Risk-Neutral Pricing and Replication Bounds in a Trinomial Model
Summary
The example shows how an incomplete one-period trinomial market can produce multiple risk-neutral probability distributions and a range of prices for a European call. With three possible stock outcomes but only the probability-sum and discounted expected-price restrictions, one probability can be treated as a free parameter. Enforcing positive probabilities then gives a family of admissible distributions, and valuing the call under that family yields a range of risk-neutral prices.
The response also formulates super-replication and sub-replication as portfolios of stock and cash whose payoffs respectively dominate or lie below the option payoff in every state. For the stated tree, it derives a cheapest dominating portfolio and a largest dominated portfolio, connecting these bounds to the extreme prices from the risk-neutral distributions. The example is limited to the specified one-period model, strike, and zero interest rate; the bounds should not be read as a general numerical result for other trees or market inputs.
Key ideas
- A trinomial model can be incomplete because risk-neutral probabilities are not uniquely determined.
- Parameterize the probability solutions and enforce their admissibility to find the full price range.
- Super-replication finds the least costly portfolio that covers the claim in every state.
- Sub-replication finds the greatest portfolio payoff that stays below the claim in every state.
- In this example, the replication bounds align with the extremes of risk-neutral prices.
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# Risk-Neutral Probabilities, Trinomial Model
# Risk-Neutral Probabilities, Trinomial Model
My professor has many grammatical mistakes and errors in his questions, so apologies ahead of time. I am just trying to understand what he wants for this question,
In trinomial model, let $S_0 = 1$, $R = 0$, $u = 2$, $m = 1$, and $l = 1/2$
a.) Find all risk-neutral probabilities and the range of prices generated by them for a call option with strike $K = 1$.
b.) Find the super-replication price and sub-replication price for this call option and compare them to the lowest and highest prices in part (a).
Let's assume the call option is a European call option.
Attempted solution for a.): We have two conditions $$\begin{cases} \pi_l\times1/2 + \pi_m\times1 + \pi_u\times2 = 1\\ \pi_l + \pi_m + \pi_u = \frac{1}{1+R} = 1 \end{cases}$$ Since we have 2 equations and 3 unknowns we have an infinite number of risk-neutral probabilities. Hence, we can set one as the free variable and then solve for the other two. So, assume $\pi_m = \pi_u$, then we have $$1/2\pi_l + 3\pi_u = 1 \ \text{and}$$ $$\pi_l + 2\pi_u = 1$$ multiplying the first equation by 2 and solving we get $\pi_u = 1/4$, $\pi_l = 1/2$, and $\pi_m = 1/4$. Therefore in the up state we have $C = \pi_u(u - K)^+ = 1/4$ and for the down state we have $C = \pi_l(l-K)^+ = 0$, and lastly for the middle state we have $C = \pi_m(m - K)^+ = 0$.
I am not sure if this is correct, any suggestions is greatly appreciated.
Note: There is nothing in my professors notes or book that even mentions sub-replication so I would ignore it. Super-replication of a contingent claim is to find the smallest value of a portfolio which has a payoff equal to or greater than the payoff of the contingent claim.
Attempted solution for b.): We cannot replicate precisely but we can find a portfolio that dominates the payoff. Since there is nothing in notes nor book that mentions sub-replication I will just focus on super=replication. Hence we have three conditions $$V_0 = aS_0 + b = \begin{cases} aS_u + b(1+R) \geq D_1\\ aS_m + b(1+R) \geq D_2\\ aS_l + b(1+R) \geq D_3 \end{cases}$$ We know that $S_0 = 1$, $R = 0$, $u = 2$, $m = 1$, and $l = 1/2$, hence $$V_0 = a + b = \begin{cases} 2a + b \geq D_1 = (2 - K)_+ = (2 - 1)_+ = 1\\ a + b \geq D_2 = (1 - K)_+ = 0\\ \frac{1}{2}a + b \geq D_3 = (\frac{1}{2} - K )_+ = (\frac{1}{2} - 1)_+ = 0 \end{cases}$$
I believe this dominates the pay off of a.) hence we are done... not sure? any suggestions is greatly appreciated.
