I have been asked to prove mathematically that a binary option close to maturity should be hedged using a call spread with the same maturity.

I understand that far from maturity, one would use delta hedging to sell or purchase the underlying asset. Yet as time to expiry tends to zero the delta profile tends towards a dirac delta function and so renders the hedge impractical. See: delta of a binary option

Other than calculus to derive delta, are there any other rigorous ways to construct hedges of this kind?

As this is a homework question, hints rather than full answers are most welcome.

Thanks in advance,

  • $\begingroup$ I think the intuition here is that a call spread has similar payoff to a binary option if the two strikes are close enough. But in general coming up with such approximations would seem to be an open ended process requiring some cleverness, not a rigorous algorithm afaik. You have to ask 'what do I already know that is similar to this'. $\endgroup$ – noob2 Mar 1 '17 at 15:40
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    $\begingroup$ Well, if you write the value of an non-path dependent option as being equal to the (discounted) expected value of the payoff at maturity using a risk neutral density and combine this with the Breeden-Litzernberger formula, then you can write it's value as an (infinite) combination of calls and puts. So if you pick a distance between call strikes, discretizing can't be too hard? $\endgroup$ – Bram Mar 1 '17 at 15:53
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    $\begingroup$ Also, for hints, google for static replication $\endgroup$ – Bram Mar 1 '17 at 15:53
  • $\begingroup$ Brief overiew of Static Replication here google.com/… $\endgroup$ – noob2 Mar 1 '17 at 16:58

The key point here is that when close to maturity a binary option should be hedged with a call spread.

Note that, for a binary option with a payoff at maturity $T$ of the form $\mathbb{1}_{S_T>K}$, the value at time $0\le t < T$ is given by $$e^{-r(T-t)}N(d_2), $$ where $$d_2 = \frac{\ln \frac{S_t}{K}+(r-\frac{1}{2}\sigma^2)(T-t)}{\sigma\sqrt{T-t}}.$$ Since \begin{align*} \frac{\partial N(d_2)}{\partial S_t} = \frac{\phi(d_2)}{\sigma S_t \sqrt{T-t}}\rightarrow \infty, \end{align*} as $t\rightarrow T$, the delta based hedging is not applicable.

However, with a call spread approximation of the form \begin{align*} \mathbb{1}_{S_T>K} \approx \frac{1}{\varepsilon}\Big[\big(S_T-(K-\varepsilon)\big)^+ - \big(S_T-K\big)^+\Big]. \end{align*} The delta at time $0\le t < T$ is given by $$\frac{1}{\varepsilon}\big[N(d_1^{-\varepsilon})-N(d_1^0)\big], $$ which is finite as $t\rightarrow T$. Here, \begin{align*} d_1^{\alpha} = \frac{\ln \frac{S_t}{K+\alpha}+(r+\frac{1}{2}\sigma^2)(T-t)}{\sigma\sqrt{T-t}}. \end{align*}

  • $\begingroup$ Do you know of a reference that derives the call spread delta from the pricing equation rather than by deducing it from the gradient of the payoff profile? $\endgroup$ – Nick Wilton Mar 3 '17 at 11:28
  • $\begingroup$ @NickWilton: I do not have a reference. The delta for a call spread is derived from the pricing formula rather than the payoff profile, by noting that the delta of a vanilla call is of the form $N(d_1)$. $\endgroup$ – Gordon Mar 3 '17 at 14:06

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