Evaluating a Stochastic Integral Exponential with Girsanov's Theorem
Summary
The document evaluates an exponential expectation involving a process with linear drift and Brownian noise. Its method changes probability measure using an exponential martingale, selecting the measure under which the drifted process becomes standard Brownian motion. This transforms the original expectation into one involving the integral of Brownian motion against itself.
Itô's formula reduces that integral to a function of the terminal Brownian value and elapsed time. The remaining expectation is computed from the Gaussian distribution, yielding a closed form for the stated parameter range, beta less than the reciprocal of the horizon. The answer assumes the change of measure is valid and presents no separate proof of the required martingale condition or discussion of extensions beyond that range. The argument is a compact stochastic calculus technique, rather than a trading strategy or empirical result.
Key ideas
- An exponential change of measure can remove the process's linear drift.
- Under the new measure, the transformed process follows standard Brownian motion.
- Itô's formula expresses the stochastic integral using the terminal value and time horizon.
- The final expectation follows from the Gaussian distribution under the stated parameter restriction.
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# Computation of Expectation
# Computation of Expectation
This question has so long preoccupied my mind.Please help me to solve it.
Question: Assume $X_t$ described by the following stochastic differential equation $$dX_t^{\,\alpha}=\alpha X_t^{\,\alpha} dt+dW_t$$ where $W_t$ is a standard wiener process and $\alpha \in R$. How do I compute
$$ E\left[exp\left((\beta-\alpha)\int_{0}^{T}X_t^{\,\alpha}\,dX_t^\alpha+\frac{\alpha^2}{2}\int_{0}^{T}X_t^{\,2\alpha}\,dt\right)\right]$$ for all $\beta<\frac{1}{T}$
## Answer by user16891 (score 4)
https://quant.stackexchange.com/a/18940
let $Y_t=(X_t)^\alpha$,then $$dY_t=\alpha Y_tdt+dW_t^P$$ we define $Q$ measure by $$\frac{dQ}{dP}=exp\left(-\alpha\int_{0}^{T}Y_t\,dW_t^p-\frac{1}{2}\alpha^2\int_{0}^{T}Y_t^2 dt\right)$$ this shows that $$W_t^Q=W_t^P+\alpha\,\int_{0}^{t}Y_s\,ds$$ is standard wiener process under $Q$ measure, thus we have $$dW_t^P=dW_t^Q-\alpha\,Y_t dt$$ and $$dY_t=\alpha Y_tdt+dW_t^P=\alpha Y_tdt+dW_t^Q-\alpha\,Y_t dt=dW_t^Q$$ This means that $\{Y_t\}_{0\leq t \leq T}$ is a standard Wiener process under $Q$ measure. Now we can compute the expectation as follow $$ E^P\left[exp\left((\beta-\alpha)\int_{0}^{T}Y_t\,dY_t+\frac{\alpha^2}{2}\int_{0}^{T}Y_t^2\,dt\right)\right]=E^Q\left[exp\left(\beta\int_{0}^{T}Y_t\,dY_t\right)\right]=E^Q\left[exp\left(\beta\int_{0}^{T}W_t^Q\,dW_t^Q\right)\right]=E^Q\left[exp\left(\frac{\beta}{2}[(W_T^Q)^2-T]\right)\right]=\frac{e^\frac{-\beta\,T}{2}}{\sqrt{1-\beta\,T}}\,\,\,\,\,\,\,\,\,\,\,\,\,$$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.