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Joris Bierkens
Joris Bierkens
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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 = \mathcal E(\theta)_t = \exp \left( \int_0^t \theta_s \ d W_s - \frac 1 2 \int_0^t \theta_s^2 \ ds \right)$$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.

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 $Z_t = \mathcal E(\theta)_t = \exp \left( \int_0^t \theta_s \ d W_s - \frac 1 2 \int_0^t \theta_s^2 \ ds \right)$ 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.

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.

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Joris Bierkens
Joris Bierkens
  • 985
  • 5
  • 19

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 = \mathcal E(\theta)_t = \exp \left( \int_0^t \theta_s \ d W_s - \frac 1 2 \int_0^t U_s^2 \ ds \right)$$Z_t = \mathcal E(\theta)_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^U \left[\int_0^{\infty} \theta_s^2 \ ds \right] \right)^{1/2} + \frac 1 2 \mathbb E^U \left[ \int_0^{\infty} \theta_s^2 \ d s \right] < \infty.\end{align*}\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.

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 $Z_t = \mathcal E(\theta)_t = \exp \left( \int_0^t \theta_s \ d W_s - \frac 1 2 \int_0^t U_s^2 \ ds \right)$ 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^U \left[\int_0^{\infty} \theta_s^2 \ ds \right] \right)^{1/2} + \frac 1 2 \mathbb E^U \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.

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 $Z_t = \mathcal E(\theta)_t = \exp \left( \int_0^t \theta_s \ d W_s - \frac 1 2 \int_0^t \theta_s^2 \ ds \right)$ 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.

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Joris Bierkens
Joris Bierkens
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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 $Z_t = \mathcal E(\theta)_t = \exp \left( \int_0^t \theta_s \ d W_s - \frac 1 2 \int_0^t U_s^2 \ ds \right)$ 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^U \left[\int_0^{\infty} \theta_s^2 \ ds \right] \right)^{1/2} + \frac 1 2 \mathbb E^U \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.