Conditional expectation on the product of squared Brownian motion with Doléans-Dade exponential (Radon-Nikodym derivatives)

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I try to compute the conditional expectation $E_{t}^{P}\left[B_{T}^{2} \ e^{-\frac{1}{2} \int_{0}^{T} \left( a + b B_{u}^{2}\right)^{2} d u+\int_{0}^{T} \left( a + b B_{u}^{2}\right) d B_{u}}\right]$ where

  • $a$ and $b$ are deterministic
  • $B_t$ is a P-Brownian motion
  • $e^{-\frac{1}{2} \int_{0}^{T} \left( a + b B_{u}^{2}\right)^{2} d u+\int_{0}^{T} \left( a + b B_{u}^{2}\right) d B_{u}}$ is a P-martingale.
  • $E_t$ means $E[...| \mathcal{F}_t]$
  • There is equivalent Q measure $B^Q_t = B_t - \int_{0}^{t} \left( a + b B_{u}^{2}\right) du$

It seems difficult to work with Q-expectation as there will always a P-Brownian motion involved, e.g.

$ E^Q_t B_t = E^Q_t B^Q_t - E^Q_t \left(\int_{0}^{t} \left( a + b B_{u}^{2}\right) du \right)$

$ E^Q_t B^2_t = E^Q_t \left( B^Q_t - \int_{0}^{t} \left( a + b B_{u}^{2}\right) du \right)^2$

My thought is to use linear approximation $e^x = 1+x$

But I don't think my approach is correct. I was wondering if there is any clever way to derive this conditional expectation? I would really appreciate if anyone could show me how to solve this question.