2
$\begingroup$

Let $(B_t)_{t}$ a Brownian motion and $F \in L^2(\Omega,\mathcal{F}_T,\mathbb{P})$. Then we know by Itô's representation theorem that there exist a process $X$ such that $F=\mathbb{E}F+\int_0^T X_s \, dB_s \tag{1}$

I was wondering whether there is formula (at least for some cases) to obtain $X$ explitictely for a given random variable $Y$. I found the following statement in this paper (section 2.1, page 3):

Let $F=f(W(h_1),\ldots,W(h_n))$ where $f$ is bounded and smooth (with bounded derivatives) and $W(h_j) := \int_0^T h_j(s) \, dB_s$ for some $h_j \in L^2$. Then the Malliavin derivative is given by $D_t F =\sum_{i=1}^n \partial_i f(W(h_1),\ldots,W(h_n)) \cdot h_i(t)$ and $X_s = \mathbb{E}(D_s F|\mathcal{F}_s)$ in (1).

So let $f$ be bounded and smooth and consider $F := f(B_T)$. Then (using the notation from above) $W(h_1) := \int_0^T \underbrace{1}_{=:h_1} \, dB_s = B_T \qquad F=f(W(h_1))$ By the given formula I obtain

$D_t F = f'(B_T) \cdot 1 \\ \Rightarrow F = \mathbb{E}F + \int_0^T f'(B_T) \, dB_s \tag{2}$

On the other hand I could also apply Itô's formula,

$\underbrace{f(B_T)}_{F}-f(0) = \int_0^T f'(B_s) \, dB_s + \frac{1}{2} \int_0^T f''(B_s) \, ds \tag{3}$

Is it correct like that? I find it really surprising that both (2) and (3) hold.

And is the boundedness of $f$ really necessary or are there more general results of this type? (For example $f(x) := x$ is not bounded, but a very well-behaving function.)

  • 0
    Ah okay, that makes sense. Thanks! (I had $D_t F = f'(B_t)$ (which is adapted) and when I corrected it, i.e. $D_t F=f'(B_T)$, I forgot that this one is not adapted anymore.)2012-12-11

1 Answers 1

2

As TheBrdige already pointed out, it should read

$F= \mathbb{E}F+ \int_0^t \mathbb{E}(f'(B_T)|\mathcal{F}_s) \, dB_s \tag{4}$ in (2).

For example applied to $f(x):=x^2$ we have $F=f(B_T)=B_T^2$ and

$\begin{align} B_T^2=F &\stackrel{4}{=} \underbrace{\mathbb{E}F}_{T} + \int_0^T \underbrace{\mathbb{E}(2 B_T|\mathcal{F}_s)}_{B_s} \, dB_s = T +2 \int_0^T B_s \, dB_s \\ &= T + 2\int_0^T B_s \, dB_s \end{align}$

which is exactly the result one would obtain by applying Itô calculus.

I am still wondering whether it is really necessary for $f$ to be bounded (as assumed in the theorem above). For $f(x):=x^2$ (clearly not bounded) it work's fine as one can see...