Enhanced Stochastic Mathematical Finance Cheatsheet
With Detailed Derivations & Interview Examples
| Model | Main Equation | Application | Key Property |
|---|---|---|---|
| Black-Scholes/GBM | \(dS = \mu S dt + \sigma S dW\) | Stock/option pricing | Lognormal distribution |
| Ornstein-Uhlenbeck | \(dX = \theta(\mu - X)dt + \sigma dW\) | Mean reversion | Gaussian stationary |
| Vasicek | \(dr = a(b - r)dt + \sigma dW\) | Interest rates | Affine term structure |
| CIR | \(dr = a(b - r)dt + \sigma\sqrt{r} dW\) | Interest rates (+) | Non-central χ² |
| Heston | \(dS_t = \mu S_t dt + \sqrt{v_t} S_t dW_t^S\) \(dv_t = \kappa(\theta - v_t)dt + \xi\sqrt{v_t} dW_t^v\) \(\langle dW_t^S, dW_t^v \rangle = \rho dt\) | Stochastic vol | Semi-closed form |
| SABR | \(dF_t = \alpha_t F_t^\beta dW_t^F\) \(d\alpha_t = \nu\alpha_t dW_t^\alpha\) \(\langle dW_t^F, dW_t^\alpha \rangle = \rho dt\) | Vol smile | CEV dynamics |
| Merton Jump | \(dS = \mu S dt + \sigma S dW + S dJ\) | Jumps | Compound Poisson |
| GARCH(1,1) | \(\sigma_t^2 = \omega + \alpha\varepsilon_{t-1}^2 + \beta\sigma_{t-1}^2\) | Vol forecasting | Discrete time |
1. Black-Scholes / GBM
Solution: \(S_t = S_0 \exp[(\mu - \sigma^2/2)t + \sigma W_t]\)
BS Call Formula:
\(d_1=\frac{\ln(100/105)+(0.05+0.04/2)(1)}{0.2}=\frac{-0.0488+0.07}{0.2}=0.106\)
\(d_2=0.106-0.2=-0.094\)
\(N(0.106)=0.542, N(-0.094)=0.463\)
\(C = 100(0.542) - 105e^{-0.05}(0.463) = 54.2 - 46.16 = \boxed{8.04}\)
Solution: Under risk-neutral measure (r=0 for simplicity):
\(\ln(S_T/S_0) \sim N(-\sigma^2T/2, \sigma^2T) = N(-0.045, 0.09)\)
\(P(S_T > 120) = P(\ln(S_T) > \ln(120)) = P(Z > \frac{\ln(1.2)+0.045}{0.3}) = P(Z > 0.757) = \boxed{0.225}\)
Alternative: \(d_2 = \frac{\ln(100/120)+(0-0.09/2)(1)}{0.3\sqrt{1}} = -0.757\), so \(N(-d_2)=0.225\)
Solution: This is the Itô correction. Apply Itô to \(Y=\ln S\):
\(\frac{\partial Y}{\partial S}=1/S, \frac{\partial^2 Y}{\partial S^2}=-1/S^2\)
\(d\ln S = \frac{1}{S}dS - \frac{1}{2S^2}(dS)^2 = \frac{1}{S}(\mu S dt + \sigma S dW) - \frac{1}{2S^2}\sigma^2 S^2 dt\)
\(= \mu dt + \sigma dW - \frac{\sigma^2}{2}dt = (\mu - \sigma^2/2)dt + \sigma dW\)
The \(-\sigma^2/2\) comes from the quadratic variation term \((dW)^2=dt\).
