Options Basics

Question 1

(a) Payoffs & breakevens (6)

Call buyer profit (premium $C=250$): $\Pi_{\text{call}}(S_T) = \max(S_T - K, 0) - C = \begin{cases} -C, & S_T \le K \\ S_T - K - C, & S_T > K \end{cases}$ (1) payoff form, (1) premium subtracted.

Put buyer profit (premium $P=180$): $\Pi_{\text{put}}(S_T) = \max(K - S_T, 0) - P = \begin{cases} K - S_T - P, & S_T < K \\ -P, & S_T \ge K \end{cases}$ (1).

Breakevens (set profit $=0$ in the ITM region):

• Call: $S_T - K - C = 0 \Rightarrow S_T = 20000 + 250 = \mathbf{20{,}250}$. (1)
• Put: $K - S_T - P = 0 \Rightarrow S_T = 20000 - 180 = \mathbf{19{,}820}$. (2)

(b) Long straddle (6)

Total cost $= C + P = 250 + 180 = 430$. (1) $\Pi_{\text{str}}(S_T) = \max(S_T-K,0) + \max(K-S_T,0) - 430 = |S_T - K| - 430.$ (2) Breakevens: $|S_T - 20000| = 430 \Rightarrow S_T = 20{,}430$ (upper) and $S_T = 19{,}570$ (lower). (2)

Convexity proof: $\Pi = |S_T-20000| - 430$. The absolute-value function is convex (composition of affine argument into $|\cdot|$; for any $\lambda\in[0,1]$, $|\lambda x+(1-\lambda)y|\le \lambda|x|+(1-\lambda)|y|$ by triangle inequality). Subtracting a constant preserves convexity. Hence $\Pi$ is convex. (1)

(c) Put–call parity — statement, proof, arbitrage check (10)

Statement: For European options same $K,T$: $C - P = S_0 - K e^{-rT}$. (1)

Proof (replication / no-arbitrage): Form two portfolios:

• A: long 1 call + cash $Ke^{-rT}$ (invested at $r$).
• B: long 1 put + 1 unit of underlying.

At expiry $T$:

• A value $= \max(S_T-K,0) + K = \max(S_T, K)$.
• B value $= \max(K-S_T,0) + S_T = \max(S_T, K)$.

Both give identical terminal payoff $\max(S_T,K)$ in every state. (3) By the law of one price / no-arbitrage, their values today must be equal: $C + K e^{-rT} = P + S_0 \;\Rightarrow\; C - P = S_0 - K e^{-rT}. \quad\blacksquare$ (2)

Numerical check: $r=0.06$, $T=0.25$, $K=20000$, $S_0=20000$. $Ke^{-rT} = 20000 \, e^{-0.015} = 20000 \times 0.9851119 = 19{,}702.24$. Parity RHS $= S_0 - Ke^{-rT} = 20000 - 19702.24 = \mathbf{297.76}$. (2)

Observed $C - P = 250 - 180 = 70$. Since $70 \ne 297.76$, an arbitrage exists (call is cheap / put+stock rich relative to parity). (1)

Arbitrage trade: Since market $C-P=70 <$ fair $297.76$, buy the cheap side (the call side, portfolio A) and sell the rich side (portfolio B):

• Buy call, invest $Ke^{-rT}=19702.24$ at $r$; short the underlying, and write (sell) the put.
• Net cash flow today $= -C + P + S_0 - Ke^{-rT} = -250 + 180 + 20000 - 19702.24 = +227.76$ received, with zero terminal net payoff (both legs $\to \max(S_T,K)$ cancel). Locked riskless profit $\approx ₹227.76$. (1)

Question 2

(a) Moneyness, intrinsic, time value (8)

Spot $S=20{,}050$. Call intrinsic $=\max(S-K,0)$; Put intrinsic $=\max(K-S,0)$; Time value $=$ LTP $-$ intrinsic.

Nearest-to-spot strike (20000, and 20100) are effectively near-ATM; strictly ATM = strike closest to spot (20000/20100). (0.5 per correct cell region — 8 total)

(b) PCR (4)

$\sum$ Put OI $= 60000 + 120000 + 70000 = 250{,}000$. (1) $\sum$ Call OI $= 45000 + 80000 + 110000 = 235{,}000$. (1) $\text{PCR} = \frac{250000}{235000} = \mathbf{1.0638}.$ (1) Interpretation: PCR $>1$ (more put OI than call OI) — often read as bearish positioning / put-heavy hedging, or contrarian bullish at extremes. Here mildly put-skewed, near-neutral. (1)

