The Greeks

Question 1

(a) Identity $S\phi(d_1)=Ke^{-r\tau}\phi(d_2)$ (5 marks)

$\phi(d_1)/\phi(d_2)=\exp\!\big(-\tfrac12(d_1^2-d_2^2)\big)$. (1) $d_1^2-d_2^2=(d_1-d_2)(d_1+d_2)=\sigma\sqrt\tau\,(d_1+d_2)$. (1) Since $d_1+d_2 = 2d_1-\sigma\sqrt\tau$ and $d_1\sigma\sqrt\tau = \ln(S/K)+(r+\tfrac12\sigma^2)\tau$: $\tfrac12(d_1^2-d_2^2)=\tfrac12\sigma\sqrt\tau(2d_1-\sigma\sqrt\tau)=d_1\sigma\sqrt\tau-\tfrac12\sigma^2\tau=\ln(S/K)+r\tau$. (2) Thus $\phi(d_1)/\phi(d_2)=e^{-\ln(S/K)-r\tau}=\tfrac{K}{S}e^{-r\tau}$, giving $S\phi(d_1)=Ke^{-r\tau}\phi(d_2)$. (1)

(b) Delta, Gamma, Vega (9 marks)

$\Delta$: $\frac{\partial C}{\partial S}=N(d_1)+S\phi(d_1)\frac{\partial d_1}{\partial S}-Ke^{-r\tau}\phi(d_2)\frac{\partial d_2}{\partial S}$. Since $\partial d_1/\partial S=\partial d_2/\partial S=\frac{1}{S\sigma\sqrt\tau}$ and by (a) $S\phi(d_1)=Ke^{-r\tau}\phi(d_2)$, the last two terms cancel: $\boxed{\Delta=N(d_1)}$. (3)

$\Gamma$: $\frac{\partial \Delta}{\partial S}=\phi(d_1)\frac{\partial d_1}{\partial S}=\boxed{\dfrac{\phi(d_1)}{S\sigma\sqrt\tau}}$. (3)

$\mathcal V$: $\frac{\partial C}{\partial\sigma}=S\phi(d_1)\frac{\partial d_1}{\partial\sigma}-Ke^{-r\tau}\phi(d_2)\frac{\partial d_2}{\partial\sigma}=S\phi(d_1)\big(\tfrac{\partial d_1}{\partial\sigma}-\tfrac{\partial d_2}{\partial\sigma}\big)$ (using a). Since $d_1-d_2=\sigma\sqrt\tau$, its $\sigma$-derivative is $\sqrt\tau$: $\boxed{\mathcal V=S\phi(d_1)\sqrt\tau}$. (3)

(c) $\mathcal V=S^2\sigma\tau\Gamma$ (6 marks)

$S^2\sigma\tau\Gamma = S^2\sigma\tau\cdot\frac{\phi(d_1)}{S\sigma\sqrt\tau}=S\phi(d_1)\sqrt\tau=\mathcal V$. ✓ (3) Interpretation: both scale with $\phi(d_1)$, maximal when $d_1\approx0$ (near ATM). Gamma carries an extra $1/\sqrt\tau$ (blows up ATM as $\tau\to0$), while Vega carries $\sqrt\tau$ (vanishes at expiry, grows with maturity). Hence Gamma is sharpest for short-dated ATM; Vega's peak in strike drifts because the $d_1=0$ strike $K=S e^{(r+\sigma^2/2)\tau}$ moves up with $\tau$. (3)

(d) PDE check + intuition (4 marks)

$\Theta = -\frac{S\phi(d_1)\sigma}{2\sqrt\tau}-rKe^{-r\tau}N(d_2)$. Substitute with $\Delta=N(d_1)$, $\Gamma=\phi(d_1)/(S\sigma\sqrt\tau)$: $\tfrac12\sigma^2S^2\Gamma=\tfrac12 S\sigma\phi(d_1)/\sqrt\tau\cdot\sigma\cdot... = \frac{S\phi(d_1)\sigma}{2\sqrt\tau}$, which cancels the first theta term; $rS\Delta=rSN(d_1)$; sum $=rSN(d_1)-rKe^{-r\tau}N(d_2)=rC$. ✓ (2) Intuition: rearranged, $\Theta=rC-rS\Delta-\tfrac12\sigma^2S^2\Gamma$. Long gamma ($\Gamma>0$) forces $\Theta<0$: the holder bleeds time value (pays theta) as the "rent" for the convexity/gamma that profits from movement. (2)

Question 2

(a) Implied vol (5 marks) Straddle $=0.8 S_0\sigma\sqrt\tau=8.00$. (2) $\sigma=\frac{8}{0.8\cdot100\cdot\sqrt{1/52}}=\frac{8}{80/\sqrt{52}}=\frac{\sqrt{52}}{10}=0.7211\approx72.1\%$. (3)

