Given that the Black-Scholes formula for a European Call is given by:


$S$ is stock price, $K$ is strike price

When I take limit as $t\rightarrow T^-$, where $\sigma>0$, what are the cases that I have to take into considerations?


Look at the values of $d_1$ and $d_2$ when $t \rightarrow T$:

$$d_1 = \frac{\ln(S/K) + \left(r - D + \dfrac{1}{2}\sigma^2\right)\tau}{\sigma \sqrt{\tau}}$$


$$d_2 = d_1 - \sigma \sqrt{\tau}$$

with $\tau = T -t$.

Therefore, $t \rightarrow T$ is equivalent to $\tau \rightarrow 0$

  • Case 1: $S > K$

$d_1 \sim \frac{\ln(S/K) }{\sigma \sqrt{\tau}} \rightarrow + \infty$

$d_2 \sim d_1\rightarrow + \infty$


$$C(S,\tau) = S e^{- D \tau} N (d_1) - K e^{- r \tau} N (d_2) \rightarrow (S-K)$$

as $N(d_i) \rightarrow 1$ for $i \in \{ 1,2\}$.

  • Case 2: $S < K$

$d_1 \sim \frac{\ln(S/K) }{\sigma \sqrt{\tau}} \rightarrow - \infty$

$d_2 \sim d_1\rightarrow - \infty$


$$C(S,\tau) = S e^{- D \tau} N (d_1) - K e^{- r \tau} N (d_2) \rightarrow 0$$

as $N(d_i) \rightarrow 0$ for $i \in \{ 1,2\}$.

  • Case 3: $S = K$

$d_1 = \frac{\left(r - D + \dfrac{1}{2}\sigma^2\right)\sqrt{\tau}}{\sigma} \rightarrow 0$

$d_2 \rightarrow 0$


$$C(S,\tau) = S e^{- D \tau} N (d_1) - K e^{- r \tau} N (d_2)$$

$$C(S,\tau) = \left( S - K \right) \dfrac{1}{2} \rightarrow 0$$

as $N(d_i) \rightarrow \dfrac{1}{2}$ for $i \in \{ 1,2\}$ and $S = K$ by assumption.

Putting everything together, you see that the price of the call tends to its payoff


when the time $t$ is "infinitely" close to the maturity of the call $T$.

  • 1
    $\begingroup$ for Case I, shall we separate $S>K$ and $S=K$? Because when $S=K$, $\ln (S/K)=\ln 1=0$, hence the limit will be $0$ instead. Am I right? $\endgroup$ – lrh09 Aug 2 '17 at 15:15
  • $\begingroup$ Good point. I have updated my answer with a 3rd case. $\endgroup$ – JejeBelfort Aug 2 '17 at 15:41
  • $\begingroup$ By the way, Case 3: $S=K$, when $N(d_1),N(d_2)\rightarrow \frac{1}{2}$, isn't $C(S,t)\rightarrow \frac{1}{2}(S-K)$? Because the power $\tau=T-t\rightarrow 0$, the terms with $e$ approach $1$ $\endgroup$ – lrh09 Aug 2 '17 at 16:04
  • $\begingroup$ Yes, but since $S = K$, I factored out by $S$ first. But your approach may be more intuitive indeed. I will update $\endgroup$ – JejeBelfort Aug 2 '17 at 16:27

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