Bochner integral of stochastic process = path by path Lebesgue integral? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T14:29:46Z http://mathoverflow.net/feeds/question/110894 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110894/bochner-integral-of-stochastic-process-path-by-path-lebesgue-integral Bochner integral of stochastic process = path by path Lebesgue integral? Hauke L. 2012-10-28T11:08:14Z 2012-11-08T01:22:08Z <p>After some helpful comments, I realized that I had to repost this question in a more systematic way. </p> <p>On a complete probability space, let $\mathcal{H}_0$ denote the Hilbert space of square integrable random variables with zero mean. A stochastic process $X$ is called a second order process if $\mathbf{E}X(t)^2 &lt; \infty$ and $\mathbf{E}X(t) = 0$, all $t \in [0,T]$. Such a process can be regarded as a map $[0,T] \rightarrow \mathcal{H}_0$. It is called q.m. continuous if this map is continuous, i.e. $X(s) \rightarrow X(t)$ in quadratic mean as $s \rightarrow t$. One can show that each q.m. continuous process has a measurable version.</p> <p>Let $X$ be a q.m. continuous second order process. We want to compute the integral $\int_0^T X(s) \mathrm{d} s$. There are two ways.</p> <p><strong>Bochner integral.</strong> Clearly, $X$ considered as a continuous map $[0,T] \rightarrow \mathcal{H}_0$ is Bochner integrable. We denote its Bochner integral by $$\text{(B-)}\int_0^T X(s) \mathrm{d}s.$$</p> <p><strong>Lebesgue integral.</strong> We may assume that $X$ considered as a map $[0,T] \times \Omega \rightarrow \mathrm{R}$ is measurable. Thus, for fixed $\omega$, the integral $\int_0^T X(s,\omega) \mathrm{d} s$ exists as a Lebesgue integral, and we denote the random variable constructed in this way by $$\text{(L-)}\int_0^T X(s) \mathrm{d}s.$$</p> <p><strong>Question.</strong> Do we have $$\text{(B-)}\int_0^T X(s) \mathrm{d}s = \text{(L-)}\int_0^T X(s) \mathrm{d}s \quad \text{a.s.?}$$</p> <p><strong>Ideas.</strong> Let $\lbrace t^n = t_0^n, \ldots t_{k_n}^n \rbrace$ be a sequence of partitions of $[0,T]$ with mesh going to zero. Define the simple functions $$\xi_n = X(t_0^n)1[t_0^n,t_1^n] + \sum_{i=1}^{k_n-1} X(t_i^n) 1[t_i^n, t_{i+1}^n).$$ Then one can show that for almost every $t$, we have $\xi_n(t) \rightarrow X(t)$ in $\mathcal{H}_0$, and $$\int_0^T \xi_n(s) \mathrm{d}s \rightarrow \text{(B-)}\int_0^T X(s) \mathrm{d}s \quad \text{in \mathcal{H}_0},$$ where the integral on the left is defined in the obvious way (we omit (B-)) (to show this, one uses the fact that the covariance function $r(s,t) = \mathbf{E}X(s)X(t)$ of a q.m. continuous process is continuous). After switching to a subsequence if necessary, we may assume that $$\int_0^T \xi_n(s) \mathrm{d}s \rightarrow \text{(B-)}\int_0^T X(s) \mathrm{d}s \quad \text{\mathbf{P}-a.s.},$$ Now, we would like to have that also $$\int_0^T \xi_n(s) \mathrm{d}s \rightarrow \text{(L-)}\int_0^T X(s) \mathrm{d}s \quad \text{\mathbf{P}-a.s.},$$ But this is tricky. The sums on the left hand side are Riemann sums, i.e. $$\int_0^T \xi_n(s) \mathrm{d}s = \sum_{i=0}^{k_n-1} X(t_i^n)(t_{i+1}^n - t_i^n ).$$ So if we knew that the paths of $X$ are a.s. Riemann integrable, we would be done. But this is not clear. I also tried to use some approximation arguments, but couldn't do it. It seems like one needs to deduce some kind of path regularity of $X$ from the assumption of q.m. continuity, but I don't know any results in this direction. </p> http://mathoverflow.net/questions/110894/bochner-integral-of-stochastic-process-path-by-path-lebesgue-integral/111037#111037 Answer by George Lowther for Bochner integral of stochastic process = path by path Lebesgue integral? George Lowther 2012-10-29T22:40:07Z 2012-11-08T01:22:08Z <p>Yes, the Bochner integral does agree with the Lebesgue integral of the sample paths of the process. We can prove this in a slightly more general situation than that asked for in the question.</p> <p>For a probability space $(\Omega,\mathcal{F},\mathbb{P})$, let $X\colon[0,T]\to L^p(\mathbb{P})$ ($1\le p\le\infty$) be Bochner integrable w.r.t the Lebesgue measure on $[0,T]$, and also jointly measurable as a map $(t,\omega)\mapsto X(t)(\omega)$ from $[0,T]\times\Omega$ to $\mathbb{R}$. Then, the Bocher integral $\int_0^T X(t)\,dt$ agrees with the pathwise Lebesgue integral $\int_0^TX(t)(\omega)\,dt$ for almost every $\omega$.</p> <p>First, this statement clearly holds for simple functions, which are finite linear combinations of terms of the form $X(t)(\omega)=1_{\lbrace t\in A\rbrace}1_{\lbrace\omega\in B\rbrace}$, for $A$ a Borel subset of $[0,T]$ and $B$ in $\mathcal{F}$. Now, by definition, if $X$ is Bochner integrable then, for each $n\ge1$, there is a simple $\xi_n$ such that $$\int_0^T\lVert X(t)-\xi_n(t)\rVert_p\,dt\le2^{-n}.$$ The Bochner integral is given by $$\int_0^T\xi_n(t)\,dt \rightarrow\text{(B-)}\int_0^T X(t)\,dt.$$ Here the limit is taken in the $L^p$ norm and, hence, also holds for convergence in probability.</p> <p>Using pathwise Lebesgue integration along the sample paths of $X$ now, we can use Fubini's theorem to commute expectation, integration and summation signs. \begin{align} \mathbb{E}\left[\int_0^T\sum_{n=1}^\infty\left\lvert X(t)-\xi_n(t)\right\rvert\,dt\right] &amp;=\sum_{n=1}^\infty\int_0^T\mathbb{E}\left[\lvert X(t)-\xi_n(t)\rvert\right]\,dt\cr &amp;\le\sum_{n=1}^\infty\int_0^T\left\lVert X(t)-\xi_n(t)\right\rVert_p\,dt\cr &amp;\le\sum_{n=1}^\infty2^{-n}=1 &lt; \infty. \end{align} In particular, $$\int_0^T\sum_{n=1}^\infty\left\lvert X(t)-\xi_n(t)\right\rvert\,dt &lt; \infty$$ with probability one. Looking at any sample path for which this is finite, $\lvert X(t)-\xi_n(t)\rvert\to0$ as $n\to\infty$ for Lebesgue almost every $t$. Also, $\lvert X(t)-\xi_n(t)\rvert$ is dominated by its sum over $n$. Therefore dominated convergence applies, $$\int_0^T\xi_n(t)\,dt\rightarrow\textrm{(L-)}\int_0^T X(t)\,dt.$$ This holds for almost every sample path of $X$, so the limit holds in probability. Hence the Lebesgue integral on sample paths agrees with the Bochner integral.</p>