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Let $W_t$ be a Brownian motion with $m$ independent components on $(\Omega,F,P)$.
Let $G(\omega,t)=[g_{ij}(\omega,t)]_{1\leq i\leq n,1\leq j\leq m}$ in $V^{n\times m}[S,T]$ such that
$$\limsup_{\omega,t \in\Omega\times[S,T]} \sum_i^m \sum_j^n | g_{ij}(\omega,t)|<\infty$$ and
$$\int_S^T E|G(\omega,t)^6|dt<\infty.$$
I have to prove that:
$$E\left|\int_S^T G(\omega,t)dW_t\right|^6 \le 15^3(T-S)^2\int_S^TE|G(\omega,t)^6|dt<\infty.$$
I've also a hint:
$$\int_S^T \int_\Omega H(\omega,t)^4 K(\omega,t)^2dtdP(\omega) \le \left\{\int_S^T \int_\Omega H(\omega,t)^6 dtdP(\omega) \right\}^{4/6} \left\{ \int_S^T \int_\Omega K(\omega,t)^6dtdP(\omega))\right\}^{2/6}.$$

My idea was to use Itō's isometry in order to pass from $dW_t$ in $dt$ but I don't know if it's possible with $6$ at the exponent. Maybe a change of variable? Anyway I can't figure out where the coefficient $15^3(T-S)^2$ came from...

Thank you for your help

EDIT: I found a very interesting article from Novikov called "On moment inequalities and identities for stochastic integrals", which analyses a very similar case. I had not time to properly study this work but the key was applying the Itō's formula to a specific function.

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The lim sup in your first display doesn't seem to make sense; is there a typo? –  Nate Eldredge Dec 22 '11 at 20:40
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1 Answer 1

I've found something...
let's define $$\eta(t) = \int_S^t G(\omega,t)dW_t$$

if we apply the Ito's formula we obtain ( for the general case with $2n=6$ ) $$ E \left[\eta(T)^{2n} \right] = E \left[ \frac{2n(2n-1)}{2}\int_S^T\eta(s)^{2n-2} G^2(s,\omega)ds \right] \le$$ using the hint $$ \le \frac{2n(2n-1)}{2} \left\{ E\int_S^T\eta(s)^{2n}ds \right\}^{\frac{2n-2}{2n}} \left\{ E\int_S^T G^{2n}(s,\omega)ds \right\}^{\frac{1}{n}} $$

then I'm stuck, I don't know how to compute $ \left\{ E\int_S^T\eta(s)^{2n}ds \right\}^{\frac{2n-2}{2n}} $

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