R19.P1.T3.FIN_PRODS_HULL_Ch26_ExoticDrivatives_Topic:Zero-Cost-Derivatives

[gargi.adhikari]

In reference to R19.P1.T3.FIN_PRODS_HULL_Ch26_ExoticDrivatives_Topic:Zero-Cost-Derivatives:-
I am trying to understand how "Any derivative can be converted into a zero-cost product by deferring payment until maturity"

If c is the cost of the option when payment is made at time zero, then A = cerT is the cost when payment is made at time T, the maturity of the option. The payoff is then max(ST - K, 0) - A or max(ST - K - A, -A).

When the strike price, K, equals the forward price,-----Not Clear what happens when K= Forward Price... :-(

 

[ShaktiRathore]

Hi gargi,

You are not paying anything when you entered into the derivative contract therefore its cost less to purchase the derivative.Here you are just converting the derivative into another derivative with a zero cost but somewhat a different payoff structure.
For (hypothetical )e.g. if you enter a call option contract which costs $ c today ,you did not pay today so 0 cost ,instead pay after some time at maturity T year the value of cost of c*exp(r*T).The payment done by you is 0 today to buy the call option that is the premium is effectively 0 ,however the payoff changes accordingly as max(ST-c*exp(r*T)-K, -c*exp(r*T)). Hence we converted the call option with payoff max(ST-K,0) and cost c into a derivative with payoff max(ST-c*exp(r*T)-K, -c*exp(r*T)) and cost 0.

thanks

 

[gargi.adhikari]

@ShaktiRathore Thanks so very much for sharing this insight buddy !