Greeks:
- \(\Delta=\partial V/\partial S = N(d_1)\) (call), \(-N(-d_1)\) (put)
- \(\Gamma=\partial^2V/\partial S^2 = \frac{N'(d_1)}{S\sigma\sqrt{T}}\)
- \(\mathcal{V}=\partial V/\partial\sigma = S\sqrt{T}N'(d_1)\)
- \(\Theta=\partial V/\partial t\) (negative for long options)
2. Ornstein-Uhlenbeck
Solution: \(X_t = X_0e^{-\theta t} + \mu(1-e^{-\theta t}) + \sigma\int_0^t e^{-\theta(t-s)}dW_s\)
Properties: \(\mathbb{E}[X_t] \to \mu\), \(\text{Var}[X_t] \to \sigma^2/(2\theta)\)
\(\mathbb{E}[X_1] = 3\%e^{-0.5} + 5\%(1-e^{-0.5}) = 1.82\% + 1.97\% = \boxed{3.79\%}\)
\(\text{Var}[X_1] = \frac{0.01^2}{2(0.5)}(1-e^{-2(0.5)(1)}) = 0.0001(0.632) = 0.0000632\)
Solution: \(\mathbb{E}[r_2] = 0.06 e^{-0.3(2)} + 0.04(1-e^{-0.6})\)
\(= 0.06(0.5488) + 0.04(0.4512) = 0.0329 + 0.0181 = \boxed{5.10\%}\)
Half-life: Time to reach halfway: \(t_{1/2} = \frac{\ln 2}{\theta} = \frac{0.693}{0.3} = 2.31\) years
Solution: O-U has additive noise (\(\sigma dW\)), so it's a linear transformation of Brownian motion, which is Gaussian. CIR has multiplicative noise (\(\sigma\sqrt{r}dW\)), making the distribution non-Gaussian (specifically, non-central chi-squared).
3. Vasicek
Solution: \(r_t = r_0e^{-at} + b(1-e^{-at}) + \sigma\int_0^t e^{-a(t-s)}dW_s\)
Properties: \(\mathbb{E}[r_t] = r_0e^{-at} + b(1-e^{-at})\), \(\mathrm{Var}(r_t) = \frac{\sigma^2}{2a}(1-e^{-2at})\)
\(\mathbb{E}[r_2] = 0.03e^{-0.4} + 0.05(1-e^{-0.4}) = 2.01\% + 1.65\% = \boxed{3.66\%}\)
Long-run mean is \(b=5\%\); half-life of mean reversion is \(\ln 2/a \approx 3.47\) years.
4. Cox-Ingersoll-Ross
Feller Condition: \(2ab \geq \sigma^2\) ensures \(r_t > 0\) always
Distribution: \(r_t \sim \frac{\sigma^2(1-e^{-at})}{4a}\chi'^2_{\nu}(\lambda)\) where \(\nu=\frac{4ab}{\sigma^2}\), \(\lambda=\frac{4ae^{-at}r_0}{\sigma^2(1-e^{-at})}\)
Check Feller: \(2(0.5)(0.05)=0.05 > (0.1)^2=0.01\) ✓
Degrees of freedom: \(\nu = \frac{4(0.5)(0.05)}{0.01} = 10\) (high → near Gaussian)
Solution:
Vasicek: Analytically simpler, closed-form for bonds/options. Use when negative rates acceptable (e.g., EUR, JPY).
CIR: Guarantees positive rates, better for credit spreads, equity volatility. Use for USD, emerging markets. Downside: more complex calibration, numerical methods often needed for derivatives.
Solution:
Mean-reverting: Yes, \(a>0\) ensures mean reversion to level \(b=6\%\).
Feller condition: \(2ab = 2(0.8)(0.06) = 0.096\) vs \(\sigma^2 = 0.0225\)
Since \(0.096 > 0.0225\), Feller holds → rates stay strictly positive ✓
5. Heston Stochastic Volatility
Variables:
- \(v_t\) = variance (not volatility!)
- \(\kappa\) = mean reversion speed
- \(\theta\) = long-term variance
- \(\xi\) = volatility of volatility
- \(\rho\) = correlation (typically \(\rho \in [-0.8, -0.5]\) for equities)
Feller for variance: \(2\kappa\theta \geq \xi^2\)
Given \(dW_S=0.08, dW_v=-0.05\), \(dt=0.01\):
\(dS = 0.05(100)(0.01) + \sqrt{0.04}(100)(0.08) = 0.05 + 1.6 = 1.65\)
\(dv = 2(0.04-0.04)(0.01) + 0.3\sqrt{0.04}(-0.05) = 0 - 0.003 = -0.003\)
\(S_1=\boxed{101.65}, v_1=\boxed{0.037}\)
Solution: Leverage effect: When stock price falls, debt/equity ratio increases, making equity riskier → volatility rises. Alternatively, negative news causes both price drop and uncertainty increase. Empirically, \(\rho \approx -0.7\) for S&P 500.
