3
$\begingroup$

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,

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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*}

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.