The Black–Scholes Call Price Converges to Its Expiry Payoff
Summary
The document examines the limit of a European call’s Black–Scholes price as time approaches expiration, assuming positive volatility. It writes the standardized variables in terms of time to maturity and considers the three relationships between the underlying price and strike: above, below, or equal. When the underlying is above the strike, both normal cumulative probabilities approach one; when below, they approach zero. At the strike, both standardized variables approach zero and the probabilities approach one half.
Together, these cases show that the call value approaches its intrinsic payoff at expiry. This is a boundary check for the pricing formula, not a trading strategy or empirical test. The explanation assumes the stated Black–Scholes setup, including finite model inputs and positive volatility; the at-the-money conclusion follows because the underlying equals the strike as the limit is taken.
Key ideas
- As time to maturity shrinks, the Black–Scholes call price approaches its expiration payoff.
- For an underlying above the strike, both standardized variables tend to positive infinity.
- For an underlying below the strike, they tend to negative infinity.
- At the strike, both variables tend to zero and the normal probabilities tend to one half.
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Full text
# Black-Scholes European call price taking limits
# Black-Scholes European call price taking limits
Given that the Black-Scholes formula for a European Call is given by:
$$C(S,t)=Se^{-D(T-t)}N(d_1)-Ke^{-r(T-t)}N(d_2)$$
$S$ 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?
## Answer by JejeBelfort (score 3, accepted)
https://quant.stackexchange.com/a/35425
Look at the values of $d_1$ and $d_2$ when $t \rightarrow T$:
$$d_1 = \frac{\ln(S/K) + \left(r - D + \dfrac{1}{2}\sigma^2\right)\tau}{\sigma \sqrt{\tau}}$$
and
$$d_2 = d_1 - \sigma \sqrt{\tau}$$
with $\tau = T -t$.
Therefore, $t \rightarrow T$ is equivalent to $\tau \rightarrow 0$
- Case 1: $S > K$
$d_1 \sim \frac{\ln(S/K) }{\sigma \sqrt{\tau}} \rightarrow + \infty$
$d_2 \sim d_1\rightarrow + \infty$
Therefore,
$$C(S,\tau) = S e^{- D \tau} N (d_1) - K e^{- r \tau} N (d_2) \rightarrow (S-K)$$
as $N(d_i) \rightarrow 1$ for $i \in \{ 1,2\}$.
- Case 2: $S < K$
$d_1 \sim \frac{\ln(S/K) }{\sigma \sqrt{\tau}} \rightarrow - \infty$
$d_2 \sim d_1\rightarrow - \infty$
Therefore,
$$C(S,\tau) = S e^{- D \tau} N (d_1) - K e^{- r \tau} N (d_2) \rightarrow 0$$
as $N(d_i) \rightarrow 0$ for $i \in \{ 1,2\}$.
- Case 3: $S = K$
$d_1 = \frac{\left(r - D + \dfrac{1}{2}\sigma^2\right)\sqrt{\tau}}{\sigma} \rightarrow 0$
$d_2 \rightarrow 0$
Therefore,
$$C(S,\tau) = S e^{- D \tau} N (d_1) - K e^{- r \tau} N (d_2)$$
$$C(S,\tau) = \left( S - K \right) \dfrac{1}{2} \rightarrow 0$$
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
$$(S-K)^+$$
when the time $t$ is "infinitely" close to the maturity of the call $T$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.