Solution: \(2\kappa\theta = 2(3)(0.04) = 0.24\) vs \(\xi^2 = 0.25\)
\(0.24 < 0.25\) → Feller violated! Variance can hit zero, causing numerical issues. Fix: Increase \(\kappa\) or \(\theta\), or decrease \(\xi\). In practice, use absorption schemes or truncation.
Characteristic Function: Semi-closed form exists, used for fast FFT pricing of European options.
6. SABR Model
Variables:
- \(F_t\) = forward rate (or asset price)
- \(\alpha_t\) = stochastic volatility
- \(\beta\) = CEV parameter (constant, chosen a priori)
- \(\nu\) = volatility of volatility
- \(\rho\) = correlation
Common β values & interpretations:
- \(\beta=0\): Normal (Bachelier) - vol independent of level
- \(\beta=0.5\): CIR-like - common for interest rates
- \(\beta=1\): Lognormal (Black) - vol proportional to price
Solution: This is a skew (downward sloping). Lower strikes have higher vol.
• High β (near 1) → steep skew → use \(\beta=0.7-1.0\)
• The negative slope suggests \(\rho < 0\)
Typical equity calibration: \(\beta=0.7, \rho=-0.3, \nu=0.4\)
Solution:
1. Smile dynamics: SABR produces realistic smile evolution; local vol has unrealistic dynamics ("sticky strike").
2. Market standard: Swaption market quotes in SABR parameters.
3. Analytic approximation: Hagan formula allows fast calibration.
4. Intuitive parameters: \(\rho\) controls skew, \(\nu\) controls curvature.
Hagan Approximation (ATM):
7. Short-Rate Models and Discount Factors
Pathwise discount factor in a short-rate model:
Interpretation: discounting is random because the future short rate path is random.
Zero-coupon bond price under \(\mathbb{Q}\):
So a short-rate model is useful when it specifies \(r_t\) directly and therefore induces discount factors and the full ZCB curve.
Vasicek under \(\mathbb{Q}\):
Concise derivation summary: start from the bond-pricing PDE under \(\mathbb{Q}\), try the affine ansatz \(P(t,T)=A(t,T)e^{-B(t,T)r_t}\), then match coefficients of \(1\) and \(r_t\). This yields ODEs for \(A\) and \(B\), with \(B\) solved first and \(A\) obtained by integration.
Dimensional analysis:
- \(r_t\), \(a\), \(b\), \(\sigma\) all carry rate/time-compatible units; \(a\) has units \(1/\text{time}\).
- \(B(t,T)\) has units of time, so \(B r_t\) is dimensionless as required in \(e^{-B r_t}\).
- \(\int_t^T r_s ds\) is dimensionless, which is why both \(D(t,T)\) and \(P(t,T)\) are pure numbers in \((0,1]\).
Solution:
\(B(0,5) = \frac{1-e^{-0.2(5)}}{0.2} = \frac{1-e^{-1}}{0.2} = \frac{1-0.368}{0.2} = 3.16\)
\(\ln A = (0.05-\frac{0.0001}{2(0.04)})(3.16-5) - \frac{0.0001(3.16)^2}{4(0.2)}\)
\(= (0.05-0.00125)(-1.84) - 0.00125 = -0.0897 - 0.00125 = -0.091\)
\(A = e^{-0.091} = 0.913\)
\(P(0,5) = 0.913 \cdot e^{-3.16(0.03)} = 0.913 \cdot e^{-0.0948} = 0.913 \cdot 0.910 = \boxed{0.831}\)
Major short-rate models:
- Vasicek: fully tractable affine benchmark; downside is possible negative rates.
- Hull-White 1F: fits today’s curve exactly via time-dependent drift; downside is extra calibration structure.
- CIR: positive-rate tendency and affine bond pricing; downside is less tractable than Vasicek.
- Black-Karasinski: lognormal short rate keeps rates positive; downside is weaker closed-form convenience.
- G2++: two-factor Gaussian model improves curve/correlation fit; downside is more parameters and model risk.
