Probability & Options — Problem Set

20 questions · Conceptual, Numerical, Open-ended · Reveal-on-click answers
Score: 0 / 0
Section A — Probability Foundations (Q1–5)
Q1
A coin is flipped 10 times and lands heads every time. What is the probability of heads on the 11th flip?
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Answer: C — with an important nuance

If the coin is known to be fair, each flip is independent and P(heads) = 50% always. Past outcomes cannot influence future independent events — this is the Gambler's Fallacy and costs people billions in casinos.

However, D has merit in practice: 10 consecutive heads is evidence the coin may be biased. A Bayesian would update their prior probability of bias upward and hedge accordingly. Markets exploit this distinction: the "fair coin" assumption is the frequentist model; a trader who suspects bias updates via Bayes' theorem.

Key concept: Independent events have no memory. But observing extreme sequences is rational evidence of non-independence.

Q2
You play a game: roll one die. Win ₹30 if you roll a 6, lose ₹5 otherwise. Calculate the expected value per roll.
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Full working

P(roll 6) = 1/6 ≈ 0.1667
P(not 6) = 5/6 ≈ 0.8333

E[X] = P(6) × ₹30 + P(not 6) × (−₹5)
     = (1/6)(30) + (5/6)(−5)
     = 5.00 − 4.167
     = ₹0.833 per roll

This is a positive edge game. Over 1,200 rolls, expected total profit = 1,200 × ₹0.833 ≈ ₹1,000. An options market maker operates identically: small positive EV per trade, enormous volume.

EV = ₹0.83 per roll (accept ₹0.80–₹0.87)

Q3
NIFTY has an annualised volatility of σ = 18%. How large a move (in %) would represent a 2-standard-deviation event over 1 trading day? (Use 252 trading days/year.)
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Full working

σ_annual = 18% = 0.18

Step 1: convert to daily vol
σ_daily = σ_annual / √252
         = 0.18 / 15.875
         = 1.134%

Step 2: 2σ move
2 × σ_daily = 2 × 1.134% = 2.27%

Interpretation: a 2σ daily move (>2.27%) should occur on roughly 5% of trading days ≈ 12–13 days per year under normality. In practice (fat tails), it occurs more often.

Answer: ±2.27% (accept 2.20%–2.35%)

Q4
Which of the following best describes why stock prices are modelled as log-normally distributed rather than normally distributed?
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Answer: C

The core reason: if S is normally distributed, it has positive probability of being negative (prices can't be negative). Instead, we model log(S_T / S_0) as normally distributed — this is the log-return. If the log-return is normal, the price level S_T is log-normally distributed and is always positive.

A is wrong: log-normal distributions actually have the same tail behaviour as normal once you transform. B is irrelevant — the choice is driven by economics, not computation. D is wrong: in a log-normal distribution, the mean exceeds the median (the distribution is right-skewed).

Log-normal ensures prices stay positive. Log-returns (not prices) are the normally distributed quantity.

Q5
Explain in your own words why variance of a random walk scales with T (time) but standard deviation scales with √T. Give a concrete example using coin flips.
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Model Answer

Starting from definition: After N coin flips (each ±1), position = X₁ + X₂ + … + Xₙ. Since flips are independent, Var(sum) = Var(X₁) + Var(X₂) + … + Var(Xₙ) = N × Var(single flip) = N × 1 = N.

So Var scales linearly with N (time). Standard deviation = √Var = √N. This is a fundamental property of sums of independent variables — variance adds linearly, std dev does not.

Concrete example: After 1 flip, std dev of position = 1. After 4 flips, std dev = √4 = 2 (not 4). After 100 flips, std dev = √100 = 10 (not 100). This means uncertainty grows much slower than time — which is why long-dated options are less than N× more expensive than short-dated ones.

Applied to markets: NIFTY σ_daily = 1.13%. Over 25 days: σ_25d = 1.13% × √25 = 5.65%. Not 1.13% × 25 = 28.25%.

