# Black-Scholes model and arbitrage free price

Consider the Black-Scholes model and the derivative asset:

$$X = \begin{cases} K, \qquad \qquad \qquad \quad S_T\leq A, \\ K+A-S_T, \qquad A\leq S_T < K+A, \\ 0, \qquad \qquad \qquad \quad S_T>K+A. \end{cases}$$

Replicate this derivative using portfolio consisting of bond, asset S and European call option. Find the arbitrage free price for X.

Can anybody explain to me how to solve this exercise? I can't seem to grasp the concept of it. Thanks!

• Graphing the payoff function given above may help. Then try to compare it to known payoffs of calls, puts, etc. – noob2 Mar 19 '17 at 20:24

## 2 Answers

To replicate this payoff, we note that \begin{align*} X &= K\,\mathbb{I}_{S_T \le A} + (K+A-S_T)\, \mathbb{I}_{A< S_T \le K+A}\\ &=K\,\mathbb{I}_{S_T \le A} + (K+A-S_T)\, \mathbb{I}_{S_T>A} - (K+A-S_T)\, \mathbb{I}_{S_T > K+A}\\ &= K - (S_T-A)\, \mathbb{I}_{S_T>A} + (S_T-(K+A))\, \mathbb{I}_{S_T > K+A}\\ &= K - (S_T-A)^+ + (S_T-(K+A))^+. \end{align*} That is, long a zero coupon bond, short a call option with strike $A$, and long a call option with strike $K+A$.

This position is equivalent to a combination of bond and bear vertical call spread.

Imagine that at time $t$ you own $Ke^{-r(T-t)}$ zero-coupon bonds. Then at time T you will receive K in cash.

The spread consists of 1 short European call option with strike A and 1 long European option with strike A+K.

If $S_T < A$ both options are out of money, so their payoffs are zero. Thus you'll receive only $K$ from your bond position.

If $A \leq S_T < K+A$, 1 short European call with generate you losses of $A - S_T$ and 1 long will generate nothing. Thus you'll receive $K$ from bond and $A - S_T$ from the spread, $K + A - S_T$ in total.

In case $S_T \geq K+A$, you will receive $A - S_T$ from the short option and $S_T - (K+A)$ from the long option or $-K$ from the spread. Together with your bond position you will receive zero in total.

By the Law of One Price, the arbitrage free price of $X$ will be : $$X(t) = Ke^{-r(T-t)} + C(K+A, t) - C(A, t)$$

where $C(L,t)$ the price at time $t$ of European option with strike $L$.

For example, if you could sell your derivative security for the price $M > X(t)$, then you could use $X(t)$ to replicate the position by purchasing the bond and the spread as desribed above and keep $M - X(t)$ as your risk-free profit. This is an arbitrage.

See also Chapter 3 "Static and Dynamic Replication" in E. Derman's "The Volatility Smile"