(c) Code (8)

def analyze_chain(spot, rows):
    total_put_oi = 0
    total_call_oi = 0
    result = {}
    for strike, c_ltp, c_oi, p_ltp, p_oi in rows:
        c_iv = max(spot - strike, 0.0)
        p_iv = max(strike - spot, 0.0)
        c_tv = c_ltp - c_iv
        p_tv = p_ltp - p_iv
 
        def moneyness(intrinsic, at_spot):
            if abs(strike - spot) < 1e-9:  # exact ATM
                return "ATM"
            return "ITM" if intrinsic > 0 else "OTM"
 
        entry = {
            "call": {
                "moneyness": moneyness(c_iv, spot),
                "intrinsic": c_iv, "time_value": c_tv,
                "arb_warning": c_tv < 0,
            },
            "put": {
                "moneyness": moneyness(p_iv, spot),
                "intrinsic": p_iv, "time_value": p_tv,
                "arb_warning": p_tv < 0,
            },
        }
        result[strike] = entry
        total_call_oi += c_oi
        total_put_oi  += p_oi
 
    pcr = total_put_oi / total_call_oi if total_call_oi else float('inf')
    return {"per_strike": result, "PCR": round(pcr, 4)}

Marks: intrinsic logic (2), time value (1), moneyness classification (2), negative-TV arbitrage flag (2), PCR return (1).

Question 3

(a) European vs American; exercise vs assignment (4)

• European: exercisable only at expiry $T$. American: exercisable any time up to and including expiry. (2)
• Exercise is the buyer's right/action: the long holder chooses to convert the option into the underlying position (call → buy at $K$; put → sell at $K$). Assignment is the seller's (writer's) obligation: when a buyer exercises, the clearinghouse assigns a short-position holder to fulfil the contract (deliver/receive underlying or cash-settle). (2)

(b) No early exercise of American call (non-dividend) (8)

Lower bound (no-arbitrage): $C \ge S_0 - K e^{-rT}$. (2) Payoff from exercising the call now $= S_0 - K$. (1) Since $r>0$ and $T>0$, $Ke^{-rT} < K$, hence $C \ge S_0 - Ke^{-rT} > S_0 - K = \text{immediate-exercise value}.$ (3) Because an American call is worth at least its European counterpart, and the European value already exceeds the exercise value, the holder does strictly better by selling the option (capturing $C$) than exercising (getting $S_0-K$). Also exercising forfeits remaining time value and pays $K$ early (losing interest). Therefore early exercise is never optimal. $\blacksquare$ (2)

(c) American put early-exercise counterexample (6)

Take $K=100$, spot $S_0 = 5$ (deep ITM put), $r=10\%$, $T=1$.

• Exercise now: receive $K - S_0 = 100 - 5 = \mathbf{95}$ immediately, which can be invested to earn interest. (2)
• Max value from waiting is bounded by $K - S_0$ but the discounted strike caps European put value near $Ke^{-rT} - S_0 = 100e^{-0.1} - 5 = 90.48 - 5 = 85.48 < 95$. (2) So holding is worth less than the ₹95 obtainable now → early exercise optimal.
• Intuition: For a put, exercising brings in cash $K$ early (earning interest), so time value of money favours early exercise deep ITM; for a call you would pay out $K$, so delaying is better. Puts and calls differ by the sign of the strike cash flow. (2)

[
  {"claim":"Long call breakeven = 20250", "code":"K=20000; C=250; result = (K + C == 20250)"},
  {"claim":"Straddle upper/lower breakevens 20430 and 19570", "code":"K=20000; cost=430; result = (K+cost==20430) and (K-cost==19570)"},
  {"claim":"Put-call parity fair C-P = 297.76 with r=6%,T=0.25", "code":"import sympy as sp; val=20000-20000*sp.exp(sp.Rational(-6,100)*sp.Rational(1,4)); result = abs(float(val)-297.76) < 0.5"},
  {"claim":"Arbitrage exists: observed C-P=70 differs from fair 297.76", "code":"import sympy as sp; fair=20000-20000*sp.exp(sp.Rational(-6,100)*sp.Rational(1,4)); obs=250-180; result = abs(float(fair)-obs) > 1"},
  {"claim":"PCR = 1.0638", "code":"put=60000+120000+70000; call=45000+80000+110000; result = round(put/call,4)==1.0638"},
  {"claim":"American put early exercise optimal: 95 > European bound 85.48", "code":"import sympy as sp; euro_bound=100*sp.exp(sp.Rational(-1,10))-5; result = 95 > float(euro_bound)"}
]