12 Quant Stochastics and Options Interview Questions and Answers

Short answer: A Wiener process (standard Brownian motion) W_t is a stochastic process with four defining properties: W₀ = 0; independent increments; W_t − W_s ~ N(0, t − s) for s < t; continuous sample paths almost surely. It is the basic source of randomness in continuous-time finance.

In depth:

1. W₀ = 0 — normalization of the starting point.
2. Independent increments — increments over disjoint intervals are independent: the past does not help predict the future (hence the Markov property).
3. Gaussian increments — W_t − W_s ~ N(0, t − s): variance grows linearly in time, so the typical range scales like √t.
4. Continuous paths — almost surely no jumps, yet the paths are maximally rough.

Bonus properties interviewers fish for:
• nowhere differentiable: |ΔW| ~ √Δt ≫ Δt as Δt → 0
• quadratic variation: Σ(ΔW)² → t  ⇒  (dW)² = dt
  (the seed from which Itô's lemma grows)
• self-similarity: W_ct ~ √c·W_t;  W_t = √t·Z with Z ~ N(0,1)

⚠️ Common mistake: offering "W_t ~ N(0, t)" as the definition. That is only a consequence: without independent increments and continuous paths, a process with those marginal distributions is not a Wiener process.

Short answer: E[W_s·W_t] = min(s, t). This is the covariance function of Brownian motion: for s < t it equals s — the length of the "shared history" of the two time points.

In depth:

1. The trick — assume s < t and decompose W_t = W_s + (W_t − W_s) to get independent pieces.
2. The computation — the cross term vanishes by independence of increments, leaving E[W_s²] = Var(W_s) = s.
3. The meaning — only the common interval [0, s] creates correlation; after time s the paths diverge independently, so the extra piece contributes nothing.

s < t:  W_t = W_s + (W_t − W_s)
E[W_s·W_t] = E[W_s·(W_s + (W_t − W_s))]
           = E[W_s²] + E[W_s·(W_t − W_s)]
           = s + E[W_s]·E[W_t − W_s]   ← independent increments
           = s + 0·0
           = s = min(s, t)

⚠️ Common mistake: writing s·t "by analogy with the variance" or silently dropping the cross term. The interviewer is listening for the justification: W_s and the increment W_t − W_s are independent, so the expectation of the product factorizes.

Short answer: A martingale is a process whose best forecast of the future value is the current one: E[X_t | F_s] = X_s for s < t (a "fair game", zero drift). W_t² is not a martingale — its mean E[W_t²] = t grows. The compensated process W_t² − t is a martingale.

In depth:

1. Definition — three conditions: adapted to the filtration F_t, E|X_t| < ∞, and E[X_t | F_s] = X_s.
2. W_t² — no — the direct computation (below) gives E[W_t² | F_s] = W_s² + (t − s) > W_s²: a submartingale drifting up at rate +1 per unit time.
3. W_t² − t — yes — subtracting the compensator t removes the drift; a special case of the Doob–Meyer decomposition.
4. Non-example — GBM with drift μ ≠ 0 is not a martingale; the discounted price becomes one under the risk-neutral measure.

E[W_t² | F_s] = E[(W_s + (W_t − W_s))² | F_s]
             = W_s² + 2W_s·E[W_t − W_s | F_s] + E[(W_t − W_s)²]
             = W_s² + 0 + (t − s)              → not a martingale
E[W_t² − t | F_s] = W_s² + (t − s) − t = W_s² − s → martingale

⚠️ Common mistake: checking only that E[X_t] is constant. That is not enough — the equality must hold conditionally on the whole history F_s, not merely on average across paths.

Short answer: On paper the martingale betting strategy wins 1 almost surely: the first win covers all prior losses. The catch is the optional stopping theorem: the equality E[X_τ] = X₀ for a fair game requires a bounded stopping time or bounded wealth. With a finite bankroll or a table limit, E[profit] = 0: the rare catastrophic loss exactly offsets the frequent small wins.

In depth:

1. The strategy — bets of 1, 2, 4, …, 2^{n−1}; the first win nets +1 on top of all accumulated losses.
2. Why it "works" — P(an infinite losing streak) = 0, so the +1 arrives "almost surely". But that needs unbounded capital and unbounded time.
3. OST — the gambler's wealth is a martingale; concluding E[X_τ] = X₀ is legitimate when τ is bounded, or the process is bounded, or uniform integrability holds. Here wealth before the first win is unbounded below — the conditions fail.
4. With an n-round limit — with probability 1 − 2^{−n} you win +1, with probability 2^{−n} you lose 2^n − 1. Total: (1 − 2^{−n})·1 − 2^{−n}·(2^n − 1) = 0.

