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# Linearestimationtheory:quadraticmeanBochner integral ofstochasticprocess = lebesguepathbypathLebesgue integral?

After some helpful comments, I realized that I had to repost this question in a more systematic way.

On a complete probability space, let $\mathcal{H}$ \mathcal{H}_0$denote all the Hilbert space of square integrable random variables and let$\mathcal{H}_0$denote all square integrable r.v.s with zero mean. A stochastic process$X$is called a second order process if$\mathbf{E}X(t)^2 < \infty$and$\mathbf{E}X(t) = 0$, all$t$. It t \in [0,T]$. Such a process can be regarded as a curve in $\mathcal{H}_0$, i.e. a map $[0,T] \rightarrow \mathcal{H}_0$. Such a process It is called q.m. continuous if this map is continuous, i.e. $\mathbf{E}(X(s)-X(t))^2 X(s) \rightarrow 0$ X(t)$in quadratic mean as$s \rightarrow t$. One can show that each q.m. continuous process has a measurable version. We define Let$\mathcal{L}(X,t)$to X$ be subspace of $\mathcal{H}_0$ generated by a q.m. continuous second order process. We want to compute the integral $\lbrace \int_0^T X(s) : s \leq t \rbrace$. For mathrm{d} s$. There are two ways. Bochner integral. Clearly,$Y X$considered as a continuous map$[0,T] \in rightarrow \mathcal{H}_0$, the projection of$Y$onto$\mathcal{L}(X,t)$mathcal{H}_0$ is called the best linear estimate of $Y$ given Bochner integrable. We denote its Bochner integral by\text{(B-)}\int_0^T X(s) \mathrm{d}s.\end{equation}

Lebesgue integral. We may assume that $X(s)$, X$considered as a map$s [0,T] \leq t$times \Omega \rightarrow \mathrm{R}$ is measurable. One can show thatThus, for fixed $\omega$, the integral $\int_0^T X(s,\omega) \mathrm{d} s$ exists as a certain class of q.m. second order processesLebesgue integral, one has $\mathcal{L}(X,t) = and we denote the random variable constructed in this way by\lbrace begin{equation}\int_0^t f(stext{(L-)}\int_0^T X(s) \mathrm{d} mathrm{d}s.\end{equation} Question. Do we have \text{(B-)}\int_0^T X(s) : f \in L_2[0,t] mathrm{d}s = \rbrace$ (this is not needed here, just for intuition)text{(L-)}\int_0^T X(s) \mathrm{d}s \quad \text{a.s.?}\end{equation}

Ideas.The problem mentioned in the title arises in connection 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 following :

Questionsimple 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). Do \end{equation}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 int_0^T \xi_n(s) \mathrm{d}s \rightarrow \text{(B-)}\int_0^T X(s) \mathrm{d} s mathrm{d}s \in quad \mathcal{L}(X,t),text{in $\mathcal{H}_0$}, where the integral on the left is defined path by path as a Lebesgue-integral?

Idea of proof: One can in the obvious way (we omit (B-)) (to show this, one uses the fact that $X$ is q.m. continuous if and only if the covariance function $r(s,t) = \mathbf{E}X(s)X(t)$ is of a q.m. continuous process is continuous). Using thisAfter switching to a subsequence if necessary, one can show we may assume thatfor every sequence $\lbrace t^n \rbrace$ of partitions of $[0,t]$, the Riemann sums\xi_n = int_0^T \sum_i X(t_i^n) ( t_{i+1}^n - t_i^n xi_n(s) \mathrm{d}s \rightarrow \text{(B-)}\int_0^T X(s) \mathrm{d}s \quad \text{$\mathbf{P}$-a.s.}, are converging in $\mathcal{H}_0$ to the same limit, say $\xi$. Now, the crucial question is: do we would like to have that also \xi = int_0^T \int_0^t xi_n(s) \mathrm{d}s \rightarrow \text{(L-)}\int_0^T X(s) \mathrm{d} s mathrm{d}s This seems very logic, but since But this is tricky. The sums on the right left hand side is defined pathwise as a Lebesgue integral, this result is not immediateare Riemann sums, and I find it difficult to show this formallyi.e.

It would be enough \begin{equation}\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, but this is not clear. More generally, the assertion follows from the following conjecture: if a Riemann sum converges and the corresponding Lebesgue integral exists, then the limit of the Riemann sum must we would be this integraldone. But this is also not clearto me. I also tried to use some approximation argument with continuous processesarguments, but failed.

I found this question quite interesting, because couldn't do itwould be really surprising if the q.m. integral has a chance of not being equal to the pathwise Lebesgue integral, but somehow, the proof It seems like one needs to be difficultdeduce some kind of path regularity of $X$ from the assumption of q.m. If it is truecontinuity, then it is also an interesting result that Lebesgue integrals can be approximated by Riemann sums but I don't know any results in quadratic meanthis direction.

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On a complete probability space, let $\mathcal{H}$ denote all square integrable random variables and let $\mathcal{H}_0$ denote all square integrable r.v.s with zero mean.

A stochastic process $X$ is called a second order process if $\mathbf{E}X(t)^2 < \infty$ and $\mathbf{E}X(t) = 0$, all $t$. It can be regarded as a curve in $\mathcal{H}_0$, i.e. a map $[0,T] \rightarrow \mathcal{H}_0$. Such a process is called q.m. continuous if this map is continuous, i.e. $\mathbf{E}(X(s)-X(t))^2 \rightarrow 0$ as $s \rightarrow t$. One can show that each q.m continuous process has a measurable version.