Section B — Black-Scholes & Greeks (Q6–12)
Q6
A call option has: S=₹100, K=₹100, r=6.5%, σ=20%, T=30 days. Using the fact that for an ATM option, d₁ ≈ σ√T/2 and N(0)=0.5, estimate the call price approximately.
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Hint: For ATM (S≈K): C ≈ S · σ · √(T/2π) — the "ATM approximation" T = 30/252 years, σ = 0.20 √(T/2π) = √(0.1190/6.2832) = √0.01894 = 0.1376

Full working

ATM approximation: C ≈ S · σ · √(T/2π)

S = 100
σ = 0.20
T = 30/252 = 0.11905 years

√(T/2π) = √(0.11905 / 6.2832) = √(0.01894) = 0.1376

C ≈ 100 × 0.20 × 0.1376 = ₹2.75

Full B-S gives ≈ ₹2.78. The approximation is accurate to within 1–2%.

This approximation is used on desks for quick mental checks. It shows that ATM option prices scale linearly with σ and √T — doubling sigma doubles the option price, but quadrupling time only doubles it.

Answer: ≈ ₹2.75 (accept ₹2.50–₹3.05)

Q7
A call option has delta = 0.65. The stock moves from ₹100 to ₹102. Approximately how much does the call option price change?
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Answer: B — ₹1.30

Delta = ∂C/∂S. For a small move ΔS, the change in option price ≈ Δ × ΔS = 0.65 × ₹2 = ₹1.30.

This is the first-order (linear) approximation. The true change will be slightly more because of gamma (the curvature) — the option actually moves a bit more than ₹1.30 when the stock moves up ₹2, because delta increases during the move. The gamma correction would add ½ × Γ × (ΔS)² on top.

ΔC ≈ Δ × ΔS = 0.65 × 2 = ₹1.30

Q8
You are long 100 call contracts (each representing 1 lot = 50 shares) with delta = 0.55. How many NIFTY futures lots must you sell to be delta-neutral?
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Full working

Total delta of option position:
= Number of contracts × shares per lot × delta
= 100 contracts × 50 shares × 0.55
= 2,750 share-equivalents

Each futures lot = 50 shares.
Lots to sell = 2,750 / 50 = 55 lots

After selling 55 futures lots, net delta = 0. You are delta-neutral — a ₹1 move in NIFTY doesn't change your P&L (to first order). But you are still long gamma: a large move in either direction will make you money (your delta drifts and you re-hedge at profit).

Sell 55 NIFTY futures lots.

Q9
Gamma is highest for which option?
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Answer: C — ATM, very near expiry

Gamma = N'(d₁) / (S·σ·√T). The N'(d₁) term is maximised when d₁ = 0, i.e. when the option is ATM. The denominator shrinks as T→0 (√T → 0). So gamma spikes for ATM options near expiry.

Why? A 2-day ATM option is on the knife-edge: a tiny price move determines whether it expires worthless or worth something. Delta swings from near 0.5 to near 0 or 1 with tiny moves. That rapid change in delta is gamma.

This is called "pin risk" on expiry day — options near the strike become extremely gamma-sensitive and are hard to hedge.

Gamma is largest for ATM options with very little time remaining.

Q10
An ATM call option is priced at ₹8.50 with 30 days to expiry. Theta is −₹0.28 per day. Assuming vol and stock price are unchanged, estimate the option price after 10 days.
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Full working — and a warning

Simple linear estimate:
Price after 10 days ≈ ₹8.50 + (−₹0.28 × 10)
                   = ₹8.50 − ₹2.80
                   = ₹5.70

More accurate (theta accelerates):
The actual value would be slightly lower than ₹5.70
because theta increases as time passes (decay accelerates).

The linear estimate is a useful approximation. More precisely, theta is not constant — it grows as expiry approaches. The actual price after 10 days (now 20 days to expiry) should be recalculated with B-S using T=20/252. The true answer would be closer to ₹5.40–₹5.60.

