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Lessons · Options

Options and the Greeks

Options questions in trading interviews are less about formulas and more about intuition: what a position pays, how its value moves, and quick estimates you can do in your head.

Payoffs first

A call with strike K pays max(S − K, 0) at expiry; a put pays max(K − S, 0). Almost every strategy is a combination of these. A straddle (call + put at the same strike) pays |S − K|, so it profits from big moves either way. A call spread (long a lower strike, short a higher one) caps both the cost and the payoff.

Put–call parity

A long call plus a short put at the same strike pays S − K in every outcome, exactly like owning the stock and owing K. Positions with identical payoffs must cost the same, so

C − P = S − K·e^(−rT)

With zero rates, C − P = S − K. If a quoted call and put break this, there's an arbitrage.

Pricing with a one-step tree

A $100 stock will be $120 or $90 next period, and rates are zero. Price a $100 call. Find the probability q that makes the stock a fair bet: 120q + 90(1 − q) = 100, so q = 1/3. The call pays $20 in the up state, so it's worth 20 × 1/3 ≈ $6.67. You can replicate it by holding ⅔ of a share, which is the option's delta: (20 − 0)/(120 − 90). This no-arbitrage idea is the heart of Black–Scholes.

The Greeks

A delta-hedged long option makes about ½ · Γ · (ΔS)² on a move and pays theta for the privilege. That's the core trade-off of options market making.

The mental-math price

For an at-the-money option with low rates:

ATM call ≈ 0.4 × S × σ × √T

The 0.4 is 1/√(2π) ≈ 0.399. A 1-year ATM call on a $100 stock with 20% volatility is about 0.4 × 100 × 0.2 = $8 (exactly $7.97). The price is roughly linear in volatility, so doubling vol roughly doubles the price.

Common traps

Practice

Next: The Kelly criterion · More options questions