I've read an article on relative entropy properties that gives a result for the relative entropy of two equivalent measures as they are found in applications of girsanovs theorem.
For two measures P, Q we define the relative entropy the following way: $H(Q;P)=\int_\Omega\frac{dQ}{dP}\log\left(\frac{dQ}{dP}\right)dP$
And the Girsanov theorem is a statement about measures related through the density $\frac{dQ}{dP}=\exp\left(\int_0^T\theta_tdB_t-\frac{1}{2}\int_0^T\theta_t^2dt\right)$
Where $B_t$ is a Standard Brownian motion (with measure P) and (according to Girsanov) $B_t-\int_0^t\theta_sds$ is a Standard Brownian motion under Q.
The claim is that
$H(Q;P)=\frac{1}{2}\int_0^T\theta^2_tdt$
which is claimed to be a straightforward calculation, but I can't quite get to the end-result. Any help would be greatly appreciated.
Here is how far i got so far:
$H(Q;P)=\int_\Omega\frac{dQ}{dP}\left(\int_0^T\theta_tdB_t-\frac{1}{2}\int_0^T\theta_t^2dt\right)dP$ $=\int_\Omega\frac{dQ}{dP}\left(\int_0^T\theta_tdB_t\right)dP$
$\quad-\left(\frac{1}{2}\int_0^T\theta_t^2dt\right)\int_\Omega \frac{dQ}{dP} dP$
$=\int_\Omega\frac{dQ}{dP}\left(\int_0^T\theta_tdB_t\right)dP$
$\quad-\frac{1}{2}\int_0^T\theta_t^2dt\cdot1$
Where I've used that the integral of a density must be one. I guess I'd have to use something like Itô-Isometry to get from the integral with respect to the Brownian motion to something deterministic with the squared integrand, but I can't quite figure out what to do with the density to get to it. If I could somehow justify something like: $\int_\Omega\frac{dQ}{dP}\left(\int_0^T\theta_tdB_t\right)dP=\int_\Omega\left(\int_0^T\theta_tdB_t\right)^2dP=\int_0^T\theta_t^2dt$
that would conclude the calculation. Any ideas?
As you say, $$\theta_{t}\text{d}B_{t} = \theta_{t}\text{d}W_{t}+\theta_{t}^2\text{d}t\ ,$$ where $W$ is a $Q$-Brownian motion. Hence $$\log\frac{\text{d}Q}{\text{d}P} = \int \theta_{t}\text{d}W_{t} + \frac{1}{2} \int \theta_{t}^2\text{d}t$$ and the $P$-expectation becomes $$H(Q;P)=\mathbb{E}_{Q}\left(\int \theta_{t}\text{d}W_{t}\right)+\frac{1}{2}\mathbb{E}_{Q}\left(\int \theta_{t}^2\text{d}t\right) = 0 + \frac{1}{2}\int \theta_{t}^2\text{d}t$$ since the stochastic integral is a $Q$-martingale starting at zero and process $\theta$ is deterministic.
Hence the relative entropy equals $$H(Q;H) = \frac{1}{2}\int \theta_{t}^2\text{d}t$$ as claimed.