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Constructing a Superhedge in an Incomplete Trinomial Model

Article Quant Q&A · Author: R3S

Summary

The document studies super-replication of a call option in a two-period trinomial model with a risky asset and a zero-rate bond. Each period, the asset can move up, stay unchanged, or move down. Because the market has more possible outcomes than traded securities, an equivalent martingale measure can exist even though the market is incomplete. The questioner derives terminal holdings that cover the call payoff in each state, then finds that the implied values at the previous time do not fit one common self-financing portfolio.

The response gives a geometric method for checking the hedge: plot the payoff points and draw lines through pairs of them. A line above the remaining point corresponds to a super-replicating position, while a line below corresponds to a sub-replicating one; equality indicates replication. This suggests evaluating the possible pairs against the third branch. The brief reply does not work through the numerical example or identify the precise flaw in the proposed calculations, so it offers an approach rather than a complete solution.

Key ideas

  • The model is arbitrage-free but incomplete because the traded assets cannot span every state payoff.
  • A superhedge must cover the option payoff across all terminal branches.
  • In the trinomial setting, candidate hedge portfolios can be assessed geometrically using lines through payoff points.
  • A line above the remaining payoff point represents super-replication, while a line below represents sub-replication.
  • The response sketches a method but does not finish the numerical solution.

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Full text
# Super Hedging in incomplete Trinomial Tree


# Super Hedging in incomplete Trinomial Tree












I have a question concerning the super-replication of a call in a trinomial tree which has the following characteristics:

Suppose we have one risky asset $S_t=2+\sum_{k=1}^tZ_i$, where $P(Z_i=0)=P(Z_i=-1)=P(Z_i=1)=\frac{1}{3}$, where $P$ denotes the objective (or real-world) probability measure and one bond $B_t=1$ for t=1,2 (so the risk-free rate is assumed to be 0 over time). The call that is supposed to be super replicated is denoted by $C_2=(S_2-1)^+$.

One can show that there exists an equivalent martingale measure in this market and therefore is is arbitrage-free but obviously not complete.

If $\Delta_t$ denotes the amount of shares and $\beta_t$ denoted the units of bonds we need to hold at time $t$ for our hedging strategy, I calculated the following for the superhedge portfolio $(\beta_t,\Delta_t)$:

On $\{Z_1=1 \}$ $$ 4\Delta_2 + \beta_2 \geq 3\\ 3\Delta_2 + \beta_2 \geq 2\\ 2\Delta_2 + \beta_2 \geq 1\\ $$ which holds for $\Delta_2=1$ and $\beta_2=-1$.

On $\{Z_1=0 \}$ $$ 3\Delta_2 + \beta_2 \geq 2\\ 2\Delta_2 + \beta_2 \geq 1\\ 1\Delta_2 + \beta_2 \geq 0\\ $$ which holds for $\Delta_2=1$ and $\beta_2=-1$.

On $\{Z_1= -1 \}$ $$ 2\Delta_2 + \beta_2 \geq 1\\ 1\Delta_2 + \beta_2 \geq 0\\ 0\Delta_2 + \beta_2 \geq 0\\ $$ and here the last inequality implies the middle one, hence for $\Delta_2=\frac{1}{2}$ and $\beta_2=0$ this holds.

So far these are the values for the super-replicating portfolio at $t=2$. In order to calculate the amounts that need to be held at $t=1$ we need to keep in mind that a superhedge needs to be self-financing, i.e.

$$ 3\Delta_1 +\beta_1 = 3\Delta_2 + \beta_2 = 3*1-1 = 2 \text{ on $\{Z_1=1 \}$ }\\ 2\Delta_1 +\beta_1 = 2\Delta_2 + \beta_2 = 2*1-1=1 \text{ on $\{Z_1=0 \}$ }\\ 1\Delta_1 +\beta_1 = \frac{1}{2}\Delta_2 + \beta_2 = \frac{1}{2}*1-0=\frac{1}{2}\text{ on $\{Z_1=-1 \}$ }\\ $$ which has no solution, since $\Delta_1$ and $\beta_1$ need to be constant. So I can't find a self-financing portfolio that is superhedging the call $C$. I am really confused as I don't see where the flaw in my logic is. I have been trying to find a solution to this since a couple of days so I would really appreciate any help. Cheers.

## Answer by Mark Joshi (score 0)

https://quant.stackexchange.com/a/34029

I don't quite follow what you are doing. However, the easy way to do it is to plot the three points. Draw straight lines through all subsets of size two.

Each of these lines will either be above or below or equal to the third point.

Above means super-replication. Below means sub-replication. Equal means replication,.

Essentially do the three cases and see what that gives you for the third branch.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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