8. Merton Jump-Diffusion
Variables:
- \(\lambda\) = jump intensity (expected jumps per year)
- \(\mu_J\) = mean of log-jump size
- \(\sigma_J\) = volatility of log-jump size
Expected jump size: \(\mathbb{E}[e^Y-1] = e^{\mu_J+\sigma_J^2/2}-1 \equiv m\)
Over 3 months, expect \(\lambda T = 0.5\) jumps.
Diffusion contribution: \(100e^{0.1(0.25)-0.5(0.15)^2(0.25)+0.15\sqrt{0.25}(0.3)} = 107.25\)
With 1 jump \(Y=-0.08\): \(e^{-0.08}=0.923\)
Final: \(S_T = 107.25 \times 0.923 = \boxed{99.00}\)
Solution:
GBM: Would require \(\sigma W_T \approx -0.05\) over short time → implies huge implied vol spike, inconsistent with VIX behavior.
Jump: Natural explanation: single jump with \(Y \approx \ln(0.95) = -0.0513\). Matches observed "jump risk premium" in options.
Empirics: SPX experiences ~2-5 jumps/year (λ≈3), with average jump size -2% to -3%.
Solution: Must compensate for jump risk. Replace \(\mu\) with \(r - \lambda m\) where \(m = e^{\mu_J+\sigma_J^2/2}-1\).
Example: \(r=5\%, \lambda=3, \mu_J=-0.05, \sigma_J=0.1\)
\(m = e^{-0.05+0.005}-1 = e^{-0.045}-1 = -0.044\)
Adjusted drift: \(\mu^{\mathbb{Q}} = 0.05 - 3(-0.044) = 0.05 + 0.132 = 0.182\) (18.2%)!
The positive adjustment compensates for expected negative jumps.
Option Pricing: Weighted sum of BS prices over jump counts
9. GARCH(1,1)
Constraints:
- \(\omega > 0\) (guarantees positive variance)
- \(\alpha, \beta \geq 0\) (non-negativity)
- \(\alpha + \beta < 1\) (stationarity/covariance-stationarity)
Long-term variance: \(\bar{\sigma}^2 = \frac{\omega}{1-\alpha-\beta}\)
Persistence: \(\alpha+\beta\) close to 1 → high persistence (slow decay)
\(\sigma_t^2 = 0.000002 + 0.08(0.015)^2 + 0.90(0.0001)\)
\(= 0.000002 + 0.000018 + 0.00009 = \boxed{0.00011}\)
\(\sigma_t = \sqrt{0.00011} = \boxed{1.05\%}\) (daily)
\(\bar{\sigma}^2 = \frac{0.000002}{1-0.98} = 0.0001\) → \(\bar{\sigma} = 1.0\%\)
Solution:
Step 1: Calculate \(\sigma_{today}^2\) using yesterday's shock:
\(\varepsilon_{yesterday} = -0.02\) (assuming μ=0)
\(\sigma_{today}^2 = 0.000001 + 0.1(0.02)^2 + 0.85(0.012)^2\)
\(= 0.000001 + 0.00004 + 0.0001224 = 0.0001634\)
\(\sigma_{today} = 1.278\%\)
Step 2: Calculate \(\sigma_{tomorrow}^2\) using today's shock:
\(\varepsilon_{today} = 0.015\)
\(\sigma_{tomorrow}^2 = 0.000001 + 0.1(0.015)^2 + 0.85(0.0001634)\)
\(= 0.000001 + 0.0000225 + 0.0001389 = \boxed{0.0001624}\)
\(\sigma_{tomorrow} = \boxed{1.274\%}\)
Solution: Persistence = \(\alpha+\beta=0.99\) (very high!)
Shock decays as \((\alpha+\beta)^n\). For half-life:
\(0.99^n = 0.5 \Rightarrow n = \frac{\ln(0.5)}{\ln(0.99)} = \frac{-0.693}{-0.01005} = \boxed{69 \text{ days}}\)
Interpretation: A volatility shock takes ~3 months to decay by half - typical for equity markets.
Detailed SDE Derivations
1. Itô's Lemma - Complete Derivation
Setup: \(dX_t = \mu(X,t) dt + \sigma(X,t) dW_t\), find \(dY_t\) where \(Y_t = f(X_t,t)\)
Taylor Expansion to 2nd order:
Calculate \((dx)^2\):
Apply multiplication rules:
\(dt \cdot dW = o(dt)\) → 0
\((dW)^2 = dt\) (quadratic variation!)
