Integrable cumulative income process

Question

I am trying to read Karatzas/Shreve "Methods of Mathematical Finance". In ch. 1, Definition 5.5, a cumulative income process $\Gamma(t)=\Gamma^{\mathrm{fv}}(t)+\Gamma^\mathrm{lm}(t)$ (a semimartingale under the original measure $P$) is defined to be integrable if $$ E_0 \int_0^T \frac{d \hat\Gamma_0^{\mathrm{fv}}(u)}{S_0(u)} < \infty, E_0 \int_0^T\frac{d \langle \Gamma_0^{\mathrm{lm}} \rangle (u)}{S_0^2(u)} < \infty $$ where $E_0$ denotes expectation with respect to $d P_0 = Z_0\, dP, Z_0 =\exp \left[ - \int_0^t \theta'(s) dW(s) - \frac 1 2 \int_0^t \| \theta(s) \|^2 ds \right]$, and $\theta (\cdot )$ being the market price of risk process.

$d \hat\Gamma^{\mathrm{fv}}(u) = \vert d \Gamma^{\mathrm{fv}} (u)\vert $ denotes the absolute variation of the finite variation part of $\Gamma$. Moreover, the authors remind that, with respect to the new measure $P_0$, $$ d \Gamma_0^{\mathrm{fv}}(t) = d \Gamma^{\mathrm{fv}}(t) - \theta'(t) d \langle \Gamma^{\mathrm{lm}}, W \rangle (t) . $$

Now the authors claim in Remark 5.8 that with $H_0(t ) := Z_0 (t) / S_0 (t)$ (the state price density process), the integrability conditions for the cumulative income process can be rewritten as $$ E \int_0^T H_0(u) d \hat\Gamma^{\mathrm{fv}}(u) < \infty , $$ if $\Gamma^{\mathrm{lm}}(\cdot ) \equiv 0$.

I don't see how this is equivalent, and I don't know if it's supposed to be equivalent to both conditions or only the first one. I tried to rewrite the proposed condition. If I'm not mistaken,

$$ E \int_0^T H_0(u) d \hat\Gamma^{\mathrm{fv}}(u) = E_0 \left[ \int_0^T \frac{\exp \left( \int_{(u,T]} \theta'(s) dW(s) + \frac 1 2 \int_{(u,T]} \| \theta(s) \|^2 d s \right) }{S_0(u)} d \hat\Gamma_0^{\mathrm{fv}}(u) \right] $$

Now what do I do with the numerator?

Answer 1

Sketch of partial solution: Let $\pi_n := \lbrace 0 = \tau^n_0\leq \tau_1^n \leq \dots \leq \tau_{m_n}^n=T\rbrace$ be a sequence of partitions s.t. $\operatorname{mesh}(\pi_n)\to 0$ a.s. We have the following limit (strictly speaking, we have to choose a subsequence, plus some steps need justification): $$ E \int_0^T \frac{Z_0(t)}{S_0(t)}\,d\widehat\Gamma^{\mathrm{fv}} (t) = \lim_{n\to \infty} E\sum_{i = 1}^{m_n - 1} \vert \widehat\Gamma^{\mathrm{fv}}\vert\,([\![\tau^n_{i}, \tau^n_{i+1} [\![) \, \frac{Z_0(\tau_i^n)}{S_0(\tau_i^n)} $$ $$ = \lim_{n\to \infty} \sum_{i = 1}^{m_n - 1} E \left[ \vert \widehat\Gamma^{\mathrm{fv}} \vert ([\![\tau^n_{i}, \tau^n_{i+1} [\![)\,\frac{Z_0(\tau_i^n)}{S_0(\tau_i^n)} \right] = \lim_{n\to \infty} \sum_i E_0 \left[ \vert \widehat\Gamma^{\mathrm{fv}} \vert ([\![\tau^n_{i}, \tau^n_{i+1} [\![)\,\frac{Z_0(\tau_i^n)}{Z_0(\tau_{i+1}^n)S_0(\tau_i^n)} \right] $$ $$ = \lim E_0 \sum \dots = E_0 \int_0^T \frac{d \widehat \Gamma^{\mathrm fv} (t)}{S_0(t)} $$ Now, $$ E_0 \int_0^T \frac{d \widehat \Gamma^{\mathrm fv} (t)}{S_0(t)}\leq E_0 \int_0^T \frac{d \widehat\Gamma_0^{\mathrm{fv}} (t)}{S_0(t)} +E_0 \int_0^T \vert \theta (t) \vert \frac{d \vert \langle \Gamma^{\mathrm{lm}}, W \rangle \vert }{S_0(t)} \leq E_0 \int_0^T \frac{d \widehat\Gamma_0^{\mathrm{fv}} (t)}{S_0(t)} +\left( \int_0^T \| \theta (t) \|^2\,dt \right)^{1/2} \left( E_0 \int_0^T \frac{d \langle \Gamma^{\mathrm{lm}} \rangle }{S_0(t)^2} \right)^{1/2} $$ Here, the last step follows from the Kunita-Watanabe inequality (followed by Cauchy-Schwarz for $L^2 (P_0)$). In the same way, one can show $E_0 \int \dots \geq E_0 \int \dots - E_0 \int \dots \geq E_0 \int\dots - T^{1/2} (E_0 \int \dots )^{1/2}$. This shows that if one assumes $E_0 \int_0^T \frac{d \langle \Gamma^{\mathrm{lm}} \rangle }{S_0(t)^2} < \infty$, the other two conditions are equivalent. (EDIT: Note that $\langle \Gamma^{lm} \rangle = \langle \Gamma_0^{lm} \rangle$)

I'm still thinking about the rest.

I demand my bounty back.^^