≈ ₹5.70 (linear) or ₹5.40–₹5.60 (accounting for theta acceleration). Accept ₹5.20–₹6.10.

Q11
Explain why a market maker who sells options (short gamma) loses money when the stock makes a large move, even if they are delta-hedged. Use specific numbers to illustrate.
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Model Answer

Setup: Sell 1 ATM call at S=₹100, K=₹100. Receive ₹5 premium. Delta=0.50, so buy 50 shares at ₹100 to delta-hedge. Net cost = ₹5,000 shares − ₹500 premium received = net ₹4,500 invested.

Stock moves to ₹110:

• Option P&L: sold call at ₹5. Call is now worth ~₹11 (ITM). Loss on option = −₹6.
• Hedge P&L: +₹500 on 50 shares × ₹10 gain.
• Net: −₹6 + ₹5 = −₹1 loss.

Why? The hedge was 50 shares when delta was 0.50. But as stock moved to ₹110, delta grew toward 0.80. The extra 30 shares of delta exposure was unhedged during the move. That unhedged delta = gamma × move = Γ × ₹10. The P&L loss ≈ −½ × Γ × (ΔS)².

The gamma P&L equation: ΔP&L ≈ −½ × Γ × (ΔS)² for a short gamma position. Larger moves → larger losses, nonlinearly. This is why short-gamma desks fear large quick moves even after re-hedging.

Compensation: The market maker earns theta (₹0.17/day on this option) to offset the expected gamma loss. If realized vol < implied vol, theta income exceeds gamma cost — they profit.

Q12
NIFTY VIX jumps from 15 to 22 overnight. Your desk holds a delta-neutral portfolio with net vega of +₹50,000 per vol point. What is the approximate P&L impact?
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Answer: C — +₹350,000

Change in vol = 22 − 15 = 7 vol points
Net vega = +₹50,000 per vol point
P&L ≈ +₹50,000 × 7 = +₹350,000

Being delta-neutral only immunises you against small directional moves. It says nothing about vega (volatility risk). You can be delta-neutral and have enormous vega exposure — which is exactly the position of someone who has bought straddles.

D is wrong because VIX didn't rise by 22 points — it rose from 15 to 22, a change of 7 points. B is a common misconception — delta-neutral ≠ risk-free.

P&L = Vega × ΔVol = ₹50,000 × 7 = +₹350,000

Section C — Advanced & Applied (Q13–20)
Q13
The volatility smirk shows that OTM puts trade at higher implied vol than OTM calls. Explain what this means in terms of probability — what probability distribution is the market implying compared to log-normal?
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Model Answer

The volatility smirk encodes the market's belief that the left tail is fatter than log-normal and the right tail is thinner.

Under log-normality (B-S), a −3σ move and a +3σ move are equally likely after adjusting for drift. But the smirk shows OTM puts (hedges against left tail moves) are more expensive than OTM calls. Translating back through B-S: the market is assigning higher probability to large downside moves than log-normal predicts.

The implied real-world distribution is: negatively skewed (left tail fatter), leptokurtic (excess kurtosis — fatter tails overall). This reflects:
1. Crash risk: equity markets can gap down violently
2. Correlation in bad times: diversification fails when you need it
3. Liquidity spirals: selling begets more selling

Risk-neutral vs physical measure: There's also a subtlety — the implied distribution is under the "risk-neutral measure" (Q-measure), not the real-world measure (P-measure). The skew partly reflects genuine crash risk, partly the premium investors pay for protection (who would sell crash insurance cheaply?).

Q14
Kelly criterion: you find a strategy with win probability p=0.55, lose probability q=0.45, and even odds (b=1, win ₹1 for every ₹1 risked). What fraction of capital should you bet per trade?
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Full working

Kelly formula: f* = (bp − q) / b
b = 1 (even odds)
p = 0.55
q = 0.45

f* = (1 × 0.55 − 0.45) / 1
   = (0.55 − 0.45) / 1
   = 0.10 = 10%

Bet 10% of capital per trade. Many desks use "half Kelly" = 5% to reduce variance while accepting ~25% less expected growth rate.

