4
$\begingroup$

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 $\beta<\frac{1}{T}$

$\endgroup$
  • $\begingroup$ Where does the condition $\beta< 1/T$ come from? $\endgroup$ – Richard Jul 20 '15 at 6:37
  • 1
    $\begingroup$ This condition is necessary; otherwise, the expectation will be infinity. $\endgroup$ – Gordon Jul 20 '15 at 12:46
3
$\begingroup$

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 $$ E^P\left[exp\left((\beta-\alpha)\int_{0}^{T}Y_t\,dY_t+\frac{\alpha^2}{2}\int_{0}^{T}Y_t^2\,dt\right)\right]=E^Q\left[exp\left(\beta\int_{0}^{T}Y_t\,dY_t\right)\right]=E^Q\left[exp\left(\beta\int_{0}^{T}W_t^Q\,dW_t^Q\right)\right]=E^Q\left[exp\left(\frac{\beta}{2}[(W_T^Q)^2-T]\right)\right]=\frac{e^\frac{-\beta\,T}{2}}{\sqrt{1-\beta\,T}}\,\,\,\,\,\,\,\,\,\,\,\,\,$$

$\endgroup$
  • 1
    $\begingroup$ Indeed, a nice solution. $\endgroup$ – Gordon Jul 20 '15 at 12:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.