1. The set $S$ of smooth random variables is the set of random variables $F : \Omega \rightarrow \mathbb R$ such that there exist a function $f$ in $ \mathcal C_p^{\infty}(\mathbb R^n)$ (for some $n \geq 1$) and elements $h_1, \cdots , h_n$ of $L^2$ such that

$$F = f\left((\int_0^t h_1\, dW_s), \cdots , (\int_0^t h_n\,dW_s)\right) \qquad (*)$$

  1. The set $\mathcal P$ denotes the set of random variables of the form (*) where $f$ is a polynomial.
  2. $S_b$ (resp. $S_0$) denotes the space of random variables of the form (*) with $f$ in $ \mathcal C_p^{\infty}(\mathbb R^n)$ (resp. $ \mathcal C_0^{\infty}(\mathbb R^n)$ ).

We define the Malliavin derivative $DF$ of $F$ as the $L^2$-valued random variable $$DF =\sum_{i=1}^n \frac{\partial f}{\partial x_i}f\left((\int_0^t h_1\, dW_s), \cdots , (\int_0^t h_n\,dW_s)\right)h_i.$$

  1. Integration by part formula: Let $F$ be a smooth random variable of the form (*) and let $h$ be an element of $L^2$. The integration by parts formula say that

$$\mathbb E[\langle DF, h\rangle_{L^2}] = \mathbb E[F(\int_0^t h\, dW_s)] \qquad (**).$$

By normalization the relation (**) we can assume that $||h||_{L^2} = 1$ and $F = f\left((\int_0^t e_1\, dW_s), \cdots , (\int_0^t e_n\,dW_s)\right) $ where $\{e_1, \cdots , e_n\}$ are ONB in $L^2$ , $e_1 = h$ and $f \in \mathcal C_p^{\infty}(\mathbb R^n)$ We have that

\begin{align} \mathbb E\left[\langle DF, h\rangle_{L^2}\right] =& \mathbb E \left[ \frac{d}{dx_1}f\left((\int_0^t e_1\, dW_s), \cdots , (\int_0^t e_n\,dW_s)\right)\right] \tag{A}\\ =& (2π)^{−n}\int_{\mathbb R^n}\frac{d}{dx_1}f(x_1, \cdots , x_n) \exp\left(-\frac{1}{2}\sum_{i=1}^nx_i^2\right)\,dx_1 · · · dx_n \tag{B} \\ =& (2π)^{−n}\int_{\mathbb R^n} x_1f(x_1, \cdots , x_n) \exp\left(-\frac{1}{2}\sum_{i=1}^nx_i^2\right)\,dx_1 · · · dx_n \tag{C} \\ =& \mathbb E[FW(e_1)] \tag{D}\\ =& \mathbb E[FW(h)] . \end{align}


  1. In A: Because $\langle e_1,h_i\rangle=0, \quad \forall i=2, \cdots, n$ then they neglect the terms $dx_2, \cdots dx_n.$. Am'I right?
  2. In B: They use the PDF of the standard law normal, but why?
  3. In C: I don't can't develop how they use the integration by part in $\mathbb R^n$.
  • $\begingroup$ I do not understand the notation $\langle DF,h\rangle_{L^2}$. What is the result of this? If you mean the scalar product on $L^2$ then it should be a scalar. I imagine it is not that, but a random variable, right? Could you please explain this point a bit better? Thanks $\endgroup$
    – Avitus
    Jun 27, 2013 at 10:30
  • $\begingroup$ @Avitus $\langle DF, h \rangle_{L^2}$ is $L^2$-valued r.v and we can write $\langle DF, h \rangle_{L^2}=\langle \sum_{i=1}^n \frac{\partial f}{\partial x_i}f\left((\int_0^t h_1\, dW_s), \cdots , (\int_0^t h_n\,dW_s)\right)h_i, h\rangle_{L^2}$ $\endgroup$
    – Zbigniew
    Jun 27, 2013 at 10:40
  • $\begingroup$ thanks. I am still not familiar with notation but I see that you wrote $\frac{d}{dx}f(x_1,\dots,x_n)\exp(...)$ in B). Does the derivative refers to the only $f$ or to the whole product? If it is referred to the single $f$, then C) is obtained by integration by parts and the extra factor $x$ comes from the derivative of $\exp(...)$. I presume that the boundary term $f\exp(...)|^{\infty}_{\infty}=0$ because $\exp(-x_1^2)\rightarrow 0$ in both limits if $f$ satisfies some extra conditions (like having compact support or being in the Schwarz space ) What is $\mathcal C^{\infty}_p(\mathbb R^n)?$ $\endgroup$
    – Avitus
    Jun 27, 2013 at 11:47
  • $\begingroup$ @Avitus. Yes the derivative refers to $f$ only. $\endgroup$
    – Zbigniew
    Jun 27, 2013 at 13:04
  • $\begingroup$ I would like to create a new tag for Malliavin calculus, but infortunately I do not have enough reputation to do this. $\endgroup$
    – Zbigniew
    Jun 27, 2013 at 13:09

1 Answer 1


Let me try to provide some insight...

A) Yes. To be more precise, you can, as you said, assume $||h||_{L^{2}}=1$.

Now, there is an ONB $\{e_{n}\}_{n\geq 1}$, such that $h=e_{1}$ and

$$F = f\left((\int_0^t e_1\, dW_s), \cdots , (\int_0^t e_n\,dW_s)\right) $$

This is because the mapping $W: h \rightarrow \int_{0}^{t}h(s)dW_{s}$ is linear.

In fact, let's denote $\int_{0}^{t}h(s)dW_{s} := W(h)$.

Now, since

$$DF =\sum_{i=1}^n \frac{\partial }{\partial x_i}f\left(W(e_{1}), \cdots , W(e_{n})\right)e_i.$$

we have that

$$\langle DF,e_{1}\rangle =\sum_{i=1}^n \frac{\partial }{\partial x_i}f\left(W(e_{1}), \cdots , W(e_{n})\right)\langle e_{1},e_{i}\rangle = \frac{\partial }{\partial x_1}f\left(W(e_{1}), \cdots , W(e_{n})\right)$$

B) Because the RV's

$$\int_{0}^{t}e_{i}(s)dW_{s}$$ are normally distributed when $e_{i}$ is a deterministic function. Therefore, the vector $(W(e_{1}),\cdots, W(e_{n}))$ has an $n$-dimensional normal distribution.

C) This is just the usual integration-by-parts formula. We have, for

$$\phi(\vec{x}) = \exp\left(-\frac{1}{2}\sum_{i=1}^nx_i^2\right)$$


$$\vec{x} = (x_{1},\cdots, x_{n})$$

$$E[\langle DF, e_{1}\rangle] = (2π)^{−n}\int_{\mathbb R^n}\frac{\partial }{\partial x_1}f(\vec{x})\phi(\vec{x}) d\vec{x} = (2π)^{−n}\int_{\mathbb R^n}f(\vec{x})\phi(\vec{x})x_{1} d\vec{x}$$


$$\lim_{\vec{x}\rightarrow \infty}\phi(\vec{x}) = \vec{0}$$

and the "boundary" element is zero.

You can think of the integration-by-parts as choosing $u$ and $v$ such that



$$dv(x_{1}) = \frac{\partial}{\partial x_{i}}f(\vec{x})dx_{1}$$

Then, you have

$$du(x_{1}) = x_{1}\exp{\left(-\frac{1}{2}x^{2}_{1}\right)}$$



Hope that helps.


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