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Suppose there is a second-order real-state stochastic process $X: \Omega \times T \rightarrow \mathbb{R}$ with $T= \mathbb{R}$ and probability space $(\Omega, \mathcal{F}, P)$. I was wondering if the following inequality holds for integrals over an interval $[a,b] \subset \mathbb{R}$:

$ \Vert \int_a^b X_t dt \Vert_2 \leq \int_a^b \Vert X_t \Vert_2 dt \ ?$

Note that the integral on LHS is a stochastic integral while the one on RHS is a deterministic one. $L_2$ norm is on $L_2$ space of squared-integrable random variables as measurable mappings from $(\Omega, \mathcal{F}, P)$ to $(\mathbb{R}, \mathcal{B})$.

Is this Jensen's inequality? I don't think it is, because although the $L_2$ norm is convex, it is not a mapping from $\mathbb{R}$ to $\mathbb{R}$, but from $L_2(\Omega, \mathcal{F})$ to $\mathbb{R}$, and the types of integrals on LHS and on RHS are not the same. So I was wondering if this inequality is true and why?

Can it be seen as a generalization of Jensen's inequality? If yes, is it only because of the similarity of their forms or the similarity of some deeper things?

Thanks and regards!

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    @Nate: Thanks! Does the "Jensen's inequality" also apply to the stochastic integral with integrand and integrator both being stochastic processes?2011-03-03

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Yes, this is true*, and extends to the $L^p$ norm for all $1\le p\le\infty$, $ \left\Vert\int_a^bX_s\,ds\right\Vert_p\le\int_a^b\left\Vert X_s\right\Vert_p\,ds. $ Consider the case where $X$ is simple, so that $X_t=\sum_{k=1}^nU_k1_{\{t\in V_k\}}$ for random variables $U_k$ and disjoint measurable subsets $V_k\subseteq\mathbb{R}$. Using $\lambda$ to denote the Lebesgue measure. $ \begin{align} \left\Vert\int_a^bX_s\,ds\right\Vert_p &= \left\Vert\sum_kU_k\lambda\left(V_k\cap(a,b)\right)\right\Vert_p\\ &\le\sum_k\Vert U_k\Vert_p\lambda\left(V_k\cap(a,b)\right)\\ &=\int_a^b\Vert X_s\Vert_p\,ds \end{align} $ Then, approximate arbitrary measurable $X$ as a limit of simple functions. The same argument will hold for integrable functions $\mathbb{R}\to B$ for any Banach space $B$, and not just for $L^p$ spaces.

* I'm assuming that $X$ is measurable as a map from $\Omega\times\mathbb{R}$ to $\mathbb{R}$.

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    @Tim: Not much. It only relies on the fact that $\Vert\cdot\Vert_p$ is a norm, so satisfes the triangle inequality. As I mentioned, it extends to arbitrary Banach spaces. Jensen's inequality can be used to show that $\Vert\cdot\Vert_p$ is a norm (i.e., satisfies Minkowski's inequality), but that's rather a weak link.2011-03-03