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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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  • $\begingroup$ I am unsure, so I will leave it to someone else more knowledgeable to answer, but I think it is related to convergence in probability and the uniqueness of coefficients of Ito Processes. That could be the missing link here. Also when I was reading the book I totally overlooked this issue since I assumed the purpose was to show how $z$ is unique and since we have a martingale $E(z' -z) = 0$ leads to the equality there. So good question and I hope this gets a decent answer $\endgroup$
    – Slade
    Mar 15, 2019 at 13:07
  • $\begingroup$ @Slade Thank you for your comment! Maybe little bit context: I am interested in the question if, in a complete market model, the replication portfolio of a Payoff is uniquely determined (independent from the model of $X$ with same $Q$) Obviously, this line in the book gives a positive answer. $\endgroup$
    – quallenjäger
    Mar 15, 2019 at 13:17
  • $\begingroup$ This can be proved by Ito's isometry. For a self-financing strategy, we require that the variance of the value process is finite, that is, Ito's isometry holds. For the question in your link, the condition for this isometry is not satisfied. $\endgroup$
    – Gordon
    Mar 15, 2019 at 19:27
  • $\begingroup$ @Gordon do you know how to connect the fact that the stochastic integral is a $Q$ martingale and that $\Delta_s = \Delta_s^{'}$? I don't know of a general property that makes that connection $\endgroup$
    – Slade
    Mar 23, 2019 at 22:12
  • $\begingroup$ @Slade I think the connection is more, that $\tilde{X}$ is a Q martingale. On this way, you can apply the Ito isometry if the integrand is square integrable. This is basically the same idea as if you proof the uniqueness part of martingale representation theorem $\endgroup$
    – quallenjäger
    Apr 11, 2019 at 15:47

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