Problem 25.5 Medium Free 4 variations

Replicating an Absolute-Value Payoff

Bonds, the stock $S$ and European calls on $S$ all expire at a common time $T$ . The stock starts at $S_0=7$ and a bond paying $1 at $T$ costs $0.9$ today. Calls trade today at the following (strike $K$ , price $C_0$ ) pairs:

(5,4.2),\quad(10,1.4),\quad(15,0.7),\quad(20,0.1).

Find the time- $0$ price of a contract paying $|S_T-15|$ at time $T$ .

Work it out, then compare with the answer.

SolutionShow the worked solution

Key idea. A piecewise-linear payoff is a portfolio. Its value at $S_T=0$ and its slope below the first kink fix the bond and stock holdings. Each kink at strike $K$ with slope change $\Delta$ then adds $\Delta$ calls struck at $K$ .

The payoff $|S_T-15|$ equals $15$ at $S_T=0$ and has slope $-1$ for $S_T<15$ . At $15$ the slope jumps from $-1$ to $+1$ , a change of $2$ . Hence

|S_T-15| = 15 - S_T + 2(S_T-15)^+ .

Check both regions:

For $S_T\le 15$ the right side is $15-S_T$ .
For $S_T>15$ it is $15-S_T+2S_T-30=S_T-15$ .

So the contract is replicated by:

long $15$ bonds;
short $1$ share;
long $2$ calls struck at $15$ .

The bond and stock legs together form a synthetic put. The quoted prices are consistent with the absence of arbitrage: call prices decrease and are convex in $K$ , with slopes $-0.56,-0.14,-0.12$ that lie in $[-0.9,0]$ .

By no-arbitrage, the contract’s price equals the cost of this portfolio:

V_0 = 15(0.9) - 7 + 2(0.7) = 13.5 - 7 + 1.4 = 7.9 .

\boxed{V_0 = 7.9}

Variations

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

Variation 25.5a

The stock starts at $S_0=21$ , and European calls trade today at

(15,4.2),\quad(20,1.4),\quad(25,0.7),\quad(30,0.1).

Find the time- $0$ price of a contract paying $\min(S_T,25)$ at time $T$ .

Work it out, then compare with the answer.

Solution

The payoff equals $S_T$ up to $25$ and is flat above it, so the slope drops by $1$ at $25$ :

\min(S_T,25)=S_T-(S_T-25)^+ .

This is a covered call: long one share, short one call struck at $25$ . No bonds are needed. Its price is

V_0 = 21-0.7 .

\boxed{V_0=20.3}

Variation 25.5b

Bonds, the stock $S$ and European calls on $S$ all expire at a common time $T$ . The stock starts at $S_0=7$ and a bond paying $1 at $T$ costs $0.9$ today. Calls trade today at

(5,4.2),\quad(10,1.4),\quad(15,0.7),\quad(20,0.1).

Find the time- $0$ price of a contract paying $\max(3S_T,15)$ at time $T$ .

Work it out, then compare with the answer.

Solution

The payoff equals $15$ for $S_T\le 5$ and has slope $3$ above $5$ :

\max(3S_T,15)=15+3(S_T-5)^+ .

Replicate it with $15$ bonds and $3$ calls struck at $5$ :

V_0=15(0.9)+3(4.2)=13.5+12.6 .

\boxed{V_0=26.1}

Variation 25.5c

A contract $V$ pays $\min(2S_T,4)$ at time $T$ and currently trades at $V_0=3.4$ . You also have access to European puts expiring at $T$ , trading today at

(K=4,\ P_0=2.4),\qquad (K=2,\ P_0=0.8),

the underlying stock $S$ with $S_0=3$ , and bonds paying $1 at $T$ costing $Z_0=0.9$ today. Find a trade that locks in a riskless profit, giving your answer as

\#\,V + \#\,\text{Put}(K=4) + \#\,\text{Put}(K=2) + \#\,\text{Bonds}.

Work it out, then compare with the answer.

Solution

Replicate $V$ . The payoff is $4$ for $S_T\ge 2$ and $2S_T$ below $2$ . Hence

\min(2S_T,4)=4-2(2-S_T)^+ ,

that is, $4$ bonds and short $2$ puts struck at $2$ . The replication costs

4(0.9)-2(0.8)=2.0 .

Exploit the mispricing. $V$ trades at $3.4>2.0$ , so it is overpriced. Sell $V$ and buy the replicating portfolio: sell one $V$ , write two $K=2$ puts and buy four bonds. This collects $3.4-2.0=1.4$ today. The payoffs cancel at $T$ for every $S_T$ , so the profit is riskless. The $K=4$ put is not used.

In the requested format this is $-1\,V+0\,\text{Put}(4)-2\,\text{Put}(2)+4\,\text{Bonds}$ . The counts sum to $1$ .

\boxed{-1\,V+0\,\text{Put}(K{=}4)-2\,\text{Put}(K{=}2)+4\,\text{Bonds}}

Variation 25.5d

A straddle with strike $K=17$ trades today at $V_0=5.4$ . A European put struck at $K=17$ trades at $P_0=4.2$ and a European call struck at the same $K=17$ trades at $C_0=1.4$ , all expiring at a common time $T$ . The underlying stock $S$ starts at $S_0=14$ , and bonds paying $1 at $T$ cost $B_0=0.9$ today. Find a trade that locks in a riskless profit, giving your answer as

\#\,\text{Stock} + \#\,\text{Call} + \#\,\text{Put} + \#\,\text{Bonds} + \#\,\text{Straddle}.

Work it out, then compare with the answer.

Solution

Exploit the straddle quote. A straddle pays $|S_T-17|=(S_T-17)^++(17-S_T)^+$ , so it should cost $C_0+P_0=1.4+4.2=5.6$ . It trades at $5.4$ , which is too cheap. Buy the straddle and sell the call and the put. This collects $0.2$ today, and the payoffs cancel exactly at $T$ . No stock or bonds are needed.

In the requested format this is $0$ stock, $-1$ call, $-1$ put, $0$ bonds and $+1$ straddle. The counts sum to $-1$ .

Remark. Put–call parity is also violated: $C_0-P_0=-2.8$ but $S_0-17B_0=14-15.3=-1.3$ . A second arbitrage is therefore available: buy the call, sell the put, short the stock and buy $17$ bonds. This earns $1.5$ today with zero payoff at $T$ . Consequently the riskless trade is not unique. The one above is the trade that exploits the straddle quote directly.

\boxed{0\,\text{Stock}-1\,\text{Call}-1\,\text{Put}+0\,\text{Bonds}+1\,\text{Straddle}}

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