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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.

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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.

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  • $\begingroup$ Thanks for your answer. I don't really understand what you mean with this.. with the method above, I was trying to find/construct the portfolio, of which the price at time 0 gives me the superreplication price, the price that is the strict upper bound of the arbitrage-free prices of the call. Does your method yield this upper bound or the smallest super-replicating portfolio? $\endgroup$ – R3S May 4 '17 at 13:53
  • $\begingroup$ the cheapest super-replicating portfolio is the strict upper bound of the arbitrage-free prices of the call! $\endgroup$ – Mark Joshi May 5 '17 at 1:24
  • $\begingroup$ Maybe we use a different terminology: By saying I was trying to construct the super-replicating portfolio, I meant that I want to find the self-financing, predictable process, of which the corresponding value process at time 0 gives the strict upper bound of the arbitrage-free prices for the call. So I need to know how many units of the share and units of the bond I need to hold in each period to super-hedge the call. Sorry for the confusion. $\endgroup$ – R3S May 5 '17 at 13:39
  • $\begingroup$ once you have the line, the slope is the number of shares, the intersect is the number of bonds. $\endgroup$ – Mark Joshi May 6 '17 at 2:01

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