Replicating a Capped Oil-Linked Note with a Call Spread
Summary
The note explains how to replicate a zero-coupon structured note that returns its principal at maturity plus an oil-linked payment. The extra payment rises in proportion to oil prices above 25, but is capped at 2,550. The proposed replication combines a zero-coupon bond paying 1,000 with 170 long calls struck at 25 and 170 short calls struck at 40. The call spread creates the desired rising payoff up to the cap and no further increase beyond it.
An algebraic derivation rewrites the capped payment as the difference between two call payoffs, each scaled by 170. This illustrates a general payoff-replication technique: represent a capped upside exposure using a long lower-strike call and a short higher-strike call, then add a bond for the principal. The discussion also notes that a cap can reduce the cost relative to uncapped call exposure, while limiting gains. It gives no option prices or valuation assumptions, and focuses on maturity payoffs rather than interim pricing or issuer risk.
Key ideas
- The principal repayment is replicated by a zero-coupon bond paying 1,000 at maturity.
- The oil-linked addition is a capped payoff that begins increasing above an oil price of 25.
- Long calls at 25 and short calls at 40 create the capped upside exposure.
- The option positions are scaled to 170 contracts each for the stated contract size.
- A payoff can be decomposed algebraically into standard option payoffs for replication.
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Full text
# Constructing payoff with options
# Constructing payoff with options
Suppose that COMPANY A has issued a special bond that does not pay any coupons. At maturity T, the bondholder receives the principal (face value) equal to 1,000 plus an additional payment equal to the maximum of zero or 170 times (O-25), where O is the oil price at the bond's maturity T. However, this additional payment cannot exceed 2,550. What positions should you take in a zero-coupon bond and/or European call and/or put options on the oil price to replicate the payoff for the holder of this special bond at maturity T? Clearly indicate the face value of the bond, the type of options (call/put), the strike price of the options, the number of options, and the position in the bond and/or options (long or short). Assume that the option contract size is 1. I am trying to solve this problem, but I would be interested in understanding the underlying logic.
## Answer by Kai (score 1)
https://quant.stackexchange.com/a/77576
This is essentially some form of commodity-linked structured note (this case being oil). In this case, such a structured is a Leveraged Participation note with Cap in industry lingo.
In terms of how to replicate this, it is pretty simple. Consider a call spread, and how many units of this call you need to make to adjust the gradient accordingly!
You need a zero-coupon bond paying 1000, plus 170 units call spread with buy call at strike 25 and sell call at strike 40 (create a max payout of 2550) to create this structure's payoff at maturity
Perhaps you can elaborate a bit more on what underlying logic it is you want to understand? In terms of why invest in these kinds of structures? There are various reasons but the main two are:
- Pure vanilla calls are expensive sometimes. Embedding a cap on the upside cheapens the option.
- You hold the view that there might be an upside but that the upside would not be far ITM, so you wouldn't want to pay for that additional optionality.
## Answer by ir7 (score 0)
https://quant.stackexchange.com/a/77583
Just to add the the previous answer, the additional payment is: $$ \min\left(\max\left(0,170(X-25) \right), 2550\right) $$ This can be algebraically manipulated too (chasing for standard option payoffs): $$ = 170 \left[ \min\left(\max \left(0,X-25 \right), 15\right) \right] $$ $$\stackrel{min(a,b)=a+b -max(a,b)}{=} 170 \left[ \max \left(0,X-25 \right) + 15 - \max\left(\max \left(0,X-25 \right), 15\right) ) \right]$$
$$ = 170\left[ \max \left(0,X-25 \right) - \max\left(\max \left(0,X-25 \right)-15, 15-15\right) \right] $$ $$= 170\left[ \max \left(0,X-25 \right) - \max\left(\max\left(-15,X-40 \right), 0\right)\right]$$
$$ = 170\left[ \max\left(0,X-25 \right) - \max \left(0,X-40 \right) \right]$$ $$=170\max\left(0,X-25 \right) - 170\max \left(0,X-40 \right) $$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.