We define $\mathcal{L}(X,t)$ to be subspace of $\mathcal{H}_0$ generated by $\lbrace X(s) : s \leq t \rbrace$. For $Y \in \mathcal{H}_0$, the projection of $Y$ onto $\mathcal{L}(X,t)$ is called the best linear estimate of $Y$ given $X(s)$, $s \leq t$. One can show that, for a certain class of q.m. second order processes, one has $\mathcal{L}(X,t) = \lbrace \int_0^t f(s) \mathrm{d} X(s) : f \in L_2[0,t] \rbrace$ (this is not needed here, just for intuition). The problem mentioned in the title arises in the proof of connection with the following question:

Question. Do we have $$\int_0^t X(s) \mathrm{d} s \in \mathcal{L}(X,t),$$ where the integral is defined path by path as a Lebesgue-integral?

Idea of proof: One can show that $X$ is q.m. continuous if and only if the covariance function $r(s,t) = \mathbf{E}X(s)X(t)$ is continuous. Using this, one can show that for every sequence $\lbrace t^n \rbrace$ of partitions of $[0,t]$, the Riemann sums $$\xi_n = \sum_i X(t_i^n) ( t_{i+1}^n - t_i^n )$$ are converging in $\mathcal{H}_0$ to the same limit, say $\xi$. Now, the crucial question is: do we have $$\xi = \int_0^t X(s) \mathrm{d} s \quad \text{a.s.?}$$ This seems very logic, but since the right hand side is defined pathwise as a Lebesgue integral, this result is not immediate, and I find it difficult to show this formally.

It would be enough if we knew that the paths of $X$ are a.s. Riemann integrable, but this is not clear. More generally, the assertion follows from the following conjecture: if a Riemann sum converges and the corresponding Lebesgue integral exists, then the limit of the Riemann sum must be this integral. But this is also not clear to me. I also tried to use some approximation argument with continuous processes, but failed.

I found this question quite interesting, because it would be really surprising if the q.m. integral has a chance of not being equal to the pathwise Lebesgue integral, but somehow, the proof seems to be difficult. If it is true, then it is also an interesting result that Lebesgue integrals can be approximated by Riemann sums in quadratic mean.

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On a complete probability space, let $\mathcal{H}$ denote all square integrable random variables and let $\mathcal{H}_0$ denote all square integrable r.v.s with zero mean.

A stochastic process $X$ is called a second order process if $\mathbf{E}X(t)^2 < \infty$ and $\mathbf{E}X(t) = 0$, all $t$. It can be regarded as a curve in $\mathcal{H}_0$, i.e. a map $[0,T] \rightarrow \mathcal{H}_0$. Such a process is called q.m. continuous if this map is continuous, i.e. $\mathbf{E}(X(s)-X(t))^2 \rightarrow 0$ as $s \rightarrow t$. One can show that each q.m continuous process has a measurable version.

We define $\mathcal{L}(X,t)$ to be subspace of $\mathcal{H}_0$ generated by $\lbrace X(s) : s \leq t \rbrace$. For $Y \in \mathcal{H}_0$, the projection of $Y$ onto $\mathcal{L}(X,t)$ is called the best linear estimate of $Y$ given $X(s)$, $s \leq t$. One can show that, for a certain class of q.m. second order processes, one has $\mathcal{L}(X,t) = \lbrace \int_0^t f(s) \mathrm{d} X(s) : f \in L_2[0,t] \rbrace$ (this is not needed here, just for intuition). The problem mentioned in the title arises in the proof of the following question:

Question. Do we have \int_0^t X(s) \mathrm{d} s \in \mathcal{L}(X,t) \ \text{?} mathcal{L}(X,t), where the integral is defined path by path as a Lebesgue-integral?

Idea of proof: One can show that $X$ is q.m. continuous if and only if the covariance function $r(s,t) = \mathbf{E}X(s)X(t)$ is continuous. Using this, one can show that for every sequence $\lbrace t^n \rbrace$ of partitions of $[0,t]$, the Riemann sums $$\xi_n = \sum_i X(t_i^n) ( t_{i+1}^n - t_i^n )$$ are converging in $\mathcal{H}_0$ to the same limit, say $\xi$. Now, the crucial question is: do we have $$\xi = \int_0^t X(s) \mathrm{d} s \quad \text{a.s.?}$$ This seems very logic, but since the right hand side is defined pathwise as a Lebesgue integral, this result is not immediate, and I find it difficult to show this formally.

It would be enough if we knew that the paths of $X$ are a.s. Riemann integrable, but this is not clear. More generally, the assertion follows from the following conjecture: if a Riemann sum converges and the corresponding Lebesgue integral exists, then the limit of the Riemann sum must be this integral. But this is also not clear to me. I also tried to use some approximation argument with continuous processes, but failed.

I found this question quite interesting, because it would be really surprising if the q.m. integral has a chance of not being equal to the pathwise Lebesgue integral, but somehow, the proof seems to be difficult. If it is true, then it is also an interesting result that Lebesgue integrals can be approximated by Riemann sums in quadratic mean.

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