The Greeks

Question 1 (12 marks)

(a) Net delta $= 200 \times 0.45 = +90$ delta. Directional exposure = long, equivalent to being long 90 shares. (1 mark computation, 1 mark long/short, 1 mark shares-equivalent) (3)

(b) To be delta-neutral, offset $+90$ delta. Each share has delta $+1$, so short 90 shares. (1 sign, 1 quantity, 1 reasoning) (3)

(c) Gamma is the rate of change of delta w.r.t. $S$: $\Delta S = +3$. New per-option delta $\approx 0.45 + \Gamma \cdot \Delta S = 0.45 + 0.03 \times 3 = 0.45 + 0.09 = 0.54$. (2) New option-leg net delta $= 200 \times 0.54 = 108$. (1) Hedge from (b) is still $-90$ (stock delta constant). Net portfolio delta $= 108 - 90 = +18$. (2) Hedge is imperfect because the option delta changed (positive gamma) while the static stock hedge did not — the portfolio has drifted long as the stock rose. Re-hedging (dynamic delta hedging) is required. (1) (6)

Question 2 (12 marks)

(a) European call: $C = S N(d_1) - K e^{-rT} N(d_2)$ where $d_1 = \frac{\ln(S/K) + (r + \tfrac{1}{2}\sigma^2)T}{\sigma\sqrt{T}}, \qquad d_2 = d_1 - \sigma\sqrt{T}.$ Symbols: $S$ spot, $K$ strike, $r$ risk-free rate, $T$ time to expiry, $\sigma$ volatility, $N$ standard normal CDF. (2 formula, 2 symbols/d1,d2) (4)

(b) $N(d_2)$ = risk-neutral probability the option expires in-the-money (finishes $S_T > K$). $N(d_1)$ = the option's delta and relates to the expected discounted value of receiving the stock conditional on exercise. $N(d_1)$ is the call delta because $\partial C/\partial S = N(d_1)$ — the terms involving the derivative of $N(d_1)$ and $N(d_2)$ cancel exactly (a known Black–Scholes identity), leaving $N(d_1)$. (2 for probabilities, 2 for delta reasoning) (4)

(c) With $r = 0$: $C = S N(d_1) - K N(d_2)$. As $\sigma \to 0$: $d_1, d_2 \to \dfrac{\ln(S/K)}{\sigma\sqrt T} \to +\infty$ when $S > K$ (numerator positive, denominator $\to 0^+$). Hence $N(d_1) \to 1$ and $N(d_2) \to 1$. So $C \to S \cdot 1 - K \cdot 1 = S - K$ = intrinsic value. (1 setup, 1 limit of d's, 1 N→1, 1 conclusion) (4)

Question 3 (10 marks)

(a) Theta = rate of change of option value w.r.t. passage of time (time decay), $\Theta = \partial V/\partial t$. For a long option it is typically negative — value erodes as expiry approaches. (2)

(b) A long option has positive gamma: its delta improves in your favour as the stock moves (you gain on large moves either way). But there is no free lunch — you "pay rent" via negative theta. The Black–Scholes PDE $\Theta + \tfrac{1}{2}\sigma^2 S^2 \Gamma + rS\Delta - rV = 0$ links them: for a delta-neutral position the theta term and the gamma term roughly offset, so positive gamma is bought at the cost of negative theta. Realised volatility above (below) implied makes the gamma gains exceed (fall short of) the theta cost. (4)

(c) Break-even: gamma P&L = |theta|: $\tfrac{1}{2}\Gamma (\Delta S)^2 = |\Theta|$ $\tfrac{1}{2}(0.04)(\Delta S)^2 = 0.05$ $(\Delta S)^2 = \frac{0.05}{0.02} = 2.5 \Rightarrow |\Delta S| = \sqrt{2.5} \approx 1.58.$ So roughly a ±$1.58 daily move (~1.6%) is needed to offset decay. (1 equation, 2 algebra, 1 answer) (4)

Question 4 (12 marks)

(a) Vega = sensitivity of option value to a 1-point (1%) change in volatility, $\nu = \partial V/\partial\sigma$. Implied volatility = the volatility figure that, plugged into Black–Scholes, reproduces the market price (forward-looking, market's expectation). Historical (realised) volatility = the actual observed standard deviation of past returns (backward-looking). (4)

(b) Before earnings, uncertainty is high, so demand for options bids IV up (inflated premiums). After the announcement, the uncertainty resolves and IV collapses ("crush"). A long-option holder can lose because the fall in IV reduces the option's value via vega — even if the stock moved favourably, the vega loss from the IV drop can exceed the intrinsic gain from the move, unless the move is large enough. (4)

(c) IV change $= 30\% - 80\% = -50$ vol-points. Vega P&L = \nu \times \Delta\sigma = 0.06 \times (-50) = -\3.00. *(2)* So IV crush alone destroys ~\3.00 of the $8.00 premium. Combined with the $8 cost, the trade needs a large stock move (well beyond the $8 premium net of decay) to break even — this is why post-earnings straddles often lose despite big moves. (2) (4)

Question 5 (8 marks)

(a) Equity-index skew: lower strikes (OTM puts) carry higher IV than higher strikes (OTM calls) — a downward-sloping curve. Reason: demand for downside crash protection and the fact that markets fall faster/harder than they rise (leverage/fear effect). (3)

(b) A "smile" is a U-shaped IV curve — IV high for both deep OTM puts and calls, lowest ATM. Commonly seen in FX (currency) options (and pre-crash equity). (2)

(c) Rho = sensitivity of option value to interest-rate changes, $\rho = \partial V/\partial r$. Long call Rho is positive. Rising rates increase a call's value because the present value of the strike paid at expiry ($Ke^{-rT}$) falls, effectively making the deferred purchase cheaper. (3)

Question 6 (6 marks)

def hedge_shares(net_delta):
    # shares to trade = -net_delta (each share has delta +1)
    return -net_delta   # negative => short, positive => buy

(3 marks for correct sign/logic)

Traders neutralise delta first (using cheap, liquid underlying shares/futures — it's the largest, most immediate directional risk), then gamma (using options, since only options carry gamma) to stabilise delta against moves, then vega (also via options). Delta comes first because it is the dominant first-order exposure and can be adjusted instantly and cheaply, whereas gamma/vega hedges require options and are set to shape how delta evolves. (3 marks) (6)

[
  {"claim":"Q1c new net delta = +18", "code":"new_opt_delta=0.45+0.03*3; net=200*new_opt_delta-90; result=(abs(net-18)<1e-9)"},
  {"claim":"Q3c breakeven move approx 1.58", "code":"import sympy as sp; dS=sp.sqrt(sp.Rational(5,100)/(sp.Rational(1,2)*sp.Rational(4,100))); result=(abs(float(dS)-1.5811388300841898)<1e-9)"},
  {"claim":"Q4c IV crush PnL = -3.00", "code":"pnl=0.06*(30-80); result=(abs(pnl-(-3.0))<1e-9)"},
  {"claim":"Q1a net delta = 90 and hedge = short 90", "code":"nd=200*0.45; hedge=-nd; result=(nd==90.0 and hedge==-90.0)"}
]