Therefore: \((dx)^2 = \sigma^2 dt\)
Final result:
Key insight: The \(\frac{1}{2}\sigma^2\frac{\partial^2 f}{\partial x^2}\) term is unique to stochastic calculus!
Given: \(dS = \mu S dt + \sigma S dW\), find closed form for \(S_t\).
Step 1: Let \(Y = \ln S\), so \(f(S) = \ln S\)
\(\frac{\partial f}{\partial S} = \frac{1}{S}, \quad \frac{\partial^2 f}{\partial S^2} = -\frac{1}{S^2}\)
Step 2: Apply Itô:
\(d(\ln S) = \frac{1}{S}(\mu S dt + \sigma S dW) + \frac{1}{2}(-\frac{1}{S^2})(\sigma S)^2 dt\)
\(= \mu dt + \sigma dW - \frac{\sigma^2}{2}dt = (\mu - \frac{\sigma^2}{2})dt + \sigma dW\)
Step 3: Integrate:
\(\ln S_t - \ln S_0 = (\mu - \frac{\sigma^2}{2})t + \sigma W_t\)
\(S_t = S_0 \exp[(\mu - \frac{\sigma^2}{2})t + \sigma W_t]\) ✓
2. Black-Scholes PDE - Full Derivation
Objective: Derive the fundamental PDE for option pricing.
Step 1: Setup
Option: \(V(S,t)\) - value depends on \(S\) and \(t\)
Bond: \(dB = rB dt\)
Step 2: Apply Itô to \(V(S,t)\)
\(= \frac{\partial V}{\partial t}dt + \frac{\partial V}{\partial S}(\mu S dt + \sigma S dW) + \frac{1}{2}\frac{\partial^2 V}{\partial S^2}\sigma^2 S^2 dt\)
\(= \left[\frac{\partial V}{\partial t} + \mu S\frac{\partial V}{\partial S} + \frac{1}{2}\sigma^2 S^2\frac{\partial^2 V}{\partial S^2}\right]dt + \sigma S\frac{\partial V}{\partial S}dW\)
Step 3: Construct hedged portfolio
Choose \(\Delta = \frac{\partial V}{\partial S}\) (delta-hedge)
Step 4: Calculate \(d\Pi\)
\(= \left[\frac{\partial V}{\partial t} + \mu S\frac{\partial V}{\partial S} + \frac{1}{2}\sigma^2 S^2\frac{\partial^2 V}{\partial S^2}\right]dt + \sigma S\frac{\partial V}{\partial S}dW - \frac{\partial V}{\partial S}(\mu S dt + \sigma S dW)\)
\(= \left[\frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2 S^2\frac{\partial^2 V}{\partial S^2}\right]dt\)
Key observation: The \(dW\) terms cancel! Portfolio is riskless.
Step 5: No-arbitrage condition
\(d\Pi = r\Pi dt = r(V - \Delta S)dt = r\left(V - S\frac{\partial V}{\partial S}\right)dt\)
Step 6: Equate growth rates
Black-Scholes PDE:
Boundary conditions:
- Call: \(V(S,T) = \max(S-K, 0)\)
- Put: \(V(S,T) = \max(K-S, 0)\)
Remarkable facts:
- Drift \(\mu\) doesn't appear! (Risk-neutral pricing)
- Volatility \(\sigma\) is the only unknown parameter
- Linear PDE → can be solved analytically
Goal: Price zero-coupon bond \(P(r,t,T)\) when short rate \(r_t\) follows SDE.
General Short Rate SDE:
Step 1: Apply Itô to \(P(r,t,T)\)
\(= \frac{\partial P}{\partial t}dt + \frac{\partial P}{\partial r}(\mu dt + \sigma dW) + \frac{1}{2}\frac{\partial^2 P}{\partial r^2}\sigma^2 dt\)
\(= \left[\frac{\partial P}{\partial t} + \mu\frac{\partial P}{\partial r} + \frac{1}{2}\sigma^2\frac{\partial^2 P}{\partial r^2}\right]dt + \sigma\frac{\partial P}{\partial r}dW\)
Step 2: Form hedged portfolio
\(\Pi = P_1 - \frac{\partial P_1/\partial r}{\partial P_2/\partial r}P_2\)
This eliminates \(dW\) (similar to stock case).
