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

  • $\begingroup$ Where does the condition $\beta< 1/T$ come from? $\endgroup$
    – Richi Wa
    Commented Jul 20, 2015 at 6:37
  • 1
    $\begingroup$ This condition is necessary; otherwise, the expectation will be infinity. $\endgroup$
    – Gordon
    Commented Jul 20, 2015 at 12:46

1 Answer 1


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

  • 1
    $\begingroup$ Indeed, a nice solution. $\endgroup$
    – Gordon
    Commented Jul 20, 2015 at 12:57

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