# Proving $\sqrt{\operatorname{Var}(\int_0^t\Phi(s)\,\mathrm ds)}\le\int_0^t\sqrt{\operatorname{Var}\Phi(s)}\,\mathrm ds$

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?

• The t is less than 1 no? – Hamza Apr 9 at 13:09

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