Options interview questions

For European options on the same underlying asset with the same strike $K$ and time to expiry $T$, put-call parity gives:

where $C$ is the call price, $P$ the put price, $S$ the spot price, and $r$ the risk-free rate. Gamma is the second derivative of the option price with respect to $S$:

Differentiating put-call parity twice with respect to $S$:

because $S$ is linear in $S$ (second derivative zero) and $K e^{-rT}$ is constant. Hence:

Given $\Gamma_C = 0.02$, the put gamma is also $0.02$.

The three calls have equally spaced strikes ($\Delta K = 10$) and prices $C_1=4$, $C_2=3.5$, $C_3=2.75$. A no-arbitrage condition for call options is that the price as a function of strike must be convex: $C_1 + C_3 \ge 2C_2$. Here $4 + 2.75 = 6.75 < 7 = 2 \times 3.5$, so convexity is violated, creating an arbitrage.

Construct a butterfly spread: buy one call at strike 1000, buy one call at strike 1020, and sell two calls at strike 1010. The net premium is

so the position generates an upfront credit of $0.25$.

The payoff at expiration $S_T$ is

Evaluating piecewise:

• $S_T \le 1000$: all options expire worthless, $V=0$.
• $1000 < S_T \le 1010$: $V = (S_T-1000) \ge 0$.
• $1010 < S_T \le 1020$: $V = (S_T-1000) - 2(S_T-1010) = 1020 - S_T \ge 0$.
• $S_T > 1020$: $V = (S_T-1000)+(S_T-1020)-2(S_T-1010)=0$.

The payoff is non-negative for all $S_T$ and strictly positive for $1000 < S_T < 1020$. Since the position was entered at a net credit of $0.25$, the guaranteed profit is $0.25$ dollars (25 cents).

High-level idea

In the Black–Scholes model, the price of a European call depends only on the risk‑free rate and volatility, not on the real‑world drift. Hence two assets with the same current price, volatility, strike, maturity, and risk‑free rate have identical European call prices, regardless of their real‑world drifts.

When one asset is also subject to random downward jumps, the comparison changes — but not in the direction naive intuition suggests. Under the risk‑neutral measure the jumps must be compensated by extra drift between jumps so that the forward price is unchanged. The compensated jumps therefore act as a mean‑preserving spread of the terminal price, and the call payoff is convex, so by Jensen's inequality the call on the jump‑exposed asset is more expensive than the call on the pure‑diffusion asset.

Derivation

1. Pure diffusion (no jumps)

Under the risk‑neutral measure $\mathbb{Q}$ the asset follows

so that

The European call price is

with

The formula contains $r$ and $\sigma$ but no real‑world drift $\mu$. Therefore, if two assets share the same $S_0$, $\sigma$, $K$, $T$, and $r$, their European call prices are identical — distinct real‑world drifts are irrelevant.

2. Adding random downward jumps to one asset

Now let one asset follow a jump‑diffusion (Merton 1976). Under a suitable risk‑neutral measure its dynamics are

where $N_t$ is a Poisson process with intensity $\lambda$, $Y_t < 1$ is the downward jump multiplier, and $\kappa = \mathbb{E}^{\mathbb{Q}}[Y_t - 1]$ is the compensator that keeps the expected instantaneous return equal to $r$. For downward jumps $\kappa < 0$, so the drift between jumps is raised above $r$: the compensation exactly offsets the jumps and preserves the forward, $\mathbb{E}^{\mathbb{Q}}[S_T] = S_0 e^{rT}$ for both assets.

Solving the SDE, the terminal price factorizes as

where $S_T^{\text{diff}}$ is the pure‑diffusion terminal price and $J$ is an independent jump factor with $\mathbb{E}^{\mathbb{Q}}[J] = 1$. Conditioning on $S_T^{\text{diff}}$ and applying Jensen's inequality to the convex payoff $x \mapsto (x - K)^{+}$,

Taking expectations and discounting shows the jump‑exposed call is worth at least as much for every strike — and strictly more whenever the jumps are genuinely random. Equivalently: at the same forward, the jumps add total risk‑neutral variance, and extra dispersion always benefits a convex payoff (a mean‑preserving spread raises the value of a convex function's expectation).

The tempting argument that "downward jumps create negative skew and therefore depress the call" is wrong because it ignores the compensator: between jumps the asset drifts upward faster than $r$, and through the convexity of the payoff this more than makes up for the heavier left tail.

As a numerical check, with $S_0 = K = 100$, $r = 0.02$, $\sigma = 0.2$, $T = 1$, $\lambda = 1$, and jump multiplier $Y = 0.8$, Monte Carlo gives a jump‑diffusion call price of about $12.45$ versus a Black–Scholes price of about $8.92$.

Consequently, the call on the asset with random downward jumps is more expensive than the call on the pure‑diffusion asset.

Final answer

Without jumps the two European calls have the same price. When one asset is also subject to random downward jumps (compensated under the risk‑neutral measure so the forward is unchanged), its call becomes more expensive than the call on the pure‑diffusion asset: the jumps act as a mean‑preserving spread of the terminal price, and the call payoff is convex.