Integration by Parts for the Ornstein–Uhlenbeck Process
Summary
The document resolves a boundary-term discrepancy when applying integration by parts to a deterministic exponential factor and a Wiener process in an Ornstein–Uhlenbeck derivation. The calculus integration-by-parts rule gives an endpoint term evaluated at both zero and time t, followed by an ordinary integral involving the process and the derivative of the exponential factor.
Because the Wiener process starts at zero almost surely, the lower endpoint contributes nothing. The upper endpoint is the process value at t, leaving the stated form: the terminal Wiener value minus the integral of the process weighted by the exponential kernel and its rate parameter. The answer addresses this specific deterministic-integrand case; it does not develop the more general stochastic integration-by-parts formula or discuss nonzero initial conditions.
Key ideas
- The integration-by-parts boundary term must be evaluated at both endpoints.
- For a Wiener process starting at zero, the lower endpoint vanishes.
- Differentiating the exponential kernel produces its rate parameter as a multiplier.
- The resulting expression contains the terminal process value and a weighted time integral.
- The derivation assumes the Wiener process has zero initial value.
Tags
Full text
# Ornstein–Uhlenbeck process – integration by parts
# Ornstein–Uhlenbeck process – integration by parts
While deriving the solution for the stochastic differential equation that models the Ornstein–Uhlenbeck process, Paul Wilmott (Paul Wilmott on Quantitative Finance, chapter 4, page 87) performs the following integration by parts ($X$ is normal):
$$\int_0^t e^{\gamma(s-t)} \, dX(s) = X - \gamma \int_0^t e^{\gamma(s-t)}X(s) \, ds$$
I was trying to replicate this result by using the following result from calculus:
$$u=e^{\gamma(s-t)} \implies du = \gamma e^{\gamma(s-t)} \, ds$$ $$dv = dX(s) \implies v = X(s)$$
Then, we could write
$$\int_0^t u\, dv = u\,v|_0^t - \int_0^t v \, du = X(1-e^{-\gamma t})-\gamma \int_0^t e^{\gamma(s-t)}X(s) \, ds$$
which is different from his original solution. Where is my mistake? Is there any other way of evaluating integration by parts in the context of stochastic calculus?
## Answer by Pleb (score 7, accepted)
https://quant.stackexchange.com/a/60271
First and foremost if $u=e^{\gamma (s-t)}$ then $\frac{du}{ds} = \gamma e^{\gamma(s-t)} \iff du=\gamma e^{\gamma(s-t)} \,ds$. Now, in the Ornstein–Uhlenbeck process $X_t$ is a Wiener process and satisfies $X_0 = 0 \: \: \text{a.s.}$ (see this and this). Then, the first term in the integration by parts formula specified above gives you:
$$uv\vert^t_0 = e^{\gamma (t-t)} X_t - e^{\gamma (0-t)} X_0 = X_t$$
In conclusion, we get: \begin{align} \int_0^t u \: dv &= uv\vert^t_0 - \int_0^t v \: du\\ &= X_t - \int_0^t X(s) \gamma e^{\gamma(s-t)} \,ds\\ &= X_t - \gamma\int_0^t e^{\gamma(s-t)} X(s) \,ds \end{align}
I believe this should be the essence of it.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.