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Question:

In general, $Z_t := \exp(\int_0^t \phi_s dBs - \frac{1}{2}\int_0^t\phi_s^2ds)$ is the exponential martingale. Is $Z_t := \exp(-\int_0^t \phi_s dBs - \frac{1}{2}\int_0^t\phi_s^2ds)$ also an exponential martingale?

Origin:

Let $(B_t)_{t \geq 0}$ be abrownian motion starting at $x>0$. Let $T_a := \inf\{t :B_t =a\}$. Compute

$$ E[exp(-\int_0^{T_a} \frac{1}{B_s^2}ds) | B_0 = x]$$ where $0 < a < x$.

The idea is to rewrite $exp(-\int_0^{T_a} \frac{1}{B_s^2}ds)$ into a martingale and use the optimal stopping theorem. Using Itô on $\ln(B_t)$ (and assuming $(B_0 \neq 0)$) yields

$$ \frac{B_0}{B_t}e^{-\int_0^{T_a} \frac{1}{B_s^2}ds} = e^{-\int_0^t \frac{1}{B_s}dBs -\frac{1}{2} \int_0^{T_a} \frac{1}{B_s^2}ds}$$

Now it would be very convenient if the right hand side was indeed the exponential martingale.

Oskar
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  • What happens to the first $Z_t$ if you replace $\phi_s$ with $-\phi_s,?$ – Kurt G. Mar 08 '24 at 17:24
  • Ah, I feel stupid! Just say $\psi_s = -\varphi_s$ and we got the same back, since the quadratic variation is not affected by the sign. Correct? – Oskar Mar 08 '24 at 17:32

1 Answers1

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I would write the Ito decomposition of the process: let $$X_t=-\int_0^t \phi_s dBs - \frac{1}{2}\int_0^t\phi_s^2ds, \quad Y_t := \exp\left(X_t\right)$$ then $dX_t=-\frac{1}{2}\phi^2_t\,dt-\phi_t\,dB_t$. Using the Ito's formula we get that \begin{align} dY_t&=Y_t\,dX_t+\frac{1}{2}Y_t\,d[X]_t\\ &=Y_t\left(-\frac{1}{2}\phi^2_t\,dt-\phi_t\,dB_t+\frac{1}{2}\phi^2_t\,dt\right)\\ &= -\phi_t\,Y_t\,dB_t \end{align} which is the decomposition of an exponential martingale, since $-\phi\in M^2_{loc}$

Davide
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