Problem 26.1 Easy Free 3 variations

Two-Step Tree for a Squared Payoff

A stock currently trades at $S_0 = 5$ and evolves over two time steps under a binomial model with zero interest rates: at each step it either doubles or halves in value. A derivative pays $S_T^2$ at expiry $T=2$ . Using the risk-neutral probabilities that make the discounted stock price a martingale, find the time- $0$ price of this contract.

Work it out, then compare with the answer.

SolutionShow the worked solution

Key idea. At zero rates the risk-neutral up-probability makes the stock a martingale, so the price is the expected payoff under that probability.

Let $u=2$ and $d=\tfrac12$ . The condition $qu+(1-q)d=1$ gives

q=\frac{1-d}{u-d}=\frac{1/2}{3/2}=\frac13 .

After two steps the stock has three possible values:

$20$ with probability $q^2=\tfrac19$ ;
$5$ with probability $2q(1-q)=\tfrac49$ ;
$\tfrac54$ with probability $(1-q)^2=\tfrac49$ .

The price is therefore

V_0=\tfrac19\cdot 400+\tfrac49\cdot 25+\tfrac49\cdot\tfrac{25}{16} =\frac{400+100+6.25}{9}=\frac{506.25}{9}=\frac{225}{4}.

Alternative. Write $S_2=S_0X_1X_2$ , where $X_1,X_2$ are i.i.d. and equal $2$ with probability $\tfrac13$ and $\tfrac12$ with probability $\tfrac23$ . Then

\mathbb{E}[X^2]=\tfrac13\cdot4+\tfrac23\cdot\tfrac14=\tfrac32,

\mathbb{E}[S_2^2]=S_0^2\,\bigl(\mathbb{E}[X^2]\bigr)^2=25\cdot\tfrac94 .

The price exceeds $S_0^2=25$ by Jensen’s inequality, because the payoff is convex.

\boxed{V_0=\tfrac{225}{4}=56.25}

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 26.1a

A stock starts at $S_0 = 4$ and, over each of two time steps, either rises by $90\%$ or falls by $30\%$ , with zero interest rates. A European binary call pays $\$1$ at $T=2$ if $S_T > 2$ , and $\$0$ otherwise. Using risk-neutral valuation, find the fair price of this option, rounded to three decimal places.

Work it out, then compare with the answer.

Solution

Here $u=1.9$ and $d=0.7$ , so

q=\frac{1-0.7}{1.9-0.7}=\frac{0.3}{1.2}=\frac14 .

The terminal prices are $4(1.9)^2=14.44$ , $4(1.9)(0.7)=5.32$ and $4(0.7)^2=1.96$ .

Only the down–down node, $1.96$ , lies at or below $2$ . The binary call therefore pays $1$ unless the stock falls twice. Its price is the risk-neutral probability of that event:

1-(1-q)^2=1-\left(\tfrac34\right)^2=\tfrac{7}{16}.

\boxed{\tfrac{7}{16}=0.4375\approx 0.438}

Variation 26.1b

A stock starts at $S_0 = 10$ and, over a single time step, either doubles or halves in value, with zero interest rates. Find the time- $0$ price $C_0$ of a European call option with strike $K = 5$ expiring after this one step.

Work it out, then compare with the answer.

Solution

As in the main problem, $q=\tfrac13$ . The call pays $20-5=15$ in the up state and $0$ in the down state. Its price is

C_0=\tfrac13\cdot 15=5 .

The value of $q$ does not actually matter here. Since $K$ equals the down-state price, the payoff is $S_1-K$ in both states, so the call is equivalent to the stock minus $\$5$ cash. Hence $C_0=S_0-K$ .

\boxed{C_0=5}

Variation 26.1c

A stock currently priced at $S_0 = 8$ moves over one period to one of three outcomes: it stays unchanged with probability $0.5$ , falls by $50\%$ with probability $0.3$ , or rises by $30\%$ with probability $0.2$ . Using these given probabilities directly as pricing weights, find the time- $0$ price of a European call option with strike $K = 8$ on this stock.

Work it out, then compare with the answer.

Solution

The terminal prices are $8$ , $4$ and $10.4$ . The call with strike $8$ pays only in the last state, where it pays $10.4-8=2.4$ . Weighting by the given probability:

C_0=0.2\times 2.4=0.48 .

These weights are not risk-neutral, since under them $\mathbb{E}[S_1]=7.28\neq 8$ . A single stock cannot pin down three state prices, so the figure above is simply the price implied by the supplied weights.

\boxed{C_0=0.48}

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