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Deriving a Two-State Perfect Hedge Price by Algebra

Article Quant Q&A · Author: Wolfy
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Summary

The document works through the time-zero value of a claim in a one-period, two-state market. A portfolio holding the risky asset and a bond is chosen so that its terminal value matches the claim in both the up and down states. The hedge ratios are obtained by solving the two payoff equations, and the initial portfolio value is then expressed using the discount factor. The answer substitutes those hedge ratios into the initial value and groups terms by the claim’s up-state and down-state payoffs. This produces the requested weighted expression, with the weights determined by the asset’s discounted initial value relative to its two possible terminal prices. The derivation is an algebraic illustration of replication and discounting in a simple complete market. It assumes the stated two outcomes and bond pricing setup, and does not discuss extensions to multiple periods, transaction costs, or markets where exact replication is unavailable.

Key ideas

  • A perfect hedge matches the claim payoff in each possible terminal state.
  • The risky-asset holding is determined by the difference between the claim’s state payoffs and the asset’s state prices.
  • The bond holding finances the residual payoff after accounting for the risky asset.
  • Substituting the hedge holdings into the initial portfolio value yields the requested discounted weighted-payoff formula.
  • The calculation applies to a one-period, two-state market.

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Derivatives pricingOptionsRisk managementStatistics
Full text
# Value of a perfect hedge


# Value of a perfect hedge












Background Information:

The price of a portfolio at time $t$ ($t = 0 ,1$) is $$V_t(\pi) = \phi S_t + \psi B_t$$ The portfolio $\pi$ is a perfect hedge for the claim $X$ if $V_1(\pi) = X$ a.s. as random variables.

Given claim $X$, to have a perfect hedge then $\phi$ and $\psi$ must satisfy \begin{equation} \phi S_1(u) + \psi B_1 = X(u) \end{equation} \begin{equation} \phi S_1(d) + \psi B_2 = X(d) \end{equation} We have solving for this that $$\phi = \frac{X(u) - X(d)}{S_1(u) - S_1(d)}$$ $$\psi = B_1^{-1}(X(u) - \phi S_1(u)) = B_1^{-1}(X(d) - S_1(d))$$ Thus the resulting value of $X$ at $t=0$ is $$V_0(X) = V_0(\pi) = \phi S_0 + \psi B_0$$

Question:

> Let $\beta = B_0 B_1^{-1}$ be the discount factor. Show that $$V_0(X) = \beta\left[\left(\frac{\beta^{-1}S_0 - S_1(d)}{S_1(u) - S_1(d)}\right)X(u) + \left(\frac{S_1(u) - \beta^{-1}S_0}{S_1(u) - S_1(d)}\right)X(d)\right]$$

I have tried this three times now and I still not getting the result what I do is this if you want to see my attempted work let me know. Otherwise it would be great if someone could give me a good start to this. There must be something that I am not seeing in regards to some algebra trick.

## Answer by Gordon (score 3, accepted)

https://quant.stackexchange.com/a/30506

Using the values for $\phi$ and $\psi$ that you have derived, \begin{align*} V_0(X) &= \phi S_0 + \psi B_0\\ &= \frac{X(u) - X(d)}{S_1(u) - S_1(d)} S_0 + B_1^{-1}\left(X(u) - \frac{X(u) - X(d)}{S_1(u) - S_1(d)}S_1(u)\right) B_0\\ &=\beta\left(\frac{X(u) - X(d)}{S_1(u) - S_1(d)} \beta^{-1}S_0 + \frac{X(d) S_1(u) - X(u)S_1(d)}{S_1(u) - S_1(d)} \right)\\ &=\beta\left(\frac{\beta^{-1}S_0 - S_1(d)}{S_1(u) - S_1(d)} X(u) + \frac{S_1(u) - \beta^{-1}S_0 }{S_1(u) - S_1(d)} X(d)\right). \end{align*} What you need is to combine terms with $X(u)$ together, and likewise for terms with $X(d)$.

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