Reflecting a Brownian Stochastic Integral at a First-Passage Time
Summary
The question defines an Itô integral with an adapted, square-integrable integrand and reflects Brownian motion after its first passage to a positive level. It asks how to express the integral in terms of the reflected process. The response instead introduces an exponential process and invokes Girsanov’s theorem to construct a Brownian motion under a changed measure, then writes the reflected path using that process and the integral.
This exchange offers a possible route through change of measure, but it does not establish a direct representation of the integral from the reflected process. The stated square-integrability condition alone does not ensure that the exponential is a true martingale suitable for the claimed measure change; an additional condition is generally needed. Also, the passage time is defined using the original Brownian motion, so the response’s changed-measure argument does not by itself validate the reflected process under the new measure. Treat the displayed derivation as tentative rather than a proved result.
Key ideas
- The reflected Brownian path agrees with the original path up to its first passage time and mirrors it afterward.
- The response proposes using an exponential change of measure to relate Brownian motion and the stochastic integral.
- Square-integrability of the integrand alone does not guarantee the required exponential is a true martingale.
- The answer does not prove that the reflection and measure-change steps yield the requested representation.
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Full text
# On the reflection of a stochastic integral
# On the reflection of a stochastic integral
Let ${(I_t)}_{t\geq 0}$ be a stochastic integral defined by $$ I_t=\int_{0}^{t}\theta_sdW_t, $$ where $W$ is a standard Brownian motion defined on $(\Omega,\mathcal{F},{(\mathcal{F}_t)}_{t\geq 0},\mathbb{P})$ and $\theta$ a stochastic process adapted to $\mathcal{F}_t$ satisfying the follows condition of integrability $$ E\left(\int_{0}^{t}\theta_s^2 ds\right)<\infty\;\;\ \forall t> 0. $$
We define the first passage time at $a$ for Brownian motion $W$ by the following random variable $$ \tau_a = \inf\{t\geq 0,W_t\geq a\}, $$ where $a>0$.
It is possible to show that $\tau_a$ is a stopping time. Moreover, By virtue of the reflection principle, we know that the following process
\begin{equation*} Z_t = \begin{cases} W_t \qquad & if \qquad 0 \leq t \leq \tau_a \\ 2a-W_t \qquad & if \qquad t > \tau_a \end{cases} \end{equation*}
also follows a standard Brownian motion under $\mathbb{P}$.
My question is as follows : Is it possible to rewrite the process $I$ in relation to the process $Z$?
I would like your opinion on this issue, thank you in advance.
## Answer by user16651 (score 1)
https://quant.stackexchange.com/a/28394
Set $$X_t=\exp\left(-\int_{0}^{t}\theta_sdW_s^{\mathbb{P}}-\frac{1}{2}\int_{0}^{t}\theta_s^{\,2}ds\right)$$ By application of Gisanov theorem , we have
- $X_t$ is a $\mathbb{P}-$ martingale.
- By changing the measure $\mathbb{P}$ to $\mathbb{Q}$ such that $$\mathbb{E^P}\left[\frac{d\mathbb{Q}}{d\mathbb{P}}\Big{|}\mathcal{F_t}\right]=\frac{d\mathbb{Q}}{d\mathbb{P}}\Big{|}_\mathcal{F_t}=X_t$$ then $$W_t^{\mathbb{Q}}=W_t^{\mathbb{P}}+I_t$$ is a $\mathbb{Q}$ standard wiener process. We have \begin{equation*} Z_t = \begin{cases} W_t^{\mathbb{Q}}-I_t \qquad & , \qquad 0 \leq t \leq \tau_a \\ 2a-W_t^{\mathbb{Q}}+I_t \qquad & , \qquad t > \tau_a \end{cases} \end{equation*} then (If I am right) \begin{equation*} dZ_t = \begin{cases} dW_t^{\mathbb{Q}}-dI_t \qquad & , \qquad 0 \leq t \leq \tau_a \\ -dW_t^{\mathbb{Q}}+dI_t \qquad & , \qquad t > \tau_a \end{cases} \end{equation*}Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.