I'd like to know the definitions of $\int_{[0,\tau]}X\,dM$ and $\int 1_{[0,\tau]}X\,dM$, where $\tau$ is a stopping time, $X\in L^2(\mathbb{R}_+\times\Omega,\mathcal{P},\mu_M)$, $M$ is a right-continuous martingale, $\mu_M$ is the measure on the predictable $\sigma$-algebra $\mathcal{P}$ which is defined (uniquely) by its behavior on the predictable rectangles: $\mu_M(1_{\{0\}\times F})=0$ for all $F\in\mathcal{F}_0$, and $\mu_M(1_{(s,t]\times F_s})=1_{F_s}(M_t-M_s)^2$ for all $F_s\in\mathcal{F_s}$. $Y_t\equiv\int_{[0,t]}X\,dM$.

In the text that presented these terms (Chung and Williams' Stochastic Integration), a very similar definition if $\tau$ is just a constant (p. 39): enter image description here

But I suspect that the exact definition is not quite the same, since Theorem 2.7.iii would be immediate:

enter image description here

In the proof of Theorem 2.7.iii, we introduce $\tau_n=2^{-n}[2^n\tau+1]$, and note we can write $Y_{\tau_n}=\sum_{k=0}^{N_n}1_{\tau\in[\frac{k}{2^n},\frac{k+1}{2^n})\times \{\tau\ge\frac{k+1}{n}\}}Y_{(k+1)2^{-n}}$. Then since $\tau_n\searrow\tau$ and $Y$ is right-continuous, we have $Y_{\tau_n}\to Y_\tau$ a.s. Is this how we define $Y_\tau\equiv\int_{[0,\tau]}X\,dM$? Or is $Y_\tau$ defined some other way already?

I am hoping for a clarification about what $\int_{[0,\tau]}X\,dM$ and $\int 1_{[0,\tau]}X\,dM$ mean. Thanks a bunch!

  • 1
    $\begingroup$ Isn't the difference between the two expressions discussed in the remark following (2.17)...? $\endgroup$ – saz Feb 22 '17 at 14:23
  • $\begingroup$ Is it apparent from the remark? It seems like if we pick $\omega\in\Omega$, we get an expression of the form $\int_{[0,t]}X\,dM$, which is by definition equal to $\int 1_{[0,t]}X\,dM$ $\endgroup$ – manofbear Feb 22 '17 at 17:30
  • $\begingroup$ Ok, I might have it now. $Y_{\tau(\omega)}$ is given by $\int_{[0,\tau(\omega)]}X\,dM$. But $\int 1_{[0,\tau]}X\,dM$ is a random variable, so we evaluate it at $\omega$ after taking the integral; i.e. $(\int 1_{[0,\tau]}X\,dM)(\omega)$, which may not be equal to $\int 1_{[0,\tau(\omega)]}X\,dM$. Is this the distinction? $\endgroup$ – manofbear Feb 22 '17 at 17:36

For a (nice) process $X$ denote by

$$(X \bullet M)(t,\omega) := \left( \int_0^t X(s) \, dM_s \right)(\omega), \qquad t \in [0,\infty] \tag{1}$$

the stochastic integral of $X$ with respect to $M$. Identity (2.17) is saying that

$$(X \bullet M)(\tau(\omega),\omega) = ((X 1_{[0,\tau]}) \bullet M)(\infty,\omega) \quad \text{a.s.}$$

The left-hand side is defined as the stochastic integral $(1)$ evaluated at $t=\tau(\omega)$. In contrast, the right-hand side is the stochastic integral of the "truncated" process $X \cdot 1_{[0,\tau]}$.


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