## Answer by Gordon (score 3, accepted)
https://quant.stackexchange.com/a/23237
For question a). From the assumptions, in particular, that $R=0$, \begin{align*} \pi_l + \pi_m + \pi_u &=1\\ \frac{1}{2}\pi_l + \pi_m + 2\pi_u&=1. \end{align*} Set $\pi_m=x$, and solve for $\pi_l$ and $\pi_u$, \begin{align*} \pi_l &= \frac{2}{3}(1-x)\\ \pi_m &= x\\ \pi_u &= \frac{1}{3}(1-x), \end{align*} where $0<x<1$. The option price is then given by \begin{align*} C &= \pi_l (l-K)^+ + \pi_m (m-K)^+ + \pi_u (u-K)^+\\ &=\frac{1}{3}(1-x), \end{align*} which is in the range of $(0,\, 1/3)$.
For question b). Consider a super-replicating portfolio with $a$ units share and $b$ units cash. Then \begin{align*} a\, l + b &\ge (l-K)^+\\ a\, m + b &\ge (m-K)^+\\ a\, u + b &\ge (u-K)^+. \end{align*} That is, \begin{align*} a\, l + b &\ge 0\\ a\, m + b &\ge 0\\ a\, u + b &\ge 1. \end{align*} Since $m > l$, we can assume that \begin{align*} a\, l + b &= 0\\ a\, u + b &= 1. \end{align*} That is, $a = 2/3$ and $b=-1/3$. This is the smallest portfolio that dominates the option payoff, and has the value \begin{align*} \pi_l(a\, l+b) + \pi_m(a\, m+b)+ \pi_u(a\, u+b) &=\frac{1}{3}. \end{align*}
For the sub-replication, we find the largest portfolio dominated by the option payoff. That is, \begin{align*} a\, l + b &\le (l-K)^+\\ a\, m + b &\le (m-K)^+\\ a\, u + b &\le (u-K)^+, \end{align*} or \begin{align*} a\, l + b &\le 0\\ a\, m + b &\le 0\\ a\, u + b &\le 1. \end{align*} Since $l < m$, we assume that \begin{align*} a\, m + b &= 0\\ a\, u + b &= 1. \end{align*} Then $a=1$ and $b=-1$. This is the largest portfolio that is dominated by the option payoff, and has the value \begin{align*} \pi_l(a\, l+b) + \pi_m(a\, m+b)+ \pi_u(a\, u+b) &=0. \end{align*}
> Proof of the dominating property for the super-replication.
Assume that there is another portfolio such that \begin{align*} a'\, l + b' &\ge (l-K)^+=0\\ a'\, m + b' &\ge (m-K)^+=0\\ a'\, u + b' &\ge (u-K)^+=1. \end{align*} Then, by our choice, \begin{align*} \frac{1}{2}a' + b' &\ge \frac{1}{2}a + b=0 \tag{1}\\ 2a' + b' &\ge 2a + b=1.\tag{2} \end{align*} We show that \begin{align*} a'+b' \ge a+b = \frac{1}{3}.\tag{3} \end{align*} Assuming that $a'+b' < \frac{1}{3}.$ Then, from (2), \begin{align*} \frac{2}{3} -b' > 2(a'+b')-b' \ge 1. \end{align*} That is, $b'<-1/3$. Consequently, \begin{align*} \frac{1}{2}a' + b' &= \frac{1}{2}(a' + b') + \frac{1}{2} b' \\ &<0, \end{align*} which is contradictory to (1) above.
## Answer by Mark Joshi (score 4)
https://quant.stackexchange.com/a/23198
Trinomial trees give incomplete markets so there is a range of possible risk neutral prices. So you have to find the possible probabilities that make the tree risk-neutral and see what prices you get.
You have the correct expressions. Now just have to parametrize the set of solutions. It is one-dimensional and all the probabilities are positive so you need find them all.
For the second part, you can't replicate precisely but you can find a portfolio that dominates the pay-off. What is the cheapest such portfolio?
(see my book Concepts and Practice etc for extensive discussion and worked solutions of similar examples.)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.