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When a hedging portfolio $X$ is used to price an asset $V$ expiring at time $T$, is it required that $X(t) = V(t)$ for all $t\in [0, T]$ or is it enough to simply require $X(T)= V(T)$?

I have always thought that the first case where $X(t) = V(t)$ for all $t\in [0, T]$ is correct. However, Shreve in his book Stochastic Calculus for Finance II seems to be claiming otherwise.

The exercise below seems to be claiming that the portfolio process $Y(t)$ of $\Delta(t)S(t)$ and money market hedges $C(t)$ because $Y(T)=C(T)$ a.s. Why is it not required to have $Y(t) = V(t)$ for all $t\in [0, T]$?

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I posted a solution to this and other questions in Shreve's second volume on my blog.

To directly answer your question:

  • First note that the process $C(t)$ represents a payout rate. That is, there is not a single terminal payoff $C(T)$ but over each time interval $\mathrm{d}t$, the contract pays $C(t)\mathrm{d}t$. The total discounted cash-flow that the contract pays is \begin{equation} \int_0^T D(u) C(u) \mathrm{d}u \end{equation}

  • We are looking for an initial wealth $Y(0)$ and a portfolio process $\Delta(t)$ such that the discounted wealth process \begin{equation} D(T) Y(T) = Y(0) + \int_0^T \Delta(u) D(u) S(u) \left( (\alpha(u) - R(u)) \mathrm{d}u + \sigma(u) \mathrm{d}W(u) \right) \end{equation} is equal to the discounted cash-flow process with probability one, i.e. \begin{equation} D(T) Y(T) = \int_0^T D(u) C(u) \mathrm{d}u \qquad \mathbb{P}\text{-a.s.} \end{equation}

  • While the discounted value processes are equal, you will generally have $Y(t) \neq C(t)$. Consider the following simplified example: \begin{equation} R(t) = 0, \quad T = 2, \quad C(t) = \begin{cases} 0 & \text{for } 0 \leq t < 1\\ 1 & \text{for } 1 \leq t \leq 2 \end{cases} \end{equation} I.e. the payout rate is constant at zero for $t \in [0, 1)$ and constant at one for $t \in [1, 2]$ (independent of $W(t)$). Then $Y(0) = 1$, $\Delta(t) = 0$ but $C(0) = 0$.

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  • $\begingroup$ Maybe I am wrong, but I thought In general an asset price process is a martingale under the risk neutral measure because there is a replicating portfolio that perfectly matches the value of the asset process for anytime. If they do not match for any time, why are the asset-price process a Martingale? $\endgroup$ – user1559897 Sep 4 '16 at 13:19
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    $\begingroup$ Page 218 is concerned with a European plain vanilla call option. This contract has a payout only at maturity $T$. Like any asset without intermediate payments, the corresponding discounted price process is a martingale under the risk-neutral probability measure. For no free lunch to exist it is easy to see that when you find an initial capital $X(0)$ and a portfolio process $\Delta(t)$ such that $X(T) = V(T)$ $\mathbb{P}$-a.s. then $X(t) = V(t)$ has to hold for all $t \in [0, T]$. $\endgroup$ – LocalVolatility Sep 4 '16 at 23:43
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    $\begingroup$ Could you explain the last sentence of your answer? If X(T) = V(T) P-a.s. why does x(t)=v(t) hold for all t $\in[0, T]$? $\endgroup$ – user1559897 Sep 5 '16 at 2:32
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    $\begingroup$ Imagine it didn't. Then you'd have two asset, $V$ and $X$, that have the same terminal payoff, no other cash-flows before maturity but different prices. Buy the cheaper, sell the more expensive and walk away with the difference.. $\endgroup$ – LocalVolatility Sep 5 '16 at 6:16
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    $\begingroup$ @user1559897 Under the fundamental financial assumption of no-arbitrage, two assets $X(t)$ and $Y(t)$ of same maturity $T$ generating the same stream of cashflows $CF_0,\cdots,CF_n$ at the same times $t_0,\cdots,t_n$ must have the same price; hence your 2 self financing strategies have to be equal for all $t$ before maturity. $\endgroup$ – Daneel Olivaw Apr 3 '17 at 17:41
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I think if $X(T)=V(T)$ but $X(t) \ne V(t)$, you would have found an arbitrage. Congrats to be on the way of riches.

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When a hedging portfolio $X(t)$ is used to price an asset $V(t)$ expiring at time $T$, it is required that a.s. $$ X(t) = V(t) \quad t\in [0, T] \tag{1} \label{one}$$ Otherwise you'll have an arbitrage, as answered already

I suspect, it is the notation (probably intentionally) used by Shreve in this exercise that causes confusion.

As correctly noted already, $C(t)$ represents a payout rate, NOT an asset value as usual. The asset hedged by portofolio $X(t)$ is actually: $$V(t) = \tilde{\mathrm{E}} \int_t^T D(u) C(u) du \tag{2} \label{two}$$

$\eqref{one}$ holds for the asset defined by $\eqref{two}$. In particular, $$X(0) = \tilde{\mathrm{E}} \int_0^T D(u) C(u) du \\ X(T) = \tilde{\mathrm{E}} \int_T^T D(u) C(u) du = 0$$

Intuitively, $X(t)$ is equal to the expected remaining amount of money an agent must pay till time $T$ as calculated by $\eqref{two}$.

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