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?

  • $\begingroup$ 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

$\def\R{\mathbb{R}}\def\ndot{{\,\cdot\,}}\def\d{\mathrm{d}}\def\Ω{{\mit Ω}}\def\peq{\mathrel{\phantom{=}}{}}$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,$$ \left( D\left( \int_0^T X_t \,\d t \right) \right)^{\tfrac{1}{2}} = \left( E\left( \left( \int_0^T Y_t \,\d t \right)^2 \right) \right)^{\tfrac{1}{2}} \leqslant \int_0^T \sqrt{\smash[b]{E(Y_t^2)}} \,\d t = \int_0^T \sqrt{\smash[b]{D(X_t)}} \,\d t. $$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.