Step 3: No-arbitrage condition
\(\frac{d\Pi}{\Pi} = r dt\)
After algebra (market price of interest rate risk \(\lambda(r,t)\) appears):
Bond Pricing PDE:
Boundary condition: \(P(r,T,T) = 1\) (bond pays $1 at maturity)
Key insight: \(\mu - \lambda\sigma\) is risk-adjusted drift under \(\mathbb{Q}\)
Example: Vasicek Model
Risk-neutral: \(dr = a(\bar{b}-r)dt + \sigma dW^{\mathbb{Q}}\) where \(\bar{b} = b - \lambda\sigma/a\)
PDE becomes: \(\frac{\partial P}{\partial t} + a(\bar{b}-r)\frac{\partial P}{\partial r} + \frac{1}{2}\sigma^2\frac{\partial^2 P}{\partial r^2} = rP\)
Solving approach - Affine Ansatz:
(Exponentially affine in \(r\) - works for Vasicek, CIR, many others)
Compute derivatives:
\(\frac{\partial P}{\partial t} = (A_t - B_t r)P\)
\(\frac{\partial P}{\partial r} = -BP\)
\(\frac{\partial^2 P}{\partial r^2} = B^2P\)
Substitute into PDE:
\((A_t - B_t r)P + a(\bar{b}-r)(-BP) + \frac{1}{2}\sigma^2 B^2P = rP\)
Divide by \(P\) and collect terms in \(r\):
\(A_t - B_t r - aB(\bar{b}-r) + \frac{1}{2}\sigma^2 B^2 = r\)
\(A_t - aB\bar{b} + \frac{1}{2}\sigma^2 B^2 + r(aB - B_t - 1) = 0\)
Since this must hold for all \(r\), separate:
Coefficient of \(r\): \(B_t = aB - 1\) with \(B(T,T)=0\)
Constant term: \(A_t = aB\bar{b} - \frac{1}{2}\sigma^2 B^2\) with \(A(T,T)=0\)
Solve ODE for \(B\):
Solution: \(B(\tau) = \frac{1-e^{-a\tau}}{a}\) where \(\tau = T-t\)
Check: At \(t=T\): \(\tau=0\), \(B=0\) ✓
As \(\tau \to \infty\): \(B \to 1/a\) (duration saturates)
Solve ODE for \(A\):
Integrate from \(t\) to \(T\) (or equivalently \(\tau\) from 0 to \(T-t\)):
\(A(\tau) = \int_0^\tau \left[a\bar{b}B(s) - \frac{\sigma^2}{2}B^2(s)\right]ds\)
After integration (lengthy but straightforward):
\(A(\tau) = (\bar{b} - \frac{\sigma^2}{2a^2})(B(\tau) - \tau) - \frac{\sigma^2 B^2(\tau)}{4a}\)
Final Vasicek Bond Price:
Interpretation:
- \(B(\tau)\): Interest rate sensitivity (duration-like)
- \(A(\tau)\): Adjustment for convexity and long-term average
- Yield: \(y(t,T) = -\frac{\ln P(r,t,T)}{T-t} = \frac{B r - \ln A}{\tau}\)
CIR Model - Similar but Harder:
Same affine ansatz works! But ODEs are more complex (Ricatti equation).
Key difference: \(\sigma(r) = \sigma\sqrt{r}\) makes \(B\) ODE nonlinear:
\(\frac{dB}{dt} = aB - 1 + \frac{1}{2}\sigma^2 B^2\) (Ricatti equation)
Solution involves \(\gamma = \sqrt{a^2 + 2\sigma^2}\) and hyperbolic functions.
3. Girsanov Theorem - Change of Measure (Extended)
Purpose: Transform between physical (\(\mathbb{P}\)) and risk-neutral (\(\mathbb{Q}\)) measures.
The Big Picture - Why do we need this?
But pricing by replication only depends on hedging, not investor preferences!
Solution: Work in a "risk-neutral world" where all assets grow at rate \(r\).
