The following is the Black-Scholes formula for the value of a call European option: $$ C(s) = N(d_1)S-N(d_2)K $$ $$ d_1 = \frac{1}{\sigma \sqrt{T}} \left[ \ln{\frac{S}{K}}+ \frac{\sigma^2}{2}T \right] $$ $$ d_2 = d_1 - \sigma \sqrt{T} $$ where

  • $N$ is the cumularive distribution function of the standard normal distribution
  • $T$ is time to expiration
  • $S$ is the spot price of the underlying,
  • $K$ is the strike price of the option
  • $\sigma$ is the volativity of the returns of the underlying

A European call option can only be exercised at expiration time. What is the option value if $K=0$?

I think answer should be $0$, not sure though

  • 1
    $\begingroup$ I have no idea what this question is about, but I know that one cannot divide by zero. $\endgroup$
    – Matti P.
    Aug 23, 2019 at 7:26
  • $\begingroup$ I edited your question to add the tag "finance" and remove "derivatives". On this site, derivatives are mathematical functions studied in calculus, not financial instruments. $\endgroup$
    – awkward
    Aug 23, 2019 at 14:25
  • $\begingroup$ This is a question that can be answered by static replication alone (so Black Scholes formula is not necessary but better agree). If this option is worth differently than the stock, buy the cheap one and sell the expensive one. Then you exercise and liquidate at maturity for net zero cash flow. $\endgroup$ Aug 23, 2019 at 14:45

1 Answer 1


You can't literally plug in zero in the given formula, since it contains the term $\ln(S/K)$, but let $K \searrow 0$ (approaching $0$ from above). Then, since $S>0$, $\ln(S/K) \rightarrow \infty$, therefore $\frac{1}{\sigma\sqrt{T}} \left( \ln(S/K) + \frac{\sigma^2T}{2} \right) = d_1 \rightarrow \infty$, and $d_2 = d_1 - \sigma \sqrt T \rightarrow \infty$. This implies $N(d_1)S - N(d_2)K = C(S) \rightarrow S$, since $\lim_{x\rightarrow \infty} N(x) = 1$.

Now, this means that the option in question is worth exactly the same as the underlying, which is a welcome sanity check for the Black Scholes formula. After all, we can buy this option and short the underlying asset. When the option matures, we exercise it to gain another copy of the underlying for free (since $K=0$), covering our short obligations. Thus, we'd have gained $S-C(S)$, so, under no-arbitrage assumptions, $S \leq C(S)$. But: The option can't possibly be worth more than $S$, since it only gives us the right to own the underlying from the time of expiry on, and buying it directly would do that right now. Thus $C(S) = S$.

  • $\begingroup$ E Greuhuber I agree with your answer assuming the stock does not pay a dividend. If the stock pays a dividend and the option is European Style (meaning you cannot exercise it early) then your answer needs to be adjusted for the dividend. If the option is American Style then it does not need to be adjusted for the dividend. $\endgroup$
    – Bob
    Mar 24, 2020 at 13:27

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