What does progressively measurable actually entail? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T15:45:44Z http://mathoverflow.net/feeds/question/62246 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62246/what-does-progressively-measurable-actually-entail What does progressively measurable actually entail? Jerry Gagelman 2011-04-19T10:47:36Z 2011-04-19T18:24:02Z <p>There is a definition that has always left nagging questions in my mind. To set it up, let <code>$(\Omega, \cal{A}, (\cal{F}_t)_{t\geq 0}, P)$</code> be a filtered probability space. From Comets &amp; Meyre's <em>Calcul stochastique et modèles de diffusions</em>,</p> <p><strong>Definition 3.2.</strong> A real-valued function $\phi$ on $\mathbb{R}^+\times \Omega$ is <em>progressively measurable</em> if, for all $t\in \mathbb{R}^+$, the mapping $(s,\omega) \mapsto \phi(s, \omega)$ on $[0, t]\times \Omega$ is $\cal{B}[0,t] \otimes \cal{F}_t$-measurable.</p> <p>Clear enough. The authors go on to define $M^2(\mathbb{R}^+)$ to define the set (space) of all progressively measurable stochastic processes $\phi$ such that</p> <p><code>$\mathbf{E}\int_{\mathbb{R}^+} \phi^2(t, \omega) dt &lt; \infty$</code>.</p> <p>(This formula is verbatim from the book.) </p> <p>My question is: does one simply interpret the expression on the left as a double integral à la the Fubini-Tonelli theorem, $\int_\Omega\int_{R^+} \ldots dt dP$, and if so, does progressive measurability ensure that $\phi$ is $\cal{B}(\mathbb{R}^+)\otimes \cal{A}$ measurable, so that this interpretation makes sense?</p> <p>My attempts to realize $\phi$ as the limit of a sequence of measurable functions (like $\phi|_{[0, n]\times\Omega}$) have yielded nothing to convince me yet. </p> http://mathoverflow.net/questions/62246/what-does-progressively-measurable-actually-entail/62311#62311 Answer by BSteinhurst for What does progressively measurable actually entail? BSteinhurst 2011-04-19T18:24:02Z 2011-04-19T18:24:02Z <p>The formula </p> <p>$$\mathbb{E} \int_{R+} \phi^{2}(t,\omega)dt$$</p> <p>is a double integral a la Fubini-Tonelli. And if you did back there is probably a condition on the filtration saying that $\mathcal{F}_t \subset \mathcal{A}$ for all $t \ge 0$. So that progressive measurability does imply that $\phi^{2}(t,\omega)$ is $\mathcal{B}(\mathbb{R}^{+}) \otimes \mathcal{A}$ measurable. But making this integrand measurable isn't the main purpose of the progressive measurability condition. The main point is so that something like $f(t,X_t)$ where $X_t$ is an adapted process is again an adapted process. The integrability condition is to give $\mathcal{M}^{2}$ a Hilbert space structure. </p>