Shouldn't (according to the Black-Scholes model) the price of a call option with a strike of an arbitrary amount away from the current asset's price, be equal to the price of a put option with the same "distance to exercise", just in the opposite direction? In other words: Shouldn't a call and put with symmetric strikes around the asset's value be priced equally? According to my calculation, they don't equate for neither ITM, nor OTM options. What's the reason for this inequality of option prices?

I'm aware that this does not apply to observed market prices. But shouldn't this hold for option prices computed with the traditional Black-Scholes OP model, as this formula assumes normally distributed returns?

  • 1
    $\begingroup$ You don't get a payoff based on the return, you get a payoff based on the dollar amount by which the price $S_T$ goes above/below the strike. So it is not a symmetric situation. The price distribution is lognormal, which is not symmetric. $\endgroup$
    – nbbo2
    Apr 27, 2020 at 18:49
  • $\begingroup$ Doesn't the N(d1) "apply" a drift/return to S_t, so that the ending value S_T is indeed lognormally distributed, but the return is normally distributed? Because we also use the normal cdf for d1. The only thing I could have thought of is, that this risk-free drift moves the expected value "upwards", therefore we get a symmetric distribution at the option's maturity T and not at time t? $\endgroup$
    – elemenope
    Apr 27, 2020 at 19:46

1 Answer 1


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.


  • $\begingroup$ Thanks for your answer, really intuitive and helpful! I did some research and found the paper "The Crash of '87: Was It Expected" by Bates (1991). He writes: "If the strike prices of the put and call are spaced symmetrically around the forward price, the symmetry or asymmetry of the risk-neutral distribution will be directly reflected in the relative prices of these out-of-the-money calls and puts. Symmetric risk-neutral distributions imply equal prices for OTM European calls and puts; skewed distributions create systematic divergences." $\endgroup$
    – elemenope
    Apr 28, 2020 at 15:23
  • $\begingroup$ So if I use the forward price as "center" with strike F(1+x) for calls and F/(1+x) for puts, shouldn't they be priced equally? I get your point of more upward potential for calls. But I don't get how Bates gets to this result? If I calculate F as S_t * exp(r * (T-t)) I still get a higher value for calls, as you suggested. It would make sense if Bates was referring to equal option prices at expiry, but he is referring to prices at t. $\endgroup$
    – elemenope
    Apr 28, 2020 at 15:29
  • $\begingroup$ The distribution is not symmetrical as it is lognormal. You may just want to give it a try and see that these statements can only be approximately true. $\endgroup$ Apr 29, 2020 at 6:58

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