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I am reading Oksendal's book "Malliavin calculus for Levy processes with application to finance". In the proof of Lemma 4.9 (page 47), the author interchanges the Malliavin derivative $D_t$ with the Lebesgue integral $ds$. $$D_t\int_0^T u^2(s)\,ds = 2\int_0^T u(s)D_tu(s)\,ds$$ Could anyone shed any light?

ting
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  • As $D_t$ is a derivative, $D_t$ is linear operator on $\mathbb D_{1,2}$ and $D_t[u^2(s)]= 2 u(s) D_t[u(s)]$. – Zbigniew Nov 28 '15 at 13:41
  • @Zbigniew but why does the linearity guarantee that we can do the interchange? Could you please be more specific? – ting Nov 28 '15 at 16:05
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    @Zbigniew why can we differentiate it inside – ting Dec 01 '15 at 13:45

1 Answers1

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We'll prove this in two steps. First, we pull the derivative operator $\mathfrak{D}$ inside the (Lebesgue) integral, and then apply a "Chain Rule". Since the book by Øksendal et al. is referenced, I won't prove the "Chain Rule", but establish the interchange of the derivative and the integral through some density arguments.

The proof is not that hard, but the notation annoying though quite intuitive; I'll provide more clarification if needed.

Consider any process $U\in\mathscr{L}^2(\mathopen{[}0,\infty\mathclose{[};\mathscr{W}^{1,2})$, where $\mathscr{W}^{1,2}$ is the usual Watanabe–Sobolev space of Malliavin-differentiable functions (Nualart's and Øksendal's $\mathbb{D}^{1,2}$, but I hate the hollow 'D', and this is what Hairer uses). Note that this is the Hilbert space of processes $U\in\mathscr{L}^2(\varOmega\times\mathopen{[}0,\infty\mathclose{[};\mathbb{R})$ for which

  • $U(t)\in \mathscr{W}^{1,2}$, for (almost) all $t\in\mathopen{[}0,\infty\mathclose{[}$;
  • there exists a version of the process $(\omega,\tau,t)\mapsto\mathfrak{D}_\tau U(\omega,t)$ in $\mathscr{L}^2(\varOmega\times\mathopen{[}0,\infty\mathclose{[}^2;\mathbb{R})$.

It can be shown that processes $U_\nu$ of the form $$U_\nu(t):= \sum_{j=1}^\nu X_j h_j(t),$$ where $X_j$ is a smooth and cylindrical random variable and $h_j\in\mathscr{L}^2(\mathopen{[}0,\infty\mathclose{[};\mathbb{R})$, are dense in $\mathscr{L}^2(\mathopen{[}0,\infty\mathclose{[};\mathscr{W}^{1,2})$. In fact, $\mathscr{L}^2(\mathopen{[}0,\infty\mathclose{[};\mathscr{W}^{1,2})$ is defined as the closure of the space of such process under a suitable norm which we will not need here.

This means that we can find a sequence of such processes $(U_\nu)_{\nu\in\mathbb{N}}$ such that $$U_\nu\to U~\text{in}~\mathscr{L}^2(\varOmega;\mathscr{L}^2(\mathopen{[}0,\infty\mathclose{[};\mathbb{R})),~\text{as}~\nu\to\infty,$$ and $$\mathfrak{D}_\cdot U_\nu\to \mathfrak{D}_\cdot U~\text{in}~\mathscr{L}^2(\varOmega;\mathscr{L}^2(\mathopen{[}0,\infty\mathclose{[};\mathbb{R})^{\otimes 2}),~\text{as}~\nu\to\infty.$$

A consequence of the above is that $$\int_0^T U_\nu(t)~\!\mathrm{d}t\to \int_0^T U(t)~\!\mathrm{d}t~\text{in}~\mathscr{L}^2(\varOmega;\mathbb{R}),~\text{as}~\nu\to\infty,$$ and $$\mathfrak{D}_\cdot\int_0^T U_\nu(t)~\!\mathrm{d}t\to \mathfrak{D}_\cdot\int_0^T U(t)~\!\mathrm{d}t~\text{in}~\mathscr{L}^2(\varOmega;\mathscr{L}^2(\mathopen{[}0,\infty\mathclose{[};\mathbb{R})),~\text{as}~\nu\to\infty.$$

And these imply, $$\mathfrak{D}_\tau\int_0^T U(t)~\!\mathrm{d}t = \int_0^T \mathfrak{D}_\tau U(t)~\!\mathrm{d}t.$$

Now, set $U(t) = u(t)^2$. Then, by the "Chain Rule" for Malliavin derivatives (Theorem 3.5 in Øksendal et al.), $$\mathfrak{D}_\tau\int_0^T u(t)^2~\!\mathrm{d}t = \int_0^T \mathfrak{D}_\tau \big(u(t)^2\big)~\!\mathrm{d}t = 2\int_0^T u(t)\mathfrak{D}_\tau u(t)~\!\mathrm{d}t.$$

sami.spricht.sprache
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