Brownian Bridge™

1. An arithmetic Asian call and a geometric Asian call have the same strike $K$ , maturity $T$ , and underlying. Which is more expensive, and why?

The arithmetic Asian is more expensive: the arithmetic mean is always ≥ the geometric mean (AM-GM inequality), so E[arith. avg] ≥ E[geom. avg], and by Jensen's inequality on the call payoff, the arithmetic call price is higher.The geometric Asian is more expensive: it has a closed-form formula whereas the arithmetic requires Monte Carlo approximation, and the premium for tractability is positive.They are equal: both average the same underlying over the same monitoring dates.The arithmetic Asian is cheaper: it has higher variance than the geometric average, which inflates the denominator in the pricing formula.

2. Under Black-Scholes with constant volatility $\sigma = 25\%$ and continuous geometric averaging over $N$ fixing dates with equal spacing, what is the approximate effective volatility $\hat{\sigma}$ of the geometric average Asian as $N \to \infty$ ?

σ/√3 ≈ 14.4% — because the continuous limit of the geometric average effective variance is σ²T/3, giving effective vol σ/√3.σ/√2 ≈ 17.7% — the averaging halves the variance.σ/2 = 12.5% — the geometric average halves the volatility.σ·√(2/3) ≈ 20.4% — from the variance of the time-averaged log-price.

3. The Demeterfi-Derman-Kamal-Zou (1999) model-free variance strike formula for a variance swap is: $K_{var} = \frac{2}{T}\left[\int_0^{F_T} \frac{P(K)}{K^2}\,dK + \int_{F_T}^{\infty} \frac{C(K)}{K^2}\,dK\right]$ Under a flat implied vol surface ($\sigma_{impl}(K) = \sigma $for all$ K$), what does this formula give?

K_var = σ²: the variance strike equals the square of the constant implied vol.K_var = σ: the variance strike equals the implied vol itself.K_var = σ²/2: the integral weights halve the variance contribution.K_var = 2σ²: the two integrals (puts and calls) each contribute σ², giving twice the variance.

4. Why is the fair strike of a volatility swap ( $K_{vol}$ , paying realised vol minus $K_{vol}$ ) always strictly less than $\sqrt{K_{var}}$ (the square root of the variance swap strike)?

Jensen's inequality: for the concave function x ↦ √x, E[√V] < √(E[V]), so K_vol = E[σ_realised] < √(E[σ²_realised]) = √K_var.The volatility swap has additional transaction costs that reduce its fair strike below the square root of K_var.The log-contract replication of a variance swap introduces a convexity bias that inflates K_var above the true expected variance.Jensen's inequality runs the other way: since x ↦ √x is convex, E[√V] > √(E[V]), so the vol swap strike is above √K_var.

5. A cliquet option pays the sum of capped annual returns: $\sum_{i=1}^{5}\min(\max(S_{t_i}/S_{t_{i-1}}-1,\,0\%),\,5\%)$ . A quant prices it using a local volatility model calibrated to today's vanilla surface. A structurer argues the price is wrong. Who is correct and why?

The structurer is correct: cliquet pricing depends on the forward smile (the implied vol surface for forward-starting options), which local vol generates too flat — the cliquet is systematically underpriced under local vol.The quant is correct: calibrating to today's surface ensures all market information is incorporated, so the cliquet price is model-consistent.The structurer is correct, but only for the first reset period — subsequent periods are correctly priced by local vol.Both are correct: the difference is only in the hedging strategy, not in the fair price.

6. A forward-starting ATM call resets its strike to $S_{t_1}$ at time $t_1 = 0.5$ years and expires at $T = 1.5$ years. Under Black-Scholes with constant $\sigma = 20\%$ , $r = 5\%$ , $q = 2\%$ , $S_0 = 100$ : what is the price of this option at $t = 0$ ?

S₀·e^{−q·t₁}·c_BS(1, 1, r, q, σ, T−t₁) = 100·e^{−0.01}·c_BS(1, 1, 0.05, 0.02, 0.20, 1.0), which equals approximately $9.81.c_BS(S₀, S₀, r, q, σ, T) = c_BS(100, 100, 0.05, 0.02, 0.20, 1.5), the vanilla ATM call expiring at T.e^{−r·t₁}·c_BS(S₀, S₀, r, q, σ, T−t₁): a discounted vanilla call with spot S₀ and maturity T−t₁.S₀·c_BS(1, 1, r, q, σ, T−t₁): an undiscounted forward-starting call without the dividend adjustment.

7. A variance swap desk observes that the model-free variance strike $K_{var}$ from the DDKZ formula is $0.0529$ (implied vol equivalent $\sqrt{K_{var}} = 23\%$ ), but the ATM implied vol is only $20\%$ . What explains the gap?

The implied vol skew: OTM puts (below forward) have higher implied vol, which inflates the variance strike via the 1/K² weighting in the DDKZ integral that overweights OTM options.The variance swap includes a jump component that the DDKZ formula adds on top of the diffusion variance.The forward price F is above S₀, shifting the integration bounds and mechanically increasing K_var.The bid-offer spread on OTM options is wide, causing the numerical integral to overestimate K_var.

8. A desk wants to hedge a short position in a digital call struck at $K = 105$ (paying $\$ 1 $if$ S_T \geq 105$). Which hedging strategy is standard practice, and what is the key trade-off?

Hedge with a tight call spread: long call at K−ε, short call at K+ε. Narrower ε = better hedge but more gamma risk near expiry; wider ε = less gamma but imperfect replication.Hedge with a vanilla call struck at K: delta-hedge the vanilla call to replicate the digital payoff dynamically.Hedge with a variance swap: the digital's gamma exposure is equivalent to short variance.Hedge with a down-and-in put: the digital's discontinuous payoff is replicated by the knock-in mechanism.

Quiz: Exotic Equity Payoffs

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