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Kermittfrog
Kermittfrog
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Without delving into the mathematics of the problem, I think we can answer your question as follows.

Please note that subsequent thoughts hold for both, zero and positive interest rates.

  1. At expiry, your statement is correct: The payoff of a put option will be $(X_P-S)^+$ and the payoff of a call is $(S-X_C)^+$. If $X_C<X_P$, and if the underlying just happens to be equal to $\frac{X_P+X_S}{2}$, i.e. exactly between the two, both have the exact same value.

  2. At any time before expiry, the put option's payoff can be expected to be at most the difference between $X_P$ and $0$, i.e. the put option has a limited upside as the underlying process cannot go below zero (at least in the model). A call option, on the other hand has unlimited upside potential, i.e. the call payoff at expiry can be more than just $X_P-0$, e.g. if $S_T=X_C+kX_P$, $k>0$, then of course the call payoff will be $(kX_P)$ which may be larger than $X_P$. Although these states have a small probability, they nevertheless add to the present value of the call option. Hence, the call should always be 'a bit' more expensive for a given distance-to-strike, compared to the put option.

HTH

Without delving into the mathematics of the problem, I think we can answer your question as follows.

Please note that subsequent thoughts hold for both, zero and positive interest rates.

  1. At expiry, your statement is correct: The payoff of a put option will be $(X_P-S)^+$ and the payoff of a call is $(S-X_C)^+$. If $X_C<X_P$, and if the underlying just happens to be equal to $\frac{X_P+X_S}{2}$, i.e. exactly between the two,.

  2. At any time before expiry, the put option's payoff can be expected to be at most the difference between $X_P$ and $0$, i.e. the put option has a limited upside as the underlying process cannot go below zero (at least in the model). A call option, on the other hand has unlimited upside potential, i.e. the call payoff at expiry can be more than just $X_P-0$, e.g. if $S_T=X_C+kX_P$, $k>0$, then of course the call payoff will be $(kX_P)$ which may be larger than $X_P$. Although these states have a small probability, they nevertheless add to the present value of the call option.

HTH

Without delving into the mathematics of the problem, I think we can answer your question as follows.

Please note that subsequent thoughts hold for both, zero and positive interest rates.

  1. At expiry, your statement is correct: The payoff of a put option will be $(X_P-S)^+$ and the payoff of a call is $(S-X_C)^+$. If $X_C<X_P$, and if the underlying just happens to be equal to $\frac{X_P+X_S}{2}$, i.e. exactly between the two, both have the exact same value.

  2. At any time before expiry, the put option's payoff can be expected to be at most the difference between $X_P$ and $0$, i.e. the put option has a limited upside as the underlying process cannot go below zero (at least in the model). A call option, on the other hand has unlimited upside potential, i.e. the call payoff at expiry can be more than just $X_P-0$, e.g. if $S_T=X_C+kX_P$, $k>0$, then of course the call payoff will be $(kX_P)$ which may be larger than $X_P$. Although these states have a small probability, they nevertheless add to the present value of the call option. Hence, the call should always be 'a bit' more expensive for a given distance-to-strike, compared to the put option.

HTH

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Kermittfrog
Kermittfrog
  • 7.1k
  • 1
  • 11
  • 31

Without delving into the mathematics of the problem, I think we can answer your question as follows.

Please note that subsequent thoughts hold for both, zero and positive interest rates.

  1. At expiry, your statement is correct: The payoff of a put option will be $(X_P-S)^+$ and the payoff of a call is $(S-X_C)^+$. If $X_C<X_P$, and if the underlying just happens to be equal to $\frac{X_P+X_S}{2}$, i.e. exactly between the two,.

  2. At any time before expiry, the put option's payoff can be expected to be at most the difference between $X_P$ and $0$, i.e. the put option has a limited upside as the underlying process cannot go below zero (at least in the model). A call option, on the other hand has unlimited upside potential, i.e. the call payoff at expiry can be more than just $X_P-0$, e.g. if $S_T=X_C+kX_P$, $k>0$, then of course the call payoff will be $(kX_P)$ which may be larger than $X_P$. Although these states have a small probability, they nevertheless add to the present value of the call option.

HTH