= Solution
Use the conditional second-moment matrix $V_t=\mathbb E[P_tP_t^\top\mid\mathcal F_{t-1}]$ from the PDF. By part (b), the filtration is finite-state at every finite time. On a time-$(t-1)$ atom let $p_1,\ldots,p_m$ be the successor price vectors and $q_1,\ldots,q_m>0$ their conditional probabilities. Then $m\leq n$ and
$$
V=\sum_{i=1}^m q_i p_i p_i^\top.
$$
Positive definiteness gives rank $n$, so $m\geq n$. Hence $m=n$, and the $n$ successor vectors have <linear independence>.
We use the finite-horizon <fundamental theorem of asset pricing> in deflator form: an arbitrage-free finite market admits a strictly positive adapted process $D$, with $D_0=1$, such that $DP$ is a <martingale>. Equivalently, on every step,
$$
\mathbb E[D_tP_t\mid\mathcal F_{t-1}]=D_{t-1}P_{t-1}.
$$
Fix a finite horizon containing the step in question. On the parent atom, write $d_i=D_t/D_{t-1}>0$ on successor $i$ and $p_0=P_{t-1}$. Then
$$
\sum_iq_i d_i p_i=p_0.
$$
The proposed values $z_i=p_i^\top V^{-1}p_0$ satisfy precisely the same equation:
$$
\sum_iq_i z_i p_i
=\left(\sum_iq_i p_i p_i^\top\right)V^{-1}p_0=p_0.
$$
Because the $p_i$ have <linear independence> and the $q_i$ are positive, this linear system has a unique solution. Thus $z_i=d_i>0$, proving
$$
\boxed{Z_t>0\quad\text{almost surely for every }t\geq1.}
$$
This is the <positive regression deflator in a complete finite market>. It also proves the suggested conclusion: with $Y_0=1$ and $Y_t=\prod_{u=1}^tZ_u$,
$$
\mathbb E[Y_tP_t\mid\mathcal F_{t-1}]
=Y_{t-1}\mathbb E[P_tP_t^\top\mid\mathcal F_{t-1}]V_t^{-1}P_{t-1}
=Y_{t-1}P_{t-1}.
$$
Hence $Y$ is a strictly positive <martingale deflator>. Finite-state structure makes these expectations integrable on each finite horizon. Positive definiteness alone would not ensure positivity of the regression factor; <market completeness> and the positive deflator are essential.
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