0
$\begingroup$

Hopefully these are acceptable questions in this forum.

I assume such options are today OTC instruments only with significant corporate risk on the miner. I have a little knowledge of interest rate modelling and derivatives et al. but none of gold or other metals.

For (gold) miners, with significant debt, there are underlying practical risk management questions . References to simple models including any related tradeable instruments would be of interest

Improve this question
$\endgroup$
4
  • $\begingroup$ What is the question?? $\endgroup$
    – dm63
    Nov 13 '16 at 12:50
  • $\begingroup$ Edit made to clarify estimation question. $\endgroup$
    – Ian Calvert
    Nov 13 '16 at 12:57
  • $\begingroup$ Are you asking what are the strikes for a 6 month costless collar on gold? You didn't mention costless in the question , I'm just guessing $\endgroup$
    – dm63
    Nov 13 '16 at 13:10
  • $\begingroup$ Yes, "at-the-money .. collar", was meant to be a synonym for costless. $\endgroup$
    – Ian Calvert
    Nov 13 '16 at 13:13
0
$\begingroup$

There is an infinite set of pairs of strikes that form a costless collar, since for each (positive) put strike there is a call strike that makes the put price equal the call price. Within each pair, the put strike is less than the forward price and the call strike is greater than the forward price. The forward price of gold can be observed from otc forward markets or futures markets, and it can be calculated from spot price, interest rates and gold storage costs. Given a put strike and a forward, the call strike for a costless collar depends on the skew for out of the money puts versus out of the money calls. For example, if the out of the money calls on gold are at a much higher implied volatility than out of the money puts, then the call strike will be further from the forward than the put strike. Skew in the gold market (like other otc markets) can move significantly depending on market conditions at any given time.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.