Replicating a Capped Payoff with Bonds and a Bear Call Spread
Summary
The document shows how to replicate a piecewise derivative payoff that pays a fixed amount below one strike, declines linearly over an intermediate price range, and reaches zero above a higher level. Its terminal payoff is decomposed into a fixed cash payment, a short call at the lower strike, and a long call at the higher strike. Discounting the fixed payment as a zero-coupon bond gives the arbitrage-free value as the bond value plus the higher-strike call price minus the lower-strike call price.
The payoff checks in the response verify the replication across the three regions, and the law of one price links the replicating portfolio's cost to the derivative's price. This is a static replication argument under the Black-Scholes setting. The quoted valuation assumes the bond and European calls are available at consistent market prices and uses the stated interest-rate discounting; transaction costs, funding differences, and other market frictions are not considered.
Key ideas
- The payoff can be written as a fixed terminal payment minus a call at the lower strike plus a call at the higher strike.
- The fixed payment is replicated by a zero-coupon bond that pays the required amount at maturity.
- The option component is a bear call spread, short at the lower strike and long at the higher strike.
- The law of one price sets the derivative value equal to the cost of its replicating portfolio.
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Full text
# Black-Scholes model and arbitrage free price
# 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!
## Answer by Gordon (score 2, accepted)
https://quant.stackexchange.com/a/33109
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 $K+A$.
## Answer by zer0hedge (score 1)
https://quant.stackexchange.com/a/33103
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"Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.