Black–Scholes Delta at Expiration and the At-the-Money Limit
Summary
The document explains how to handle the Black–Scholes delta in a European option replication strategy as time approaches expiration. Although the usual formula contains a term that becomes zero at maturity, its limiting behavior gives delta of one for an in-the-money call and zero for an out-of-the-money call. At the strike, the limit is one-half under the stated model, including the dividend-yield adjustment before the limit is taken.
At expiration itself, the call payoff has a kink at the strike, so its derivative is undefined there; away from the strike, delta is one or zero according to whether the option finishes in or out of the money. The response notes that an exact finish at the strike has probability zero under the model, so this undefined point does not ordinarily affect replication. These conclusions rely on the Black–Scholes assumptions and concern the limiting hedge behavior at maturity, rather than a regular formula evaluated by dividing by zero.
Key ideas
- Take the limit of the Black–Scholes delta as time to maturity approaches zero.
- For a call, limiting delta is one above the strike and zero below it.
- At the strike, the pre-expiration limit is one-half in the stated model.
- At expiration, delta is undefined exactly at the strike because the payoff is not differentiable there.
- Under the model, finishing exactly at the strike has probability zero.
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Full text
# Black-Scholes Delta value at maturity?
# Black-Scholes Delta value at maturity?
Having to implement a replication strategy for European options, I encounter the following problem:
- Delta tells me how many shares to hold at time t in my replication strategy. To do so, I simply iterating through times t up to T, and there comes the last portfolio rebalancing. At t=T, in the delta formula, $\Delta=\Phi(d_1)$, $d_1$ has as denominator $\sigma*\sqrt{T-t}$.
Given it is the denominator, I cannot divide by 0, so I dont know the $\Delta$ value at maturity. So my question is, what values does delta take at maturity given the divide by 0 constraint?
Thank you
## Answer by Kevin (score 3, accepted)
https://quant.stackexchange.com/a/52987
You simply take limits. Recall that in the Black-Scholes world $$d_1=\frac{\ln\left(\frac{S_t}{K}\right)+\left(r-q+\frac{1}{2}\sigma^2\right)(T-t)}{\sigma\sqrt{T-t}}.$$
As $t\to T $, we have $d_1\to\begin{cases} \infty & \text{if } S_t> K \\ 0 & \text{if } S_t=K \\-\infty & \text{if } S_t<K \end{cases}$.
Thus, $\Delta=\Phi(d_1)e^{-q(T-t)} \to \begin{cases} 1 & \text{if } S_t> K \\ \frac{1}{2} & \text{if } S_t= K \\ 0 & \text{if } S_t<K \end{cases}$.
Financially, this means if you're in the money at maturity, your replicating strategy is to be long the stock and if the stock is out of the money, you don't need to hold the stock (the option expires worthless).
## Answer by user34971 (score 1)
https://quant.stackexchange.com/a/52996
Just to add to @KeSchn's answer:
At $T$ the option has price $(S_T - K)_+$. This is non-differentiable at $S_T = K$. Hence the delta is 1 when $S_T > K$, 0 when $S_T <K$ and not defined when $S_T = K$. That is not a problem since the probability that $S_T = K$ is 0 almost surely.
The limit behaviour, i.e. when $t \rightarrow T$, is as in KeSchn's answer.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.