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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.

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  • $\begingroup$ I recall that Freeport MacMoran used to have several series of preferred equity whose dividends were linked to the prices of various commodities and precious metals. However they were linear, no floors / caps / optionality - each share paid periodically something like $c\times$ the price of gold. $\endgroup$ Commented Dec 1, 2023 at 23:11

2 Answers 2

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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! enter image description here

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.
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  • $\begingroup$ Thank you very much for your clear answer. To be able to master these concepts would you suggest learning the graph of the main strategies (covered call, protective put, collar, spreads, straddles and strangles) "by hearth" or you have a precise logical procedure (or mental shortcuts) you follow (my point is I kind of need each time to write down the relevant states of the world, see the payoff of each option in those and then draw the payoff, and I am not immediately able to "visualize" the total payoff when the portfolio is made of two or more options). $\endgroup$ Commented Dec 2, 2023 at 10:05
  • $\begingroup$ The basic strats provide some framework on how to replicate these structures with vanillas. If you are unable to visualise it immediately, I would suggest you start practising by looking at various basic strategies (I am sure you can find plenty of unique structures online) and try to break them down into their corresponding structures before moving on to more exotic ones. Draw out the graph! I think very quickly you will develop a sense for it. It is pretty simple. Just imagine the shape of the payoff and then think "what vanilla options will get me close to this shape" will get you far. $\endgroup$
    – Kai
    Commented Dec 4, 2023 at 18:53
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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):
$$ = 170 \left[ \min\left(\max \left(0,X-25 \right), 15\right) \right] $$ $$\stackrel{min(a,b)=a+b -max(a,b)}{=} 170 \left[ \max \left(0,X-25 \right) + 15 - \max\left(\max \left(0,X-25 \right), 15\right) ) \right]$$

$$ = 170\left[ \max \left(0,X-25 \right) - \max\left(\max \left(0,X-25 \right)-15, 15-15\right) \right] $$ $$= 170\left[ \max \left(0,X-25 \right) - \max\left(\max\left(-15,X-40 \right), 0\right)\right]$$

$$ = 170\left[ \max\left(0,X-25 \right) - \max \left(0,X-40 \right) \right]$$ $$=170\max\left(0,X-25 \right) - 170\max \left(0,X-40 \right) $$

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