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This is a follow-up of a previous question I asked.

Suppose that $f \colon \mathbb R^n \to \mathbb R^n$ is a bounded vector field (not necessarily the gradient of another function), that $(W_t)_{0\leq t \leq T}$ is a standard Brownian motion in $\mathbb R^n$, and let $X$ denote the stochastic integral $$ X = \int_0^T f(W_t) \cdot d W_t. $$ Is it possible in general to bound conditional expectations of the following form in terms of $x \in \mathbb R^n$? $$ \mathbb E [X | W_T = x] $$ In particular, is the conditional expectation finite?

Roberto Rastapopoulos
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  • It seems that $f$ depends only on $1$ single BM and $f(W_t) = (f_1(W_t), f_2(W_t),...,f_n(W_t))' $, right? And also $X \in \mathbb{R}^n$? It suffices then to separately compute the $n$ conditional expectation $$\mathbb{E} \left( \left. \int_0^T f_{\color{red}i} (W_t)dW_t\right| W_T = x \right)$$ for $\color{red}{i} = 1,..,n$ with the method in the previous answer. And of course, there is no need to use gradient here. – NN2 Jan 10 '24 at 12:03
  • No, there are $n$ one-dimensional Brownian motions. – Roberto Rastapopoulos Jan 10 '24 at 14:50
  • The technique to solve the problem is to retrieve a scalar function $F\in \mathbb{R}$ from its gradient $\nabla F = f \in \mathbb{R}^n$. And it is possible to do that via this method. – NN2 Jan 10 '24 at 16:33
  • Thanks! Unfortunately, this works only when the vector field is rotational-free (in 3d, but this generalizes), I think. – Roberto Rastapopoulos Jan 10 '24 at 16:36
  • Perhaps do you have a case that the method doesn't work? – NN2 Jan 10 '24 at 16:52
  • Yes, take for example $\nabla f = y \mathbf i - x \mathbf j$. – Roberto Rastapopoulos Jan 10 '24 at 17:15
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    I think there is an answer to my question in this (surprisingly little cited) paper. https://www.sciencedirect.com/science/article/pii/0022247X69901826 – Roberto Rastapopoulos Jan 10 '24 at 17:18

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