Notice: even a genuine 55% edge strategy (which is excellent in practice) only warrants 10% bet size. This is why professional risk management is conservative — even with edge, over-betting leads to ruin.

f* = 10% of capital per trade

Q15
Put-call parity states: C − P = S − K·e^(−rT). An ATM call (K=100) is priced at ₹5.50, S=₹100, r=6.5%, T=30 days. What must the ATM put be worth?
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Answer: C — ₹5.03

Put-call parity: C − P = S − K·e^(−rT)

K·e^(−rT) = 100 × e^(−0.065 × 30/252)
           = 100 × e^(−0.007738)
           = 100 × 0.99229
           = ₹99.23

C − P = S − K·e^(−rT) = 100 − 99.23 = ₹0.77

P = C − 0.77 = 5.50 − 0.77 = ₹4.73

(Note: exact calculation gives ≈₹4.73; accept ₹4.60–₹5.00)

D is wrong — put-call parity holds regardless of the model used. It's an arbitrage relationship derived purely from no-arbitrage, not from any distributional assumption about returns. If the put deviated from this price, you could lock in risk-free profit.

Put = ≈₹4.73 (accept ₹4.60–₹5.00). The put is cheaper because the interest you earn on the strike price offsets the put value.

Q16
A trading strategy has Sharpe ratio of 1.5 and annualised volatility of 12%. Using the continuous Kelly formula f* = Sharpe/σ, calculate the optimal leverage. Why might a real desk use far less than this?
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Model Answer

Calculation: f* = Sharpe / σ = 1.5 / 0.12 = 12.5× leverage.

Why real desks use much less:

1. Parameter uncertainty: The Sharpe of 1.5 and σ of 12% are estimates from historical data. If the true Sharpe is 1.0 and true σ is 18%, the optimal leverage is 5.5× — and you were betting 12.5×. Overconfidence in parameters leads to over-leveraging.

2. Regime changes: Kelly maximises expected log-wealth in a stationary environment. Markets have regime changes (2008, COVID). Estimated parameters from calm periods don't apply in crises.

3. Regulatory capital: Exchanges require margin. Capital constraints impose a maximum leverage regardless of Kelly's recommendation.

4. Ruin aversion: Kelly maximises median long-run wealth, but not short-run survival probability. A single run of bad luck at 12.5× leverage could trigger a margin call before the law of large numbers kicks in.

5. Model risk: The "edge" may not exist in reality — the returns may be from data mining. Desks apply a discount factor to estimated Sharpe to account for this.

Common practice: Half Kelly at most. Many use ¼ Kelly. The return reduction is modest (~25% for half Kelly) but the drawdown reduction is substantial.

Q17
Implied volatility (IV) for a NIFTY option is 22%. Realized volatility (RV) over the past month was 18%. A vol seller's book profits when:
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Answer: D

A vol seller collects premium by selling options (IV=22%). They delta-hedge to remove directional risk. Their P&L is determined by: Theta income − Gamma hedging cost.

The gamma hedging cost is approximately ½ × Γ × (ΔS)² × days — which depends on realized volatility. If RV (18%) stays below IV (22%), the daily hedging costs are lower than the theta income, and the vol seller profits.

A is partially right but not the primary mechanism — IV falling benefits vol sellers via mark-to-market, but their real edge is the IV−RV spread over time. B is the scenario that hurts them. C describes a bad scenario for short calls.

Vol seller profits when RV < IV — the "volatility risk premium" they are paid to bear crash risk.

Q18
Design a simple experiment you could run with a spreadsheet to demonstrate the Central Limit Theorem. Describe: inputs, process, what you'd plot, and what result you'd expect to see and why it matters for options pricing.
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Model Answer

Experiment design:

1. Single roll: Simulate 10,000 rolls of a fair die (values 1–6, uniform distribution). Plot histogram — expect approximately flat (uniform), mean≈3.5, not bell-shaped.

