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      • 1Path-independent exotics

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        • Hedging exotics: static replication, and reserving for what you cannot hedge
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  1. Curriculum
  2. /Derivatives and options
  3. /Exotics and structured products
  4. /Hedging exotics

Hedging exotics: static replication, and reserving for what you cannot hedge

EXO · Chapter 4·13 min read·Asked at Optiver, SIG, Citadel Securities, Akuna

Assumes Multi-asset exotics: baskets, best-of, spreads and quantos.

After this lesson you should be able to

  • Build a static hedge for a barrier option and say why it is preferred.
  • Explain why exotic hedging problems are usually discontinuity problems.
  • Describe how a desk reserves against model risk.

Pricing an exotic is arithmetic; hedging it is the business. A dynamic hedge on a discontinuous payoff fails in exactly the scenario the payoff was written for, so the working answer is to replicate statically with vanillas wherever possible, and to reserve capital against the part you cannot.

Proposition 4.1

Static replication

Find a portfolio of vanillas, set once and left alone, whose value matches the exotic wherever it matters. For a knock-out you choose vanillas so that their combined value is zero along the barrier — then if the barrier is hit you unwind into a worthless portfolio, and if it is not you hold something that matches the payoff at expiry.

Holds when

  • A static hedge does not need rehedging, so it does not care about transaction costs or gaps.
  • It is only exact under assumptions about the dynamics, so it degrades rather than failing outright.
  • Where an exact static hedge exists it is always preferred to a dynamic one, even if it costs more up front.

Derivation 4.2

The digital, done properly

The canonical example, and the template for everything else.

  1. Digital at K≈12ε[C(K−ε)−C(K+ε)]\text{Digital at } K \approx \frac{1}{2\varepsilon}\big[C(K-\varepsilon) - C(K+\varepsilon)\big]Digital at K≈2ε1​[C(K−ε)−C(K+ε)]

    The call spread converges to the digital as ε→0\varepsilon \to 0ε→0.

  2. Sell the digital, buy the spread with ε>0\text{Sell the digital, buy the spread with } \varepsilon > 0Sell the digital, buy the spread with ε>0

    A finite width deliberately over-hedges.

  3. cost of the over-hedge=price of the risk\text{cost of the over-hedge} = \text{price of the risk}cost of the over-hedge=price of the risk

    The spread width is a commercial decision, not a modelling one.

a bounded, known cost in place of an unbounded delta\text{a bounded, known cost in place of an unbounded delta}a bounded, known cost in place of an unbounded delta

Why dynamic hedging fails here specifically. Dynamic hedging works when the delta changes gradually, so that the rebalancing you missed between trades is small. At a barrier or a digital strike the delta changes by an enormous amount over a tiny price range, so the error between rebalances is not small — it is the whole position. Worse, the failure correlates with the event: the gap that jumps through the barrier is the same gap that made the option pay. A static hedge sidesteps all of it by never needing to be adjusted, which is why desks accept a worse price to get one.

Example 4.3

You are short a digital paying $1,000,000\$1{,}000{,}000$1,000,000 above $100\$100$100. You hedge with a call spread from $99\$99$99 to $101\$101$101. What is your maximum loss against a perfect hedge?

Show the worked solutionHide the worked solution

Worked solution

  1. Formula
    notional=payoutstrike width\text{notional} = \frac{\text{payout}}{\text{strike width}}notional=strike widthpayout​
  2. Substitute
    =1,000,0002=500,000 per point= \frac{1{,}000{,}000}{2} = 500{,}000 \text{ per point}=21,000,000​=500,000 per point
  3. Solve
    Above 101: hedge pays 1,000,000, digital pays 1,000,000\text{Above } 101: \text{ hedge pays } 1{,}000{,}000, \text{ digital pays } 1{,}000{,}000Above 101: hedge pays 1,000,000, digital pays 1,000,000
  4. At 100: hedge pays 500,000, digital pays 1,000,000\text{At } 100: \text{ hedge pays } 500{,}000, \text{ digital pays } 1{,}000{,}000At 100: hedge pays 500,000, digital pays 1,000,000
  5. shortfall=500,000\text{shortfall} = 500{,}000shortfall=500,000
  6. Answer
    worst case $500,000, at exactly the strike\text{worst case } \$500{,}000, \text{ at exactly the strike}worst case $500,000, at exactly the strike

Sanity check. Narrowing the spread to 99.599.599.5–100.5100.5100.5 halves the shortfall and roughly doubles the hedge cost. That trade-off is the entire quoting decision, and it is why a digital is quoted wide: the spread you charge has to cover the over-hedge.

Proposition 4.4

Static hedges for barriers

An up-and-out call can be replicated by the vanilla call minus a portfolio of options struck above the barrier, chosen so the combination is worth nothing when spot sits at the barrier. Under a symmetric model the construction is exact and remarkably simple — reflect the payoff about the barrier — and under a more realistic model it is approximate but still far better behaved than a dynamic hedge.

Holds when

  • The reflection construction assumes zero drift and a symmetric distribution; skew degrades it.
  • Practitioners hedge the residual dynamically but only need to, which is the point.

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On this page
  • Static replication
  • The digital, done properly
  • Worked example
  • Static hedges for barriers

QuantMax · 141 lessons · 1342 questions · c5c0caa

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