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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.)

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are you sure about your calculation of Malliavin derivative? Shouldn't it be $f'(B_T).1$ ? Best regards. –  TheBridge Dec 10 '12 at 22:03
    
Ah yes, you are right... but still it's very surprising, in my oppinion. –  saz Dec 11 '12 at 6:47
1  
@ saz : please note that you integrand is not adapted in your equation (2) (it is correct but in the language of anticipative stochastic integration), to get an adapted integrand and fall back into Itô theory you must take $X_s= E[f'(B_t)|\mathcal{F}_s]$, this is this process that can be splited it 2 components giving (3). Best regards –  TheBridge Dec 11 '12 at 11:46
    
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.) –  saz Dec 11 '12 at 16:46

1 Answer 1

up vote 0 down vote accepted

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...

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