# 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.

• The answer below by Mark is fine. For probabilities, you can set one as the free variable and then solve the other two; you will have infinitely many of them. For the range of the option price, you need to find the maximal and minimal prices from the free variable you have set. Feb 10, 2016 at 14:21
• @Gordon I made an edit to my post but I am not sure where to go from here Feb 10, 2016 at 18:29
• @Gordon Is the price $C = \pi_u(u - K)$? for the up state? Feb 10, 2016 at 18:39
• @Gordon Please see edited question need your guidance, for the force is indeed strong with you Feb 11, 2016 at 18:09
• The price for the up state is correct. Set $\pi_m = x$, and try to express $\pi_l$ and $\pi_u$ in terms of $x$. Your option price is then a function of $x$. The maximum and minimum of this function will respectively be $C^+$ and $C^-$. Feb 11, 2016 at 18:14

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.

• Hence in part b.) we have a higher pay off correct? by the way it was constructed Feb 11, 2016 at 19:19
• @MorganWeiss: full answer is now provided. Feb 11, 2016 at 19:38

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.)

• What is the process of finding the probabilities that made the tree risk-neutral? Feb 10, 2016 at 0:54
• well that have to add up to 1 and they have to make the expected value of the stock today's value (since R is zero). That's it. So it's just solving an undetermined system. Feb 10, 2016 at 0:55
• thats what I thought here I am going to make an edit of my question Feb 10, 2016 at 0:57
• Please see the re-edit to my question, I am confused where to go from here Feb 10, 2016 at 18:29