⚠️ Common mistake: looking for the catch in the coin being "unfair". The coin is fair — what breaks is the applicability of optional stopping when intermediate losses are unbounded.

Short answer: For a fair coin, P(reach N before 0) = i/N. For a biased coin (p is the up-step, q = 1 − p, p ≠ q): P = (1 − (q/p)^i) / (1 − (q/p)^N).

In depth:

1. Fair case — via a martingale — the wealth X_t is a martingale; optional stopping gives E[X_τ] = X₀ = i, while X_τ takes only the values 0 and N.
2. Biased case — X_t itself is no longer a martingale, but M_t = (q/p)^{X_t} is (one-step check: p·(q/p) + q·(p/q) = 1). The same stopping argument yields the formula.
3. Sanity check — as p → 1/2 the ratio q/p → 1, and by expansion the biased formula tends to i/N.

Fair:    i = E[X_τ] = N·P + 0·(1 − P)   ⇒  P = i/N
Biased:  E[(q/p)^{X_τ}] = (q/p)^i
(q/p)^N·P + (q/p)^0·(1 − P) = (q/p)^i
⇒  P = (1 − (q/p)^i) / (1 − (q/p)^N)

⚠️ Common mistake: quoting i/N for the biased coin. Even a small edge changes the answer radically for large N: the odds against the disadvantaged player decay exponentially in the distance, not linearly.

Short answer: P(hit +a before −b) = b/(a + b) — the probability is inversely proportional to the distance to the barrier. The expected exit time is E[τ] = ab. Both results follow from optional stopping applied to the martingales W_t and W_t² − t.

In depth:

1. Probability — W_t is a martingale and τ is the exit time of the interval; E[W_τ] = W₀ = 0, with W_τ ∈ {a, −b}.
2. Time — W_t² − t is a martingale, so E[W_τ²] = E[τ]; the value W_τ² equals a² or b² with the probabilities just found.
3. Sanity check — for a symmetric interval a = b: p = 1/2 and E[τ] = a²; double the corridor width and you wait four times as long.

Probability:  0 = E[W_τ] = a·p + (−b)·(1 − p)  ⇒  p = b/(a + b)
Time:         E[W_τ²] = E[τ]
E[W_τ²] = a²·p + b²·(1 − p)
        = a²b/(a + b) + ab²/(a + b) = ab(a + b)/(a + b) = ab
⇒  E[τ] = ab

⚠️ Common mistake: flipping it to p = a/(a + b). Check the limiting case: as a → 0 the +a barrier sits right at the start and p → 1 — so the numerator must be b, the distance to the opposite barrier.

Short answer: For a smooth function f of an Itô process dX = μ dt + σ dW, Itô's lemma adds the term ½f″σ² dt to the ordinary chain rule. The reason is the non-vanishing quadratic variation of Brownian motion: (dW)² = dt, so the second-order Taylor term is of order dt and does not disappear.

In depth:

1. Taylor to second order — df = f′ dX + ½f″(dX)²; in classical calculus (dX)² = O(dt²) and gets discarded.
2. Brownian motion is different — the increment ΔW has size √Δt, hence the Itô multiplication table: (dW)² = dt, dW·dt = 0, (dt)² = 0.
3. The canonical example — the log of GBM: a −σ²/2 correction to the log-return drift.

df = (∂f/∂t + μ·∂f/∂x + ½σ²·∂²f/∂x²) dt + σ·∂f/∂x dW

GBM: dS = μS dt + σS dW,  f = ln S
f′ = 1/S,  f″ = −1/S²
d ln S = (μ − σ²/2) dt + σ dW      ← the Itô correction

⚠️ Common mistake: dropping the −σ²/2 and claiming the log-return drift equals μ. At σ = 30% the correction is −4.5% per year: average log growth is noticeably below μ, and the median path grows more slowly than the mean.

Short answer: The price follows GBM (lognormal distribution) with constant volatility; the rate r is constant; there are no frictions (fees, taxes, short-sale constraints) and no dividends; delta hedging is continuous; exercise is European. The most violated assumption is constant volatility — that is exactly why the volatility smile exists.

In depth:

1. The key violation — constant σ — the market quotes implied volatility (the volatility smile): σ differs across strikes and maturities, which a single-constant model cannot explain.
2. Fat tails and jumps — real returns are leptokurtic; lognormality underprices crashes → out-of-the-money puts trade richer than the model says, hence the skew.
3. Discrete hedging — continuous rebalancing is impossible; gaps and transaction costs leave hedging error.

⚠️ Common mistake: naming the constant rate as the main problem. At typical maturities it is second-order; the very existence of the volatility smile is direct empirical refutation of constant σ.

