Why a Driftless American Digital Call Can Be Twice a European Digital
Summary
The document explains a result for American and European digital calls with the same strike and expiry, when the underlying follows Brownian motion without drift and interest rates are zero. A digital pays when its exercise condition is met. The American contract may pay as soon as the underlying reaches the strike, while the European contract pays only if it finishes at or above the strike at expiry.
If the price reaches the strike before expiry, symmetry of the subsequent driftless Brownian paths gives an equal chance of ending above or below that level at expiry. Thus, conditional on that earlier touch, the European digital’s payout probability is half the American digital’s, which can explain the stated value relationship under the exercise’s assumptions. This reasoning is conditional on a strike touch and does not establish a general pricing rule for other models. Drift, rates, payoff definitions, or different exercise conditions can change the comparison.
Key ideas
- The comparison concerns digital calls with matching strike and expiry under driftless Brownian motion and zero rates.
- An American digital can pay when the underlying reaches the strike before expiry.
- After a strike touch, path symmetry gives an even chance of finishing on either side of the strike.
- The half-probability argument depends on the stated assumptions and is not universal.
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Full text
# Reason why a European binary call should be worth half of its American counterpart when driftless and out-of-the-money # Reason why a European binary call should be worth half of its American counterpart when driftless and out-of-the-money Exercise 11 of chapter 8 of Mark Joshi's "The concepts and practice of mathematical finance", asks to compare prices of an American and a European digital (binary) calls when out-of-the-money. The options have same strike and expiry, and are cast on an asset following Brownian motion with no interest rates. Before peeking at the solution, I had argued that due to the fact we are out-of-the-money, American optionality plays less of a role and the two would be worth about the same. However, the answer to the exercise states that, given $r = 0$ and the symmetry of paths of a Brownian motion, when the American call pays off there is a 50% chance the European also will, so the latter is worth about half the former. Now, I don't understand the logic of this answer: since we have no knowledge of how bad out-of-the-money we are, I'll concede there is a 50% chance that the asset will end up being worth more than its current value, yes, and also that the value of the European digital is its current probability to end up in-the-money, but how can it be justified that the European is worth half of the American digital? ## Answer by Robert (score 3, accepted) https://quant.stackexchange.com/a/63331 Lets assume the price of the underlying equals the strike at some point prior to expiry. Then the probability of the price being still greater or equal the strike at expiry is 0.5. So the probability of the European option paying out is exactly half of the probability for the American option.
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