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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

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    $\begingroup$ Delta is the sensitivity of your price to a move on the underlying. At maturity, the product is $\max(S(T) - K, 0)$ so if $S(T) \leq K$, it's zero, otherwise it's 1... However, at maturity, why would you want to compute the delta? there's nothing to replicate anymore as your option has expired. $\endgroup$
    – byouness
    Apr 2, 2020 at 9:16
  • $\begingroup$ @byouness yes, as you say the delta at maturity doesn't change anything to the replication but I simply wanted continuity in my implementation. I had errors due to this divide by 0 problem in my python file. $\endgroup$
    – Syle
    Apr 2, 2020 at 9:37

2 Answers 2

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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).

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  • $\begingroup$ Thank you for the mathematical clarification. Did not think of taking limits! $\endgroup$
    – Syle
    Apr 2, 2020 at 8:21
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    $\begingroup$ Added a bit to your answer below. $\endgroup$
    – user34971
    Apr 2, 2020 at 13:04
  • $\begingroup$ The function $d1$ (namely $\ln(S_t/K)$) is undefined for $S_t=K$. I assume your reasoning uses the call options payoff function $\max(S_T - K; 0)$ which is not differentiable for $S_T = K$ but has deriviative wrt. $S_T$ given by $1$ if $S_T > K$ and $0$ if $S_T < K$. The midpoint is thus $1/2$ as in your limit of $\Delta$. This is obtained if and only if the argument of the standard normal distribution is 0, i.e. $d_1 = 0$ for $S_t = K$. $\endgroup$
    – Landscape
    Mar 26 at 10:43
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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.

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    $\begingroup$ Should definitely be $S_t$ and not $S_T$! Thanks! Why I included the limit for ATM options: If you keep $S_t\equiv K$ fixed and plot $d_1$ (and $d_2$), the functions will converge to $\frac{1}{2}$ as $t\to T$ but yes, the final payoff is not differentiable at $S_T=K$ in the classical sense. $\endgroup$
    – Kevin
    Apr 2, 2020 at 13:20
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    $\begingroup$ This is why everyone hates pin-risk. Way out of this: never hold an option until maturity if the stock hovers around its strike! :) $\endgroup$
    – user34971
    Apr 2, 2020 at 13:22
  • $\begingroup$ I hadn't heard this term before. Thank you, I learnt something new (: $\endgroup$
    – Kevin
    Apr 2, 2020 at 13:27

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