Itô Product Rule for a Stochastic Exponential and Process
Summary
The document works through an Itô-calculus derivation involving a stochastic exponential and another process. It first differentiates the exponential, then computes its quadratic covariation with the second process, using the stated orthogonality of one martingale component to simplify the expression. Applying the Itô product rule yields a drift contribution from that covariation term.
Taking expectations of the integrated product differential gives an identity relating the expected terminal product to the initial expectation and an integral involving the exponential, a function of quadratic variation, and the second process’s coefficient. The derivation illustrates how stochastic product rules and covariation terms enter such expectation calculations. It relies on the setup and assumptions from a larger proof that are not fully included here, so the displayed steps alone do not establish all conditions needed for the expectations, integrals, or limiting terminal values to be well defined.
Key ideas
- The stochastic exponential is differentiated using Itô’s formula.
- The quadratic covariation with the second process contributes an additional term to the product differential.
- Orthogonality of the martingale components simplifies the covariation calculation.
- Taking expectations of the integrated product rule produces an identity involving a weighted integral.
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# stochastic calculus - Itô formula?
# stochastic calculus - Itô formula?
I encounter a problem in the proof below:
- I don't know how to proove the first line in yellow (cf below): it makes me think about the Itô formula a lot
- I don't undertand the deduction (ok $\gamma^{\chi}$ has the constant law of a brownian motion but what does it tell us ?)
Thank you
## Answer by Gordon (score 3, accepted)
https://quant.stackexchange.com/a/17753
For the first question, since by definition, \begin{align*} \varepsilon_t^{if} = e^{i \int_0^{t}f\big(\frac{1}{\xi}\langle M\rangle_s\big)\frac{dM_s}{\sqrt{\xi}} + \frac{1}{2}\int_0^t f\big(\frac{1}{\xi}\langle M\rangle_s\big)\frac{d\langle M\rangle_s}{\xi}}, \end{align*} then, \begin{align*} d\varepsilon_t^{if} = i \varepsilon_t^{if} f\Big(\frac{1}{\xi}\langle M\rangle_t\Big)\frac{dM_t}{\sqrt{\xi}}. \end{align*} Moreover, \begin{align*} \langle \varepsilon_t^{if}, H_t \rangle = \int_0^t i \varepsilon_s^{if} f\Big(\frac{1}{\xi}\langle M\rangle_s\Big) h_s \frac{d\langle M\rangle_s}{\sqrt{\xi}}, \end{align*} as $\langle M_t, R_t \rangle = 0$. Consequently, \begin{align*} d\big(\varepsilon_t^{if} H_t\big) &= H_t d\varepsilon_t^{if} + \varepsilon_t^{if} dH_t + d \langle \varepsilon_t^{if}, H_t \rangle\\ &= H_t d\varepsilon_t^{if} + \varepsilon_t^{if} dH_t + i \varepsilon_t^{if} f\Big(\frac{1}{\xi}\langle M\rangle_t\Big) h_t \frac{d\langle M\rangle_t}{\sqrt{\xi}}. \end{align*} That is, \begin{align*} \mathbb{E}\big(\varepsilon_{\infty}^{if} H_{\infty}\big) -\mathbb{E}(H_0) &= \mathbb{E}\bigg(\int_0^{\infty}\!\!\!\! H_t d\varepsilon_t^{if} + \int_0^{\infty} \!\!\!\!\varepsilon_t^{if} dH_t + i\int_0^{\infty}\!\!\!\!\varepsilon_t^{if} f\Big(\frac{1}{\xi}\langle M\rangle_t\Big) h_t \frac{d\langle M\rangle_t}{\sqrt{\xi}} \bigg) \\ &=i \mathbb{E}\bigg(\int_0^{\infty}\!\!\!\!\varepsilon_t^{if} f\Big(\frac{1}{\xi}\langle M\rangle_t\Big) h_t \frac{d\langle M\rangle_t}{\sqrt{\xi}} \bigg), \end{align*} or \begin{align*} \mathbb{E}\big(\varepsilon_{\infty}^{if} H\big) = \mathbb{E}(H)+i \mathbb{E}\bigg(\int_0^{\infty}\!\!\!\!\varepsilon_t^{if} f\Big(\frac{1}{\xi}\langle M\rangle_t\Big) h_t \frac{d\langle M\rangle_t}{\sqrt{\xi}} \bigg). \end{align*}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.