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Reading the book by Andrea Pascucci "PDE and Martingale Methods in Option Pricing", pp. 84, I found something that appears inconsistent to me. It concerns the construction of the optimal portfolio for an American derivative, whose pay-off a time $t_n$ is indicated by $X_n$. I resume here the main steps. Let $H_n^*$ be the discounted price of the derivative, that is $H_n^*=H_n/(1+r)^n$. No-arbitrage arguments define the discounted price as

$$ H_n^*=\left\{\begin{array}{ll} X_N^* & \text{ if }n=N\\ \max\left(X_n^*,\mathbb{E}^{\mathbb{Q}}\left[H_{n+1}^*\mid\mathcal{F}_n\right]\right)& \text{ if }n=0,...,N-1 \end{array} \right., $$ so let $E_n$ be defined as

$$ E_n=\left\{\begin{array}{ll} -1 & \text{ if }n=N\\ \frac{1}{1+r}\,\mathbb{E}^{\mathbb{Q}}\left[H_{n+1}\mid\mathcal{F}_n\right]& \text{ if }n=0,...,N-1 \end{array} \right., $$

whence $H_n=\max\left(X_n,E_n\right)$. Using the Dobb's decomposition theorem we can write

$$ H_n^*=M_n+A_n, $$

where $M$ is a martingale such that $H_0=H_0^*=M_0$ and $A$ is a predictable decreasing process. Let $\nu_M$ be defined as the maximal time in which it is still optimal to execute the option. It can be shown that

$$ \nu_M=\min\left\{n\mid X_n>E_n\right\} $$

and that

$$ H_n^*=M_n\text{ in }\left\{n\leq\nu_M\right\}.\quad (1) $$

The hedging strategy is defined as the replicating strategy for the European derivative with pay-off $M_N$. It is not need to calculate it for $n>\nu_M$ since it is not optimal to exercise the option after $\nu_M$. So it is enough to replicate $M_n$ for $\left\{n\leq\nu_M\right\}$ and hence, thanks to equation (1), it is enough to replicate $H_n^*$ for $\left\{n\leq\nu_M\right\}$. Nevertheless in the book it is said that the
the components of the replicating portfolio are determined by

$$ \alpha_{n,k}=\frac{H_{n,k+1}-H_{n,k}}{(u-d)\,S_{n-1,k}},\quad\beta_{n,k}=\frac{u\,H_{n,k}-d\,H_{n,k+1}}{(1+r)^n\,(u-d)} $$

(which are equations (2.168) of the book at page 84), where $S_{n,k}$ is the price of the process at time $n$ when there have been $k$ upward jumps and $n-k$ downward jumps, $u$ and $d$ are the amplitude of the upward and downward jumps respectively, $H_{n,k}$ is the price of the option in the scenario with $k$ upward jumps and $r$ is the risk-free rate. The inconsistency comes from the fact that $H_n^*=M_n$ for $n\leq\nu_M$, but $\alpha_n$ and $\beta_n$ are computed replicating $H_n$, while I expected that they should replicate $H_n^*$. So I expected to find

$$ \alpha_{n,k}=\frac{H^*_{n,k+1}-H^*_{n,k}}{(u-d)\,S_{n-1,k}},\quad\beta_{n,k}=\frac{u\,H^*_{n,k}-d\,H^*_{n,k+1}}{(1+r)^n\,(u-d)} $$

Where is the flaw in my logic?

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    $\begingroup$ What is $G_t$? The book does not have $E_t$. It is better to use the notations exactly the same as in the book, then we may be able to locate the confusions you have. Otherwise, we need to spend time compare your notations with those the book use. $\endgroup$ – Gordon Dec 14 '15 at 19:11
  • $\begingroup$ I do not see any problem with the book, as it is the same as Formula (2.69) in the European case. By the way, you $E_n$ should have the factor $\frac{1}{1+r}$ in front of the expectation. $\endgroup$ – Gordon Dec 15 '15 at 19:17
  • $\begingroup$ The problem comes from the fact that we have to find the replicating strategy of for $H^*$ not for $H$ while the formula (2.69), as you pointed out, is as for the European case when one has to replicate $H$. $\endgroup$ – AlmostSureUser Dec 16 '15 at 10:58

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