= Solution
The usual replication argument needs two hypotheses implicit in the pricing conclusion: the strategy must replicate the payoff, $V_N=X$, and its discounted gains must be integrable. Bounded <predictable> risky holdings suffice for the latter. Under these conditions, the <equivalent martingale measure> $\mathbb Q$ and part (c) give
$$
\mathbb E_{\mathbb Q}[\widetilde V_n-\widetilde V_{n-1}\mid\mathcal F_{n-1}]
=\phi_n\mathbb E_{\mathbb Q}[\widetilde S_n-\widetilde S_{n-1}\mid\mathcal F_{n-1}]=0.
$$
Thus discounted wealth is a <martingale>, and repeated conditioning gives
$$
\boxed{V_n=B_n\mathbb E_{\mathbb Q}[X/B_N\mid\mathcal F_n].}
$$
For a normalized initial <bank account>, $B_0=1$, deterministic terminal bank value, and trivial initial information, this reduces to the intended <risk-neutral pricing> formula
$$
\boxed{V_0=B_N^{-1}\mathbb E_{\mathbb Q}[X].}
$$
If $B_N$ is random it must remain inside the expectation. If a claim is not replicable, the existence of an <equivalent martingale measure> alone does not make its arbitrage-free price unique.
The <discounted wealth martingale integrability condition> cannot simply be dropped. For a concrete counterexample, set $B_n=1$, $S_0=3/2$, $S_1=1+U$, and $S_2=1+U+\varepsilon/2$, where $U$ is uniform on $(0,1)$ and $\varepsilon$ an independent fair sign. These positive bounded prices form a <martingale> in the natural filtration. Start with zero holdings; at time one buy $1/U$ shares financed by borrowing $S_1/U$. This is <previsible> and <self-financing>, but its terminal wealth is $\varepsilon/(2U)$, whose absolute expectation is infinite. It is therefore not a <martingale>. The stated pricing proof applies to the standard integrable replication class, rather than all unconstrained previsible strategies.
Back to article page