Setup: Under physical measure \(\mathbb{P}\), stock follows
Goal: Find measure \(\mathbb{Q}\) where discounted stock \(e^{-rt}S_t\) is a martingale.
Step 1: The Radon-Nikodym Derivative (Measure Change Tool)
\(Z_t = \exp\left(-\int_0^t \theta_s dW_s^{\mathbb{P}} - \frac{1}{2}\int_0^t \theta_s^2 ds\right)\)
where \(\theta_t\) = market price of risk (to be determined)
Why this form? The \(-\frac{1}{2}\theta^2\) term ensures \(\mathbb{E}^{\mathbb{P}}[Z_T]=1\) (valid probability measure).
\(d\ln Z_t = -\theta_t dW_t^{\mathbb{P}} - \frac{1}{2}\theta_t^2 dt\)
\(dZ_t = Z_t d\ln Z_t + \frac{1}{2}Z_t(d\ln Z_t)^2\)
\(= Z_t[-\theta_t dW_t^{\mathbb{P}} - \frac{1}{2}\theta_t^2 dt] + \frac{1}{2}Z_t\theta_t^2 dt\)
\(= -Z_t\theta_t dW_t^{\mathbb{P}}\) (drift cancelled!)
So \(Z_t\) is a \(\mathbb{P}\)-martingale with \(\mathbb{E}^{\mathbb{P}}[Z_t]=Z_0=1\) ✓
Step 2: Girsanov's Fundamental Result
the process
\(\tilde{W}_t = W_t^{\mathbb{P}} + \int_0^t \theta_s ds\)
is a \(\mathbb{Q}\)-Brownian motion.
Equivalently: \(dW_t^{\mathbb{Q}} = dW_t^{\mathbb{P}} + \theta_t dt\)
Intuition: Adding drift \(\theta_t dt\) to BM changes the "center" but preserves randomness!
Step 3: Transform Stock Dynamics to \(\mathbb{Q}\)
\(dS_t = \mu S_t dt + \sigma S_t dW_t^{\mathbb{P}}\)
Substitute \(dW_t^{\mathbb{P}} = dW_t^{\mathbb{Q}} - \theta_t dt\):
\(dS_t = \mu S_t dt + \sigma S_t (dW_t^{\mathbb{Q}} - \theta_t dt)\)
\(= \mu S_t dt - \sigma\theta_t S_t dt + \sigma S_t dW_t^{\mathbb{Q}}\)
\(= (\mu - \sigma\theta_t)S_t dt + \sigma S_t dW_t^{\mathbb{Q}}\)
Step 4: Determine \(\theta\) - Make Discounted Stock a Martingale
Apply Itô's lemma with \(f(S,t) = e^{-rt}S\):
\(\frac{\partial f}{\partial t} = -re^{-rt}S, \quad \frac{\partial f}{\partial S} = e^{-rt}, \quad \frac{\partial^2 f}{\partial S^2} = 0\)
\(dX_t = -re^{-rt}S_t dt + e^{-rt}dS_t\)
\(= -re^{-rt}S_t dt + e^{-rt}[(\mu-\sigma\theta_t)S_t dt + \sigma S_t dW_t^{\mathbb{Q}}]\)
\(= e^{-rt}S_t[(\mu-\sigma\theta_t-r)dt + \sigma dW_t^{\mathbb{Q}}]\)
For martingale: drift must be zero!
\(\mu - \sigma\theta_t - r = 0\)
\(\boxed{\theta_t = \frac{\mu - r}{\sigma}}\) (Market Price of Risk)
Interpretation of \(\theta = (\mu-r)/\sigma\):
- Numerator \(\mu-r\): excess return per unit time
- Denominator \(\sigma\): risk per unit time
- Ratio: Sharpe ratio! Compensation per unit risk
Step 5: Final Risk-Neutral Dynamics
\(dS_t = (\mu - \sigma \cdot \frac{\mu-r}{\sigma})S_t dt + \sigma S_t dW_t^{\mathbb{Q}}\)
\(= (\mu - \mu + r)S_t dt + \sigma S_t dW_t^{\mathbb{Q}}\)
Beautiful result: Under \(\mathbb{Q}\), ALL assets grow at risk-free rate \(r\)!