2. Sum of 2 rolls: Simulate 10,000 trials of (Die₁ + Die₂). Plot histogram — expect a triangular distribution, starting to centralise.

3. Sum of 10 rolls: Plot histogram — should look distinctly bell-shaped, even though each die is uniform.

4. Sum of 30 rolls: Very close to normal distribution. Calculate sample mean and std dev — should cluster tightly around theoretical values (30×3.5=105, σ=√(30×35/12)≈9.35).

Expected result: The histograms converge to a bell curve as N increases, despite each die being uniformly distributed. This is the CLT in action.

Why it matters for options: A stock's daily return is the sum of thousands of individual trades and news reactions throughout the day. Each individual event has an unknown distribution. But by CLT, the daily return (sum of many effects) approaches normality. This is the statistical foundation for why log-normal returns are a reasonable first approximation — and also why it breaks down in crashes (when all effects suddenly become correlated, the independence assumption fails).

Q19
A delta-neutral straddle (long call + long put, same strike and expiry) has: Gamma = ₹450 per ₹1 move, Theta = −₹320/day, Vega = +₹28,000 per vol point. Over 1 day, NIFTY moves +2.5% (₹250 on a ₹10,000 index). What is the approximate P&L?
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Hint: Gamma P&L = ½ × Gamma × (ΔS)² Theta P&L = Theta × 1 day Assume vol unchanged. ΔS = 250 (₹250 move on ₹10,000 index)

Full working

Gamma P&L = ½ × Γ × (ΔS)²
          = ½ × 450 × (250)²
          = ½ × 450 × 62,500
          = ₹140,625

Theta P&L = −₹320 × 1 day = −₹320

(Vega P&L = 0, vol assumed unchanged)

Total P&L ≈ 140,625 − 320 = ₹140,305

The move was large enough that gamma profits massively overwhelmed theta decay. This is exactly why people buy straddles before earnings — they expect a large move (gamma wins) to exceed the theta they pay each day while waiting.

If the move had been only ₹50: Gamma P&L = ½×450×2,500 = ₹562,500 — wait, that's ½×450×50² = ₹562,500. No: ½×450×(50)² = ½×450×2,500 = ₹562,500. Actually ₹562,500. But theta cost is ₹320. So even a small move is profitable! The breakeven daily move is where ½×Γ×(ΔS)² = Θ → ΔS = √(2Θ/Γ) = √(640/450) = ₹1.19.

P&L ≈ ₹140,305 (accept ₹135,000 – ₹145,000)

Q20
"An options desk with positive EV on every trade can still lose money over a month." Give three distinct mechanisms that could cause this, using the mathematical concepts from today.
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Model Answer — Three mechanisms

1. Variance — the law of large numbers needs large N
Even with EV = +₹100 per trade and σ = ₹2,000 per trade, after 20 trades: E[total] = +₹2,000 but std dev = ₹2,000 × √20 = ₹8,944. The P&L after 20 trades is approximately N(2,000; 8,944²) — there is significant probability of being negative for the month. You need hundreds of trades for EV to dominate variance.

2. Correlation / regime change — independence fails
Kelly and EV calculations assume trades are independent. In a crisis (COVID March 2020), ALL positions move against simultaneously: short vol, long equities, carry trades. Correlation between positions spikes to near 1. The diversification benefit evaporates and true variance of the portfolio is far larger than the sum of individual trade variances. A month of 100 positive-EV trades can produce catastrophic losses if they're all correlated.

3. Estimation error — the EV was wrong
The desk estimated EV = +₹100 using historical data. But if implied vol was 20% and the desk assumed realized vol would be 17% (as it often is — the vol risk premium), but actual realized vol came in at 24% (a turbulent month), then gamma hedging costs exceeded theta income. The true EV was negative. Every Bayesian estimate has uncertainty — the vol risk premium disappears in certain months and the desk, not knowing this, keeps trading its model.