Girsanov Theorem, Radon-Nikodym Derivative backward

Question

Given a filtered probablity space $(\Omega,\mathcal{F},{\mathcal{F}}_t,\mathbb{P})$ and a standard Brownian motion $W_t$.

Normally, in Girsanov Theorem, we use the exponential martingale $Z_t=\exp(-\int_0^tH_sdW_s -\frac{1}{2}\int_0^tH_s^2 d_s)$ as the Radon-Nikodym Derivative to find an equivalent martingale measure, i.e. to define a probability measure $\mathbb{Q}$, s.t. $\frac{\mathrm{d}\mathbb{Q}}{\mathrm{d}\mathbb{P}}=Z_T$.

Then $W_t^{\mathbb{Q}}=W_t+\int_0^tH_sds$ is a standard Brownian motion under $\mathbb{Q}$.

Now, my question is, since $\mathbb{P}$ and ${\mathbb{Q}}$ are equivalent, by Radon-Nikodym Theorem, there exists a $\mathcal{F}_T$-measurable random variable $\Lambda$, s.t. $\frac{\mathrm{d}\mathbb{P}}{\mathrm{d}\mathbb{Q}}=\Lambda$, can we compute $\Lambda$ when $Z_T$ is known?

Answer 1

The result you're looking for is $$ \left. \frac{d\Bbb{P}}{d\Bbb{Q}}\right\vert_{\mathcal{F}_t} = \left( \left. \frac{d\Bbb{Q}}{d\Bbb{P}}\right\vert_{\mathcal{F}_t} \right)^{-1} $$ This is a result from measure theory but since you mention it, let's see how we can show it based on Girsanov theorem.

Starting from the definitions you provide and introducing some notations $$ Z_t = \left. \frac{d\Bbb{Q}}{d\Bbb{P}}\right\vert_{\mathcal{F}_t} = \exp\left( -\int_0^t H_s dW_s^\Bbb{P} - \frac{1}{2}\int_0^t H_s^2 ds \right) := \mathcal{E} \left( -\int_0^t H_s dW_s^\Bbb{P} \right) $$ where $\mathcal{E}(X_t)$ figures the stochastic exponential of the process $X_t$ i.e. $$ \mathcal{E}(X_t) = \exp\left( X_t - \frac{1}{2} \langle X \rangle_t \right) $$ Similarly let's define the stochastic logarithm $\mathcal{L}$ of a process $X_t$ such that: $$ \mathcal{L}(\mathcal{E}(X_t)) = X_t $$

What Girsanov theorem says, is that the process on the LHS of the following equation is a Brownian motion under $\Bbb{Q}$ \begin{align} W_t^\Bbb{Q} &= W_t^\Bbb{P} - \left\langle W_s^\Bbb{P}, \mathcal{L}\left( \left. \frac{d\Bbb{Q}}{d\Bbb{P}}\right\vert_{\mathcal{F}_s} \right) \right\rangle_t \\ &= W_t^\Bbb{P} - \left\langle W_s^\Bbb{P}, -\int_0^s H_u dW_u^\Bbb{P} \right\rangle_t \\ &= W_t^\Bbb{P} + \int_0^t H_s ds \tag{1} \end{align} Now turning this on its head gives \begin{align} W_t^\Bbb{P} &= W_t^\Bbb{Q} - \int_0^t H_s ds \\ &:= W_t^\Bbb{Q} - \left\langle W_t^\Bbb{Q}, \mathcal{L}\left( \left. \frac{d\Bbb{P}}{d\Bbb{Q}}\right\vert_{\mathcal{F}_s} \right) \right\rangle_t \end{align} which shows that ('reverse' Girsanov) $$ \left. \frac{d\Bbb{P}}{d\Bbb{Q}}\right\vert_{\mathcal{F}_t} = \mathcal{E}\left( \int_0^t H_s dW_s^\Bbb{Q} \right) \tag{2} $$ Starting from $(2)$ using the definition of the stochastic exponential and differentiating $(1)$ to plug it in the resulting expression then yields \begin{align} \left. \frac{d\Bbb{P}}{d\Bbb{Q}}\right\vert_{\mathcal{F}_t} &= \exp\left( \int_0^t H_s dW_s^\Bbb{Q} - \frac{1}{2} \int_0^t H_s^2 ds \right) \\ &= \exp\left( \int_0^t H_s (dW_s^\Bbb{P} + H_s ds) - \frac{1}{2} \int_0^t H_s^2 ds \right) \\ &= \exp\left( \int_0^t H_s dW_s^\Bbb{P} + \frac{1}{2} \int_0^t H_s^2 ds \right) \\ &= Z_t^{-1} \\ &= \left( \left. \frac{d\Bbb{Q}}{d\Bbb{P}}\right\vert_{\mathcal{F}_t} \right)^{-1} \end{align}