In Delta of binary option, I do not see how to prove that the limit of $\partial C_t/\partial S_t$ is equal to $+\infty$ as $t \rightarrow T$. Can someone help ?

  • 3
    $\begingroup$ surely this is only true at the money? away from the money, the delta will be zero $\endgroup$
    – Mark Joshi
    Jan 29 '17 at 22:51

The value of a bond binary call in the Black-Scholes model is given by

\begin{equation} B_t = e^{-r (T - t)} \mathcal{N} \left( d_- \right), \end{equation}


\begin{equation} d_- = \frac{1}{\sigma \sqrt{T - t}} \left( \ln \left( \frac{S_t}{K} \right) + \left( r - \frac{1}{2} \sigma^2 \right) (T - t) \right). \nonumber \end{equation}

The delta is

\begin{equation} \frac{\partial B_t}{\partial S_t} = e^{-r (T - t)} \mathcal{N}' \left( d_- \right) \frac{1}{S_t \sigma \sqrt{T - t}}. \end{equation}

We now want to take the limit as $t \rightarrow T$. First note that

\begin{equation} \lim_{t \rightarrow T} d_- = \begin{cases} -\infty & \text{if } S_t < K\\ 0 & \text{if } S_t = K\\ +\infty & \text{if } S_t > K \end{cases}. \end{equation}


\begin{equation} \lim_{t \rightarrow T} \mathcal{N}' \left( d_- \right) = \begin{cases} 0 & \text{if } S_t \neq K\\ 1 / \sqrt{2 \pi} & \text{if } S_t = K \end{cases} \end{equation}


\begin{equation} \lim_{t \rightarrow T} \frac{\partial B_t}{\partial S_t} = \begin{cases} 0 & \text{if } S_t \neq K\\ +\infty & \text{if } S_t = K \end{cases}. \end{equation}

In the last step we used that the exponential in $\mathcal{N}' \left( d_- \right)$ approaches zero faster than the $1 / \sqrt{T - t}$ approaches plus infinity in the limit when $S_t \neq K$.

  • $\begingroup$ Ok, I did not know that "at-the-money" means $S_t=K$. $\endgroup$
    – Georges
    Jan 29 '17 at 23:52

Alternatively to LocalVolatility's already very nice answer, here's an approach to see that this result does not only hold under the Black-Scholes dynamics.

The $t$-value of a binary call expiring at $T$ can be written as $$ C_t = \Bbb{E}_t^\Bbb{Q} \left[ e^{-r(T-t)} {\bf{1}}\{S_T \geq K \} \right] $$

Its "delta" is defined as $$\Delta = \frac{\partial C_t}{\partial S_t}$$ Under some light conditions (discussed in e.g. Monte Carlo Methods in Financial Engineering, Glasserman, 2004), you can permute the expectation and differential operators to write: \begin{align} \Delta_t &= \frac{\partial}{\partial S_t} \Bbb{E}_t^\Bbb{Q} \left[ e^{-r(T-t)} {\bf{1}}\{S_T \geq K \} \right] \\ &= \Bbb{E}_t^\Bbb{Q} \left[ e^{-r(T-t)} \frac{\partial}{\partial S_t}{\bf{1}}\{S_T \geq K \} \right] \\ &= \Bbb{E}_t^\Bbb{Q} \left[ e^{-r(T-t)} \delta(S_T-K) \frac{\partial S_T}{\partial S_t} \right] \tag{1} \end{align} where we've used the chain rule ($S_T$ functionally depends on $S_t$) and the fact that the derivative of the Heaviside function ${\bf{1}}(x \geq a)$ is a Dirac impulse at $a$, i.e. $\delta(x-a)$.

It should then clear that: $$ \lim_{t \to T} \Delta_t = \delta(S_t-K) $$ hence the result.


You're short a digital call struck at 100. Your payoff : -\$1 above 100, \$0 below.

1 second before expiry, spot is 99.9999. If its stays there you owe nothing, if it goes a touch higher you owe $1 to the option's buyer.

You need to replicate this payoff via delta hedging : how much of the underlying do you need to hold to generate a \$1 gain, and offset your \$1 loss, when the spot moves from 99.9999 to 100.0?

Answer : a lot of it.


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