1
$\begingroup$

Suppose the function double bsCall(double S0, const double &K, double T, double r, double sigma) computes analytically the Black-Scholes price of a call option and double impVolCall(double S0, const double &K, double T, double r, double C) calculates the implied volatility. Using the put-call parity, one can define the function that returns the put implied volatility in this way:

double ImpliedVolPut(double S0, const double &K, double T, double r, double C)
{
  double x = impVolCall(S0, K, T, r, C + S0 - K*exp(-r*T));
  return x;
}

Moreover, I have a function that computes European Call/Put option price for the Heston model semi-analytically:

hestonClosedPrice(double lambda, double vbar, double eta, double rho, double v0, double r, double tau, double S0, double K, char optionType)

My question is why the volatility surface obtained for different values of K and T for the call and put options look the same:

std::vector<double> hestonPrice(std::vector<double> k, std::vector<double> t)
{
   if(optionType == 'Call'){
       HestonPrice(k[i],t[j]) = hestonClosedPrice(lambda, vbar, eta, rho, v0, r, t[j], S0, k[i], 'Call');
   }
   else{
       HestonPrice(k[i],t[j]) = hestonClosedPrice(lambda, vbar, eta, rho, v0, r, t[j], S0, k[i], 'Put');
   }
   if (optionType == "Call") {
       hestonIVS = impVolCall(S0, k, t, r, HestonPrice(k[i],t[j]));
   } else if (optionType == "Put") {
       hestonIVS = impVolPut(S0, k, t, r, HestonPrice(k[i],t[j])); 
   }

   ...
}

Intuitively, one should have that the lower the strike, the higher the call implied vol and the lower the put implied vol. The option pricer for both call and put is correct.

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ I believe the code and that the underlying model is Heston is irrelevant to your question, which is why European puts and calls have the same IV (in theory). Maybe you can edit your question and leave only the real question in the text? $\endgroup$
    – user34971
    Jul 10, 2019 at 18:12
  • $\begingroup$ @ilovevolatility yes I see your point! And actually I get this result because I use the put-call parity, which works for the same level of volatility. I just find a bit counterintuitive that for strikes less than ATM values, ITM calls and OTM puts have the same implied volatility. An ITM option has a positive intrinsic value compared to an OTM that has zero intrinsic value. Therefore, I expected the value (or implied volatility) of an ITM call to be higher than then value (or implied volatility) of an OTM put. $\endgroup$
    – FunnyBuzer
    Jul 10, 2019 at 20:08
  • 2
    $\begingroup$ I think I understand your point, but the answer to that, however counter intuitive it may seem is that if puts and calls satisfy put-call parity with different implied volatilities then there would be arbitrage. If you want me to show that put call parity implies IVs must be the same let me know and I'll put it in a proper answer. $\endgroup$
    – user34971
    Jul 11, 2019 at 6:50
  • $\begingroup$ @ilovevolatilit, thanks! I'm aware of the result but somehow it seems like I forgot about its main implication. Thanks for pointing it out. $\endgroup$
    – FunnyBuzer
    Jul 11, 2019 at 9:25

1 Answer 1

Reset to default
1
$\begingroup$

A call and a put option with the same strike and same expiry should have the same implied volatility by put-call parity indeed. A call and a put are essentially the same when you hedge the initial delta in terms of greek exposures

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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