# 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. – Gordon Feb 10 '16 at 14:21
• @Gordon I made an edit to my post but I am not sure where to go from here – Wolfy Feb 10 '16 at 18:29
• @Gordon Is the price $C = \pi_u(u - K)$? for the up state? – Wolfy Feb 10 '16 at 18:39
• @Gordon Please see edited question need your guidance, for the force is indeed strong with you – Wolfy Feb 11 '16 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 $C^-$. – Gordon Feb 11 '16 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 – Wolfy Feb 11 '16 at 19:19
• @MorganWeiss: full answer is now provided. – Gordon Feb 11 '16 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? – Wolfy Feb 10 '16 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. – Mark Joshi Feb 10 '16 at 0:55
• thats what I thought here I am going to make an edit of my question – Wolfy Feb 10 '16 at 0:57
• Please see the re-edit to my question, I am confused where to go from here – Wolfy Feb 10 '16 at 18:29