When Alternative Assets Can Replicate an Option Payoff

# Is Trading in the Underlying Necessary for Replication?


# Is Trading in the Underlying Necessary for Replication?












In a simple one-period binomial model we have two possible payoffs: $f(S^u)$ and $f(S^d)$. To replicate this we must trade in two assets, usually the stock $S$ and the money market account (assumed to initially be 1), and then solve the equations $$ \phi S^u + \psi e^{rT} = f(S^u), \\ \phi S^d + \psi e^{rT} = f(S^d) $$ to get our hedging strategy $\phi = \frac{f(S^u) - f(S^d)}{S^u - S^d}$, $\psi = e^{-rT} (f(S^u) - \phi S^u)$. Then for a multi-period model we use backward induction and repeat this program at each node.

Why must we trade in the underlying to replicate this payoff? It seems like absolutely any asset would do as long as they follow our binomial model. For example, let $C$ be the current price of a cow. Then, trading in cows (and still assuming $S$ follows the binomial model) I can still solve the equations $$ \hat{\phi} C^u + \hat{\psi} e^{rT} = f(S^u), \\ \hat{\phi} C^d + \hat{\psi} e^{rT} = f(S^d) $$ to get my hedging strategy $\hat{\phi} = \frac{f(S^u) - f(S^d)}{C^u - C^d}$, $\hat{\psi} = e^{-rT}(f(S^u) - \hat{\phi} C^u)$.

It just seems that we don't actually need to trade in the underlying at all to replicate an option payoff.

However, letting $V = \phi S_0 + \psi$ and $\hat{V} = \hat{\phi} C_0 + \hat{\psi}$, we have $V \neq \hat{V}$ in general. Indeed, \begin{align*} V = \hat{V} & \iff \phi S_0 + \psi = \hat{\phi} C_0 + \hat{\psi} \\ & \iff \frac{f(S^u) - f(S^d)}{S^u - S^d}S_0 + e^{-rT} \left(f(S^u) - \frac{f(S^u) - f(S^d)}{S^u - S^d} S^u\right) \\ & \qquad = \frac{f(S^u) - f(S^d)}{C^u - C^d}C_0 + e^{-rT} \left(f(S^u) -\frac{f(S^u) - f(S^d)}{C^u - C^d} C^u\right) \\ & \iff \frac{f(S^u) - f(S^d)}{S^u - S^d}S_0 - e^{-rT} \left(\frac{f(S^u) - f(S^d)}{S^u - S^d} S^u\right) \\ & \qquad = \frac{f(S^u) - f(S^d)}{C^u - C^d}C_0 - e^{-rT} \left(\frac{f(S^u) - f(S^d)}{C^u - C^d} C^u\right) \\ & \iff \frac{S_0}{S^u - S^d} - \frac{S^ue^{-rT}}{S^u - S^d} = \frac{C_0}{C^u - C^d} - \frac{C^u e^{-rT}}{C^u - C^d}, \end{align*}

which doesn't seem necessary. So then the question becomes, what's the correct asset to trade in to price the option?

## Answer by AFK (score 2, accepted)

https://quant.stackexchange.com/a/18035

There is an implicit assumption in your model. Namely that the price of the cow is perfectly correlated with the stock: they always move in the same direction. In this case you can indeed hedge your risk using cows. I let you be the judge of the validity of that assumption.

More likely the moves of cow prices are independent which means that you should consider 2 binomial models or a quadrinomial model and you cannot hedge your derivative using cows only.

But in practice you can often hedge your option using something else than the underlying e.g. a forward contract on the stock because it is linear in the stock price just like the cow price was in your model.

## Answer by emcor (score 1)

https://quant.stackexchange.com/a/18065

The fact that you can solve the second set of equations means that you can hedge the option through another asset aswell, in a simple binomial world this may indeed be true. Note that your strategy must replicate $f(S)$ through $C$ in all states at all time steps. E.g. it is possible that you don't get a solution for $\phi$ if $C^u-C^d=0$.

However, the hedging weights would differ based on the assets you used to replicate the option payoff.

Your question is related to incomplete markets, where an asset may be replicated through a range of trading strategies such that no unique price exists. In complete markets, $\mathbb{Q}$ is unique and $f(S)$ can be replicated through only one strategy.