Options Strategies

Question 1

(a) Net credit = $2.10 - 1.00 + 2.30 - 1.20 = \mathbf{2.20}$. (2)

Payoff (intrinsic of legs + credit). Long put(85) − short put(90) − short call(110) + long call(115), plus credit $2.20$:

(Below 85 both puts pay, net loss 5−credit; between 85–90 only short put in-money; flat middle keeps full credit; symmetric on call side.) (4) — 1 mark per correct region logic.

(b) Max profit $=+2.20$ (in $[90,110]$). Max loss $=-2.80$. (2)

Break-evens where $P=0$:

• Lower: $S_T - 87.80 = 0 \Rightarrow S_T = 87.80$.
• Upper: $112.20 - S_T = 0 \Rightarrow S_T = 112.20$. (2)

Wing width $= 90-85 = 5$. Identity: max profit + max loss magnitude $= 2.20 + 2.80 = 5 =$ wing width. (1) Interpretation: risk capital on a defined-risk vertical equals width − credit; the credit collected can never exceed the width, else free money (arbitrage). Hence $\text{credit} + \text{max loss} = \text{width}$ is the risk-conservation identity. (1)

(c) With $S_T=100e^X$, $X\sim\mathcal N(0,\sigma^2)$, density $f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-x^2/2\sigma^2}$:

with $P(\cdot)$ the piecewise function from (a). (3)

Qualitative: higher entry IV inflates all option premiums; a net seller (condor) collects a larger credit → wider break-evens, higher probability of the flat middle, more favourable expected P&L. So iron condor is a high-IV, neutral-view strategy; you want IV to be rich (and ideally to fall — short vega). This matches 5.4.13: neutral view + elevated/mean-reverting IV ⇒ sell premium (condor). (2)

(d) Legs regrouped:

• Short strangle = short put(90) + short call(110).
• Long strangle (wider) = long put(85) + long call(115). Sum = exactly the four condor legs ⇒ condor = short inner strangle + long outer strangle. (2)

Alternatively group by side:

• Bull put spread = short put(90) + long put(85) (credit $1.10$).
• Bear call spread = short call(110) + long call(115) (credit $1.10$). Sum of two credit verticals = condor, total credit $2.20$. (2)

Payoff identity: $P_{\text{condor}}(S_T) = P_{\text{bull put}}(S_T) + P_{\text{bear call}}(S_T)$ region-by-region (each vertical flat outside its own strike pair, so they add linearly). (1)

Question 2

(a) Let stock P&L $=S_T-50$ (per share).

Protective put payoff per share:

• $S_T\le48$: $(S_T-50)+(48-S_T)-1.50 = -3.50$ (max loss).
• $S_T>48$: $S_T-51.50$ (unlimited upside). (3)

Collar payoff per share:

Net premium paid $=0.60$.

• $S_T\le48$: $-3.50+0.90=-2.60$ (max loss).
• $48<S_T<54$: $S_T-50-0.60=S_T-50.60$.
• $S_T\ge54$: capped: $(S_T-50)-(S_T-54)-0.60 = 3.40$ (max gain). (3)

(b) Upside cap: for $S_T\ge54$ the short call offsets each further $1 of stock gain, so payoff frozen at$3.40$→ capped at$K_c=54$. (2)

Zero-cost collar: net premium $= P_{\text{put}} - C_{\text{call}} = 0 \Rightarrow P_{\text{put}}=C_{\text{call}}$. (2)

Put–call parity ($r=0$): $C(K)-P(K)=S_0-K$. For chosen $K_p$ (put) and $K_c$ (call): $P(K_p)=C(K_c)$. Using parity to express each around ATM, a symmetric zero-cost collar picks strikes equidistant such that the call and put premiums match; formally $C(K_c)=C(K_c)$ and $P(K_p)=C(K_c)$ gives, with $r=0$, the strike-linkage where the two OTM legs have equal time value → for a symmetric IV smile, $S_0-K_p = K_c - S_0$ (equidistant strikes). (2)

(c) Python:

def collar_pnl(S_T, S0, Kp, Kc, put_prem, call_prem, shares):
    stock  = (S_T - S0)
    put    = max(Kp - S_T, 0)
    call   = -max(S_T - Kc, 0)      # short call
    net_prem = -(put_prem - call_prem)  # pay put, receive call
    per_share = stock + put + call + net_prem
    return per_share * shares
 
for S in [40, 48, 50, 54, 60]:
    print(S, collar_pnl(S, 50, 48, 54, 1.50, 0.90, 100))

Table (per-share × 100): (marks: 4 code, 4 table)

Question 3

(a) $P(S_T)=\max(S_T-200,0)+\max(200-S_T,0) - 12.50 = |S_T-200| - 12.50.$ (2) Break-evens: $|S_T-200|=12.50 \Rightarrow S_T = 187.50$ or $212.50$. Loses money for $187.50 < S_T < 212.50$. (2)

(b) Profit requires $|S_T-200| > 12.50$; minimum absolute break-even move $= 12.50 =$ total premium. (2) Percentage move $=12.50/200 = 6.25\%$. (2) Long-vol/long-gamma: payoff convex in $S_T$ (kink at strike) ⇒ positive gamma; value rises with realised movement and with IV (long vega). Time decay erodes both premiums ⇒ short theta. Prefer straddle over strangle when expecting a large move and wanting maximum gamma at the money (strangle is cheaper, wider break-evens, chosen when a very large move is expected or to reduce cost). Ties to 5.4.13: big-move view + low/rising IV ⇒ buy volatility. (2)

(c) $\mathbb E[\text{gross}]=\mathbb E|Z| = \tau\sqrt{2/\pi}$. (3) Set equal to premium: $\tau\sqrt{2/\pi}=12.50$. (2)