(b) Event move & break-even (5 marks) One-day priced move (expected absolute move) $\approx S_0\sigma\sqrt{\tau_{day}}$ with $\tau_{day}=1/252$: 100\cdot0.7211\cdot\sqrt{1/252}=100\cdot0.7211\cdot0.06300=\4.54\approx4.5%$. **(3)** Equivalently the straddle break-even is$\pm$8$move by expiry ($\pm8%$). Buyer profits only if realised move exceeds premium:$|S_T-100|>8$. (2)

(c) IV crush outcome (5 marks) New straddle \approx0.8\cdot104\cdot0.35\cdot\sqrt{1/52}=0.8\cdot104\cdot0.35\cdot0.13868=\4.04$. **(2)** Buyer bought at 8.00, now worth ≈4.04 plus intrinsic: the$4%$($=$4$) gap gives one leg$\approx$4$intrinsic. Rough total value ≈ intrinsic + remaining extrinsic. Using ATM re-strike estimate ≈$$4.04$time value, but the realised$$4$move <$$8$paid ⇒ **buyer lost money**. **(2)** The stock moved only 4$8) was priced; the IV crush from 72% to 35% destroyed extrinsic value. Realised move failed to beat the "vol premium." (1)

(d) Skew (3 marks) Equity index/name puts trade richer (higher IV) — crash-o-phobia / leverage effect: down moves raise volatility, so OTM puts carry a skew premium. (2) A trader expecting post-event skew flattening could sell the expensive OTM put IV (or put spread) and buy relatively cheaper call-wing IV, netting the skew normalisation. (1)

Question 3

(a) Net Greeks (6 marks) Short 100 calls ×100 shares = short 10,000 option-shares.

• Delta: $-10000(0.55)+3000 = -5500+3000=\boxed{-2500}$ (2)
• Gamma: $-10000(0.04)+0 = \boxed{-400}$ (2)
• Vega: $-10000(0.12)= \boxed{-1200}$ ⇒ short volatility (negative vega). (2)

(b) Delta+Gamma neutral system (7 marks) Underlying has $\Gamma=0$. Let $x$=option-shares of hedge instrument, $y$=underlying shares. Gamma: $-400 + 0.05x = 0 \Rightarrow x = 8000$. (3) Delta: $-2500 + 0.40x + y = 0 \Rightarrow -2500 + 0.40(8000)+y=0 \Rightarrow -2500+3200+y=0 \Rightarrow y=-700$. (3) So buy 8000 option-shares (=80 contracts) and short 700 underlying shares. (1)

(c) Pseudocode (5 marks)

import numpy as np
def hedge(book_greeks, instrument_greeks):
    # book_greeks = [delta, gamma]; instrument matrix columns = [opt, underlying]
    A = np.array([[instrument_greeks['opt_delta'], 1.0],
                  [instrument_greeks['opt_gamma'], 0.0]])
    b = -np.array([book_greeks['delta'], book_greeks['gamma']])
    x = np.linalg.solve(A, b)   # [opt_shares, underlying_shares]
    return x

(3) If $A$ is singular (det = 0) — e.g. the option has zero gamma, or hedging instruments are linearly dependent — np.linalg.solve raises LinAlgError; no simultaneous neutrality exists and you need an independent second instrument (least-squares / different maturity or strike). (2)

[
  {"claim":"Vega = S^2 sigma tau Gamma identity holds symbolically","code":"S,sig,tau=symbols('S sigma tau',positive=True); d1=symbols('d1'); phi=exp(-d1**2/2)/sqrt(2*pi); Gamma=phi/(S*sig*sqrt(tau)); Vega=S*phi*sqrt(tau); result=simplify(Vega-S**2*sig*tau*Gamma)==0"},
  {"claim":"Implied vol from straddle = sqrt(52)/10 approx 0.7211","code":"sig=8/(0.8*100*sqrt(Rational(1,52))); result=abs(float(sig)-0.72111)<1e-3"},
  {"claim":"Gamma-neutral option shares x=8000 and delta hedge y=-700","code":"x=400/0.05; y=-(-2500+0.40*x); result=(x==8000) and (y==-700)"},
  {"claim":"Net book Greeks: delta=-2500, gamma=-400, vega=-1200","code":"d=-10000*0.55+3000; g=-10000*0.04; v=-10000*0.12; result=(d==-2500) and (g==-400) and (v==-1200)"},
  {"claim":"Post-event straddle estimate approx 4.04","code":"val=0.8*104*0.35*sqrt(Rational(1,52)); result=abs(float(val)-4.04)<0.05"}
]