The following is the Black-Scholes formula for the value of a call European option:

Question

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

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 $C(S) = S$.