le T$. Moreover, $$(\Phi\cdot W)_t:=\int_0^t\Phi\:{\rm d}W\;\;\;\text{for }t\in[0,T]\;.$$ Now, $$\mathcal E\ni\Phi\mapsto\Phi\cdot W\tag3$$ is a linear isometry into the space of square-integrable continuous $\mathcal F$-martingales $M_c^2(\mathcal F,\operatorname P)$ equipped with the usual norm. $\mathcal E$ is a dense subset of $$\mathcal I^2:=\left\{\Phi\in\mathcal L^2\left(\operatorname P\otimes\left.\lambda^1\right|_{[0,\:T]}\right):\Phi\text{ is }\mathcal F\text{-predictable}\right\}$$ and hence there is a unique isometric and linear extension $\Xi$ of $(3)$ to $\mathcal I^2$. If $\Phi\in\mathcal I^2$, then $$\Phi\cdot W:=\Xi(\Phi)$$ and $$\int\Phi\:{\rm d}W:=(\Phi\cdot W)_T\;.$$

If $0\le a\le b\le T$, how should we define $$\int_a^b\Phi\:{\rm d}W\tag4$$ for $\Phi\in\mathcal I^2$?

The definition found in the most books is, $$\int_a^b\Phi\:{\rm d}W:=(\Phi\cdot W)_b-(\Phi\cdot W)_a\;.\tag5$$ But there is an obvious problem with $(4)$: If $X:\Omega\to\mathbb R$ is bounded and $\mathcal F_a$-measurable, shouldn't the object $$\int_a^bX\Phi\:{\rm d}W$$ be well-defined? However, with $(4)$ this is not the case, since $X\Phi$ is obviously not $\mathcal F$-predictable and hence $X\Phi\cdot W$ is undefined. So, $(4)$ is a bad choice. Another option would be $$\int_a^b\Phi\:{\rm d}W:=\int1_{(a,\:b]}\Phi\:{\rm d}W\;.\tag6$$ But with $(5)$ we encounter an other problem: The processes $\Phi\cdot W$ and $$\left(\int_0^t\Phi\:{\rm d}W\right)_{t\in[0,\:T]}$$ should be indistinguishable; but with $(5)$ they are only modifications.

So, do we need to mimic the described construction of the stochastic integral for any possible initial value $a\in[0,T]$ and define $(4)$ in terms of a restriction, i.e. $$\int_b^T\Phi\:{\rm d}W:=\int_b^T\left.1_{(b,\:T]}^{[a,\:T]}\Phi\right|_{\Omega\times[b,\:T]}\:{\rm d}W$$ for $\Phi:\Omega\times[a,T]\to\mathbb R$ suitable in the obvious manner?

• As to (5), isn't it the case that two a.s. continuous modifications of one another are actually indistinguishable? Being modifications, a.s. they agree at all rationals, and on that event, intersect the event that both are continuous, they agree everywhere. – Nate Eldredge Oct 12 '17 at 0:25
• @NateEldredge What you say is surely true, but do you mean $(6)$ instead of $(5)$? If you mean $(5)$, as I said, $X\Phi$ is not even defined, but the object $\int_a^bX\Phi\:{\rm d}W$ clearly should be defined. If you mean $(6)$, how do you prove that $\left(\int_0^t\Phi\:{\rm d}W\right)_{t\in[0,\:T]}$ is (almost surely) (right-)continuous? – 0xbadf00d Oct 12 '17 at 0:33

Conventionally, the stochastic integral $\Phi\cdot W$ is defined for $\Phi\in\mathcal I^2$ as a continuous martingale via the Ito isometry, and then $\int_a^b \Phi\,dW$ is just notation for $(\Phi\cdot W)_b-(\Phi\cdot W)_a$. The subtlety you raise is basically our sloppy identification of $(\Phi\cdot W)_{t\wedge b}$ with $\int_0^t 1_{(0,b]}\Phi\,dW$. For a fixed $b\in[0,T]$ these two (viewed as martingales in $t$) are indistinguishable. For fixed $t$, the former is a.s. continuous in $b$ while the latter is merely continuous in probability (and in $L^2$) in $b$. Of course, $(b,t)\mapsto (\Phi\cdot W)_{t\wedge b}$ provides an a.s. jointly continuous modification of $\int_0^t 1_{(0,b]}\Phi\,dW$. (The paper "Calcul stochastique dépendant d'un paramètre" by C. Stricker and M. Yor [Z. Wahrsch. Verw. Gebiete 45 (1978) 109–133] addresses such issues in a broader context; see https://link.springer.com/article/10.1007%2FBF00715187.)
• That's clear to me, we've even got $(\Phi\cdot W)_{\tau\:\wedge\:t}=\int_0^t1_{(0,\:\tau]}^{[0,\:T]}\Phi\:{\rm d}W$ for all $t\in[0,T]$ almost surely, for any almost surely finite stopping time $\tau$. But what's your solution for the definition of $\int_a^bX\Phi\:{\rm d}W$? As you've said, one commonly sets $\int_a^bX\Phi\:{\rm d}W:=(X\Phi\cdot W)_b-(X\Phi\cdot W)_a$, but as $X\Phi$ is not even adapted, this is not well-defined. (BTW, your paper is written in French, is there an English version?) – 0xbadf00d Oct 15 '17 at 19:19
• As your integrand is not predictable, you are on your own here. One approach is to replace $X$ by the predictable process $(\omega,t)\mapsto 1_{\{a<t\}}X(\omega)$, which yields $\int_a^b X\Phi\,dW=X\int_a^b\Phi\,dW$ for all $b\in[a,T]$, a.s. (Of course $a$ is fixed in advance.) – John Dawkins Oct 16 '17 at 20:03