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. $\endgroup$ 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 $\endgroup$ Aug 2 '17 at 16:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.