Short answer: Form the portfolio Π = V − ΔS with Δ = ∂V/∂S: this choice kills the dW term, the portfolio becomes locally riskless, and by no-arbitrage a riskless portfolio must grow at the rate r. Equating the two gives the Black–Scholes PDE. The two key points: hedging kills the randomness, and μ drops out of the equation entirely.

In depth:

1. Itô's lemma for V(S, t) — dV contains drift terms and a dW term via dS.
2. Choosing Δ = ∂V/∂S — in dΠ = dV − Δ·dS the coefficient of dW vanishes: being short Δ shares exactly cancels the option's local randomness.
3. No arbitrage — a deterministic return must equal the riskless one: dΠ = rΠ dt; otherwise you arbitrage against the money-market account.
4. μ is gone — the stock's expected return never enters the equation: the price does not depend on risk preferences → the bridge to risk-neutral pricing.

dV = (∂V/∂t + μS·∂V/∂S + ½σ²S²·∂²V/∂S²) dt + σS·∂V/∂S dW
Π = V − Δ·S,  Δ = ∂V/∂S:
dΠ = (∂V/∂t + ½σ²S²·∂²V/∂S²) dt      ← dW and μ cancel
dΠ = rΠ dt = r(V − S·∂V/∂S) dt
⇒  ∂V/∂t + ½σ²S²·∂²V/∂S² + rS·∂V/∂S − rV = 0

⚠️ Common mistake: reciting the pricing formula instead of the logic. The interviewer is listening for two things: why hedging removes the randomness, and why μ vanishing means risk preferences are irrelevant so you may price in the risk-neutral world.

Short answer: For European options on a non-dividend-paying stock: C − P = S − K·e^{−rT}. The relation is model-free — no assumptions about the price distribution; it is enforced by conversion/reversal arbitrage, because a call minus a put is a synthetic forward.

In depth:

1. Replication — the portfolios "call + K·e^{−rT} in bonds" and "put + stock" are both worth max(S_T, K) at time T, so they must also agree in price today.
2. Arbitrage if violated — buy the cheap side, sell the rich one: long call + short put against the forward locks in a riskless profit.
3. Consequences — a call and a put at the same strike carry the same implied volatility; knowing one premium recovers the other with no model at all.

Portfolio A: C + K·e^{−rT}       Portfolio B: P + S
At T:  A = max(S_T − K, 0) + K = max(S_T, K)
       B = max(K − S_T, 0) + S_T = max(S_T, K)
A = B at T  ⇒  A = B today:
C + K·e^{−rT} = P + S  ⇔  C − P = S − K·e^{−rT}

⚠️ Common mistake: saying parity "follows from Black–Scholes". It is the other way around — this is pure no-arbitrage with no model; all you need is European exercise and a dividend adjustment (with dividends, S becomes S − PV(div)).

Short answer: Delta = N(d₁) ∈ (0, 1), slightly above 0.5 for an ATM call; gamma and vega are positive, theta is negative — a long option pays time decay. The link: from the Black–Scholes PDE, Θ + ½σ²S²Γ ≈ r(V − SΔ) — long gamma is financed with theta.

In depth:

1. Delta — N(d₁); slightly above 0.5 ATM because of the +σ√T/2 term in d₁; tends to 1 deep in the money and 0 out of the money.
2. Gamma — largest ATM and sharpens toward expiry: delta flips 0 → 1 ever more abruptly.
3. Vega — largest ATM and grows with time to expiry.
4. Theta — negative, most painful ATM near expiry — the mirror image of gamma.
5. The gamma-theta tradeoff — re-hedging a long-gamma book "buys low, sells high" and earns realized volatility, but on average that income is exactly paid for by theta.

⚠️ Common mistake: "ATM delta is exactly 0.5". It is slightly above — d₁ carries a +σ√T/2 term. And remember: gamma and theta almost always have opposite signs — there is no free gamma.

Short answer: The risk-neutral measure Q is the probability measure under which discounted asset prices are martingales: every asset drifts at the riskless rate r. A derivative's price = e^{−rT}·E_Q[payoff], and the existence of Q is equivalent to the absence of arbitrage (the first fundamental theorem). No: Q-probabilities are not a forecast — they are real-world probabilities reweighted by the price of risk.

In depth:

1. Where it comes from — the Black–Scholes derivation showed μ drops out: since the price is independent of risk preferences, you may compute in the convenient world where everything grows at r.
2. The reweighting — Q overweights bad states (where the marginal utility of money is higher) and underweights good ones; technically a Radon–Nikodym derivative, and by Girsanov's theorem only the drift changes: μ → r.
3. Example — N(d₂) in the Black–Scholes formula is the Q-probability the call is exercised; when μ > r it sits below the real-world one.

⚠️ Common mistake: reading N(d₂) as "the real chance the option finishes in the money". It is a probability under Q; mistaking risk-neutral probabilities for a forecast is the classic trap.