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I am trying to prove that for any process $\{\Phi(s); s\in[0,t]\}$ such that $\sqrt{\operatorname{Var}\Phi(s)}$ is integrable on $[0,t]$ we have $$\sqrt{\mbox{Var}\left(\int_0^t\Phi(s)\,\mathrm ds\right)}\le\int_0^t\sqrt{\operatorname{Var}\Phi(s)}\,\mathrm ds.$$

I tried first assuming that $\mathbb{E}(\int_{0}^t\Phi(s)\,\mathrm ds)=0$ and using Jensen inequality but even in this case I don't get the desired result. I tried also using approximation but no result so far.

I wonder if there is some missing hypothesis like square integrable process.

Any ideas?

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  • $\begingroup$ The t is less than 1 no? $\endgroup$ – Hamza Apr 9 at 13:09
  • $\begingroup$ @Hamza yes sorry $\endgroup$ – TTL Apr 9 at 13:15
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$\def\R{\mathbb{R}}\def\ndot{{\,\cdot\,}}\def\d{\mathrm{d}}\def\Ω{{\mit Ω}}\def\peq{\mathrel{\phantom{=}}{}}$Lemma: Suppose that $(U, \mathscr{F}, μ)$ and $(V, \mathscr{G}, ν)$ are measurable spaces. If $f: U × V → \R$ is an $\mathscr{F} × \mathscr{G}$-measurable function such that$$ \int_U |f(x, y)| μ(\d x) < +∞ $$ for almost every $y \in V$, then$$ \left( \int_V \left( \int_U f(x, y) μ(\d x) \right)^2 ν(\d y) \right)^{\tfrac{1}{2}} \leqslant \int_U \left( \int_V (f(x, y))^2 ν(\d y) \right)^{\tfrac{1}{2}} μ(\d x). $$

Proof: By the Cauchy-Schwarz inequality,\begin{align*} \text{RHS}^2 &= \left( \int_U \left( \int_V (f(x, y))^2 ν(\d y) \right)^{\smash{\tfrac{1}{2}}} μ(\d x) \right)^2\\ &= \left( \int_U \left( \int_V (f(x_1, y))^2 ν(\d y) \right)^{\smash{\tfrac{1}{2}}} μ(\d x_1) \right) \left( \int_U \left( \int_V (f(x_2, y))^2 ν(\d y) \right)^{\smash{\tfrac{1}{2}}} μ(\d x_2) \right)\\ &= \iint_{U^2} \left( \int_V (f(x_1, y))^2 ν(\d y) \right)^{\tfrac{1}{2}} \left( \int_V (f(x_2, y))^2 ν(\d y) \right)^{\tfrac{1}{2}} μ(\d x_1) μ(\d x_2)\\ &\geqslant \iint_{U^2} \left( \int_V f(x_1, y) f(x_2, y) ν(\d y) \right) μ(\d x_1) μ(\d x_2)\\ &= \int_V \left( \iint_{U^2} f(x_1, y) f(x_2, y) μ(\d x_1) μ(\d x_2) \right) ν(\d y)\\ &= \int_V \left( \int_U f(x_1, y) μ(\d x_1) \right) \left( \int_U f(x_2, y) μ(\d x_2) \right) ν(\d y)\\ &= \int_V \left( \int_U f(x, y) μ(\d x) \right)^2 ν(\d y) = \text{LHS}^2. \end{align*}


Now return to the question and suppose that $X: [0, T] × \Ω → \R$ is a measurable process such that$$ \int_0^T E(|X_t|) \,\d t < +∞. \tag{$*$} $$ Since$$ \int_\Ω \int_0^T |X_t(ω)| \,\d t \d ω = \int_0^T \int_\Ω |X_t(ω)| \,\d t \d ω = \int_0^T E(|X_t|) \,\d t < +∞, $$ then $\displaystyle \int_0^T |X_t| \,\d t < +∞$ almost surely, $\displaystyle E\left( \int_0^T X_t \,\d t \right) = \int_0^T E(X_t) \,\d t$ exists, and thus $E(X_t)$ exists for almost every $t \in [0, T]$. For such $t$, define $m(t) = E(X_t)$, $Y_t = X_t - m(t)$, then $D(X_t) = E(Y_t^2)$ and$$ E\left( \int_0^T Y_t \,\d t \right) = E\left( \int_0^T X_t \,\d t \right) - \int_0^T m(t) \,\d t = 0, $$ which implies$$ D\left( \int_0^T X_t \,\d t \right) = D\left( \int_0^T Y_t \,\d t + \int_0^T m(t) \,\d t \right) = D\left( \int_0^T Y_t \,\d t \right) = E\left( \left( \int_0^T Y_t \,\d t \right)^2 \right). $$ By the lemma,$$ \left( D\left( \int_0^T X_t \,\d t \right) \right)^{\tfrac{1}{2}} = \left( E\left( \left( \int_0^T Y_t \,\d t \right)^2 \right) \right)^{\tfrac{1}{2}} \leqslant \int_0^T \sqrt{\smash[b]{E(Y_t^2)}} \,\d t = \int_0^T \sqrt{\smash[b]{D(X_t)}} \,\d t. $$

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  • $\begingroup$ Oh very nice, I tried to use C-S but didn't get the idea to use introduce $Y_t.$ thanks $\endgroup$ – TTL 2 days ago

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