# Example of Girsanov change of density with finite relative entropy, but with infinite integral over squared changed drift

Let $(\Omega, (\mathcal F_t), \mathbb P)$ denote the usual Wiener space where $\Omega = C[0,\infty)$, etc., and where $(W_t)_{t \geq 0}$ denotes the Wiener process. Let $Z \in L^1(\mathbb P)$ with $Z > 0$, $\mathbb P$-almost surely. Let $Z_t = \mathbb E [ Z | \mathcal F_t ]$ denote the density process, which is a positive UI martingale.

By e.g. [Kallenberg, 2002, Lemma 18.21], there exists a process $\theta_t$ such that $Z_t = \exp \left( \int_0^t \theta_s \ d W_s - \frac 1 2 \int_0^t \theta_s^2 \ ds \right)$ for all $t \geq 0$.

Let $\mathbb Q$ be defined, as usual, by $d \mathbb Q/d \mathbb P = Z$. If $\mathbb E^{\mathbb Q} \left[ \int_0^{\infty} \theta_t^2 \ d t \right] <\infty$, it is not hard to show (see below) that the relative entropy $\mathcal H(\mathbb Q;\mathbb P) := \mathbb E^{\mathbb Q} \left[ \ln Z \right]$ is finite and is in fact given by $\mathcal H(\mathbb Q;\mathbb P) = \frac 1 2 \mathbb E^{\mathbb Q} \left[ \int_0^{\infty} \theta_t^2 \ d t \right]$.

I do not succeed in showing the reverse direction, i.e. that finite relative entropy implies square integrability of $(\theta_t)$, (unless $Z$ is bounded from above and below by a constant). I expect that there exists a counterexample to the general statement.

Can you provide such a counterexample?

Proof of $\mathcal H(\mathbb Q;\mathbb P) = \frac 1 2 \mathbb E^{\mathbb Q} \left[ \int_0^{\infty} \theta_t^2 \ d t \right]$ if $\mathbb E^{\mathbb Q} \left[ \int_0^{\infty} \theta_t^2 \ d t \right] <\infty$:

\begin{align*} \mathbb E^{\mathbb Q} | \ln Z | & = \mathbb E^{\mathbb Q} \left| \int_0^{\infty} \theta_s \ d W_s - \frac 1 2 \int_0^{\infty} \theta_s^2 \ d s \right| \\ & \leq \left( \mathbb E^{\mathbb Q} \left[\int_0^{\infty} \theta_s^2 \ ds \right] \right)^{1/2} + \frac 1 2 \mathbb E^{\mathbb Q} \left[ \int_0^{\infty} \theta_s^2 \ d s \right] < \infty.\end{align*} In particular, $\ln Z$ is $\mathbb Q$-integrable, so that $\mathcal H(\mathbb Q;\mathbb P) = \mathbb E^{\mathbb Q} \ln Z < \infty$. Furthermore, \begin{align*} \mathcal H(\mathbb Q;\mathbb P) - \frac 1 2 \mathbb E^{\mathbb Q} \left[\int_0^{\infty} \theta_t^2 \ d t \right]& = \mathbb E^{\mathbb Q} \left[ \ln Z - \frac 1 2 \int_0^{\infty} \theta_t^2 \ d t \right] \\ & = \mathbb E^{\mathbb Q} \left[ \int_0^{\infty} \theta_t \ d W_t - \int_0^{\infty} \theta_t^2 \ d t \right] \\ & = \mathbb E^{\mathbb Q} \left[\int_0^{\infty} \theta_t \ d W_t^{\mathbb Q} \right] = 0, \end{align*} where the expectation of the stochastic integral with respect to $W^{\mathbb Q} = W - \int_0^{\cdot} \theta_t \ d t$ vanishes since, by $\mathbb Q$-square integrability of $\theta$, this is in fact the expectation of a martingale that is null at zero.

• @ Joris Bierkens : Maybe you should look at strict local martinagles that can be expressed as stochastic exponential. You might be able to find example where their entropies are finite and such that they are not $\theta$-square integrables. You might be interested in this direction by paper "On the martingale property of certain local martingales" by Mijatovic et al (and the reference therein). Best regards. – The Bridge Oct 8 '13 at 8:45

I can now answer my own question. There exists no such counterexample. I can show that finite relative entropy in this setting implies $\mathbb E^{\mathbb Q} \int_0^{\infty} \theta_t^2 \ d t < \infty$. This can be proven by a localization argument using stopping times $(\tau_n)$, such that $Z^{\tau_n} \geq \frac 1 n$ and $[Z]^{\tau_n} \leq n^2$.