Fix $T>0$. Write $C$ for the space of all continuous functions $f:[0,T] \to \mathbb{R}$. For a standard Brownian motion $B = \{B_t\}_{t \in [0,T]}$ defined on probability space $(\Omega,F,P)$, let $W$ be the classical Wiener measure. That is, we treat $B$ as a random variable $B: \Omega \to C$ and $B(\omega)\in C$ is the continuous sample path, so $W$ is the distribution law of $B$: if $E \subseteq C$ is a measurable set, then $W(E) = P(\{\omega \in \Omega | B(\omega) \in E\})$.

Now fix $t \in [0, T]$. I am asked to compute this integral:

\begin{equation*} \int_{C}\phi(t)dW(\phi) \end{equation*}

How do I even start? I have no idea how to approach such a problem. In a special case which I integrate on a finite set $E=\{\phi_1,\cdots,\phi_n\} \subseteq C$, I can write:

\begin{equation*} \int_{E}\phi(t)dW(\phi) = \sum_{i=1}^n\phi_i(t)W(\{\phi_i\}) = \sum_{i=1}^n\phi_i(t) P\{B(\omega) = \phi_i\} \end{equation*}

Even in such a special case I don't know how to calculate $P\{B(\omega) = \phi_i\}$ .

  • $\begingroup$ What is $\phi$? Is it a set of continuous functions? If so, what is $\phi(t)$? $\endgroup$ Commented Mar 11, 2022 at 16:35
  • $\begingroup$ $W$ is a probability measure on $C$. $ \phi$ is an element of $C$ so $ \phi(t) \in \mathbb{R}$ $\endgroup$
    – 温泽海
    Commented Mar 11, 2022 at 16:39
  • $\begingroup$ for a fixed $t$ you have $W|_t$ is just a normal distrtibution with parameters $\mu = 0$, $\sigma^2 = t$, so that integral $\int_C \phi(t)W(\mathrm d \phi) = 0$. You can understand it also as follows: for every $\phi$ you are as likely to get $-\phi$ $\endgroup$
    – SBF
    Commented Mar 11, 2022 at 16:49

1 Answer 1


This all may become simpler to grasp if you add some formalism. You have a measure $W$ on $C$ and you are asked to compute an integral $$ \int_C e_t(\phi)W(\mathrm d\phi) $$ where $e_t:C \to \Bbb R$ is the evalution map that assigns to each function $\phi\in C$ its value $\phi(t) \in \Bbb R$. So you can treat $(C,\mathcal C, W)$ as a probability space and hence $e_t$ would just be a random variable, whose expectation you are asked to find. You very well know the distribution of this random variable: $$ W(\phi:e_t(\phi) \in A) = P(\omega: B(t, \omega) \in A) = \mathcal N(0, t)(A). $$ The rest should be easy.


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