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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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@Shai: Let me guess. In deterministic case, X is a function of two variables, and the L2 norm is an integral of the squared of the function inside the norm relative to the variable other than t. Are you saying there is also an inequality for deterministic X? –  Tim Mar 3 '11 at 0:10
    
Can we interpret that inequality as ${\rm E}[\int_a^b {X_t dt} ]^2 \le \int_a^b {{\rm E}[X_t^2 ]dt}$? –  Shai Covo Mar 3 '11 at 0:29
    
@Shai: Almost. I think taking square root is needed after taking expectation. L2 norm of a function is first taking point-wise square of the function, then taking integral/expectation, and then taking square root. Right? –  Tim Mar 3 '11 at 0:37
    
So, you consider $\sqrt {{\rm E}[\int_a^b {X_t dt} ]^2 } \le \int_a^b {\sqrt {{\rm E}[X_t^2 ]} dt} $. Right? –  Shai Covo Mar 3 '11 at 0:48
    
@Shai: Yes, I think that is what the inequality means. –  Tim Mar 3 '11 at 0:49

1 Answer 1

up vote 2 down vote accepted

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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Thanks! I think you are right. Does this have some relation with Jensen's inequality? Not necessarily one implying the other, but if the two are based on some common ground. I dont see some. –  Tim Mar 3 '11 at 1:29
    
@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. –  George Lowther Mar 3 '11 at 1:30

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