When a Self-Financing Hedge Is Unique

Summary

The document questions a proof that an attainable option payoff has a unique initial value and hedging strategy. The proof equates two candidate strategies and concludes that their stochastic integral being a constant forces their integrands to match. The question points out that a stochastic integral can vanish even when its integrand is nonzero, so the conclusion needs more than the martingale property alone.

The exchange provides no resolution of the uniqueness issue; it only states the setup and the concern. In general, uniqueness requires conditions ensuring the asset’s stochastic integral representation is identifiable, such as suitable properties of the price process and admissible strategies. The document therefore serves as a prompt about assumptions behind replication arguments rather than a complete proof or practical hedging method.

Key ideas

• A martingale integral equal to a constant does not by itself establish that its integrand is zero.
• The uniqueness of a hedging strategy depends on assumptions about the price process and admissible strategies.
• The document raises a mathematical question but does not provide a definitive resolution.

Tags

# Uniqueness of the Hedging strategy


# Uniqueness of the Hedging strategy












I am currently reading the book "Nonlinear Option Pricing" by Julien Guyon. In the book they defined an attainable payoff $F_T$ as a $\mathcal{F}_T$ measurable random variable for which there exists an admissible portfolio and a real number $z$ such that $$z+\int_0^T \Delta_s\mathrm{d}\tilde{X}_s+D_{0,T}F_T=0$$ and $\int_0^T\Delta_s\mathrm{d}\tilde{X}_t$ should be a true $Q$-martingale.( $D_{0,T}$ is the discount factor and $\tilde{X}$ is the discounted stock price.)

Next, they claim that the pair $z,\Delta_s$ is unique because suppose there is a $z',\Delta'_s$, then $$\int_0^T(\Delta_s-\Delta_s')\mathrm{d}\tilde{X}_s=z'-z$$.

Now since $\tilde{X}$ is a $Q$-Martingale, $\Delta_s=\Delta'_s$ and thus $z=z'$.

Question: I don't understand the last argument. Why is $\Delta_s=\Delta_s'$ necessarily? This question on Math.stachexchange shows one can find a non-trivial previsible process, such that the stochastic integral is almost surely equal to zero. Or do I miss something?

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.