Substitution of a random variable into a stochastic integral

Question

I am faced with the following fact while reading a paper:

Let $\{ \mathcal{F}_t \}_{t \in [0,T]} $ be the filtration generated by a one-dimensional Brownian motion $\{ B_t \}_{t \in [0,T]} $ defined on some probability space. Let $\{X_t (y) \}_{t \in [s,T]}$ be an adapted process, for each $y \in \mathbb{R}.$ Then, for any random variable $\eta$ that is $ \mathcal{F}_s$-measurable, the author claims that for every $t \in [s,T],$

$$ \bigg\{ \int_s^t X_r (y) \, dB_r \bigg\} \bigg|_{y= \eta} = \int_s^t X_r (\eta) \, dB_r.$$

The argument is that the stochastic integral $\int_s^t X_r (y) \, dB_r$ is $\sigma \big\{ B_r - B_s, r \in [s,T] \big\}$-adapted and is therefore independent of $ \mathcal{F}_s$, and in particular, independent of $\eta$. Therefore, direct substitution is allowed.

I am wondering if there is any result in the literature that guarantees direct substitution of a random variable into a stochastic integral, given its independence? I cannot show it directly from its definition.

Answer 1

I think that the assertion is, in general, wrong. The problem is, essentially, the following: If $F(y)$ is for each $y \in \mathbb{R}$ a random variable which is only defined up to a null set, then the expression

$$F(y) \bigg|_{y=\eta}$$

is ill-defined. There are null sets building up; depending on which representation we choose for $F(y)$ we get totally different results.

The same problem pops up when one tries to generalize the pull out property of the conditional expectation (see this question).

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Here is an illustrating example for the problem your are considering:

Fix $0 < s \leq T <\infty$ and consider

$$X_r(y) := 1_{\{y\}}(B_s) \qquad \eta := B_s.$$

Since $\mathbb{P}(B_s=y)=0$ for any $y \in \mathbb{R}$, we have by Itô's isometry

$$\mathbb{E} \left( \left| \int_s^t X_r(y) \, dB_r \right|^2 \right) ,$$

and therefore

$$F(y) := \int_s^t X_r(y) \, dB_r = 0.$$ (The stochastic integrals is only defined up to null set and therefore we can choose $F(y,\omega)=0$, $\omega \in \Omega$ as a representative.) Hence, $F(\eta) =F(B_s)=0$ almost surely. On the other hand, we have

$$\int_s^t X_r(\eta) \, dB_r = \int_s^t 1_{\{B_s\}}(B_s) \, dB_r = B_t-B_s.$$ Thus,

$$0 = F(\eta) \neq \int_s^t X_r(\eta) \, dB_r = B_t-B_s.$$