Step 6: Risk-Neutral Pricing Formula
\(e^{-rt}S_t = \mathbb{E}^{\mathbb{Q}}[e^{-rT}S_T|\mathcal{F}_t]\)
For derivative with payoff \(g(S_T)\):
Why is this powerful?
- No investor preferences (\(\mu\) disappeared!)
- Only depends on hedgeable quantities: \(S_0, r, \sigma, T\)
- Expectation easier than solving PDE
- Works for any payoff function \(g\)
Payoff: \(S_T - K\). Under \(\mathbb{Q}\): \(\mathbb{E}^{\mathbb{Q}}[S_T] = S_0 e^{rT}\)
Value: \(V = e^{-rT}(S_0 e^{rT} - K) = S_0 - Ke^{-rT}\)
Forward price: \(F = S_0 e^{rT}\) ✓
Payoff: $1 if \(S_T > K\), else $0
\(V = e^{-rT}\mathbb{Q}(S_T > K)\)
Under \(\mathbb{Q}\): \(\ln(S_T/S_0) \sim N((r-\sigma^2/2)T, \sigma^2 T)\)
\(\mathbb{Q}(S_T > K) = N(d_2)\) where \(d_2 = \frac{\ln(S_0/K)+(r-\sigma^2/2)T}{\sigma\sqrt{T}}\)
\(\boxed{V = e^{-rT}N(d_2)}\)
• \(\mathbb{P}\): Real-world probabilities (for forecasting)
• \(\mathbb{Q}\): Risk-neutral probabilities (for pricing)
Under \(\mathbb{Q}\), stock expected return = \(r\), not realistic!
But prices are correct because hedging adjusts for risk.
Connection to Fundamental Theorem of Asset Pricing:
Complete market: \(\mathbb{Q}\) is unique ⟺ every claim is replicable
Incomplete market: Multiple \(\mathbb{Q}\) exist → pricing bounds, not unique prices
4. Feynman-Kac Formula
Link between PDEs and expectations.
Theorem: If \(u(x,t)\) satisfies the PDE:
with terminal condition \(u(x,T) = g(x)\), then:
where \(dX_s = \mu(X_s,s)ds + \sigma(X_s,s)dW_s\)
\(\frac{\partial V}{\partial t} + rS\frac{\partial V}{\partial S} + \frac{1}{2}\sigma^2 S^2\frac{\partial^2 V}{\partial S^2} = rV\)
with \(V(S,T) = (S_T-K)^+\) gives
\(V(S,t) = \mathbb{E}^{\mathbb{Q}}[e^{-r(T-t)}(S_T-K)^+ | S_t=S]\)
This is how we get from PDE to expectation formula!
5. Martingale Representation Theorem
Statement: Any \(\mathcal{F}_t\)-martingale \(M_t\) can be written as
for some adapted process \(\phi_t\) with \(\int_0^T \phi_s^2 ds < \infty\).
Financial Interpretation:
- \(M_t\) = value of derivative
- \(\phi_t\) = replicating strategy (units of stock)
- The derivative can be replicated by dynamic trading!
Key Implication: In a complete market (enough Brownian motions), every contingent claim can be hedged.
\(\phi_t = \frac{\partial V}{\partial S} = N(d_1)\) (the delta!)
6. Ornstein-Uhlenbeck Solution - Integrating Factor
Goal: Solve \(dX_t = \theta(\mu - X_t)dt + \sigma dW_t\)
Step 1: Rewrite as
Step 2: Integrating factor method
\(e^{\theta t}dX_t + \theta e^{\theta t}X_t dt = \theta\mu e^{\theta t}dt + \sigma e^{\theta t}dW_t\)
Step 3: Recognize as product rule
\(d(e^{\theta t}X_t) = e^{\theta t}dX_t + \theta e^{\theta t}X_t dt\)
Step 4: Integrate from 0 to \(t\)
\(= \mu(e^{\theta t} - 1) + \sigma\int_0^t e^{\theta s}dW_s\)
Step 5: Solve for \(X_t\)
Properties:
- \(\mathbb{E}[X_t] = X_0 e^{-\theta t} + \mu(1-e^{-\theta t}) \to \mu\) as \(t \to \infty\)
- \(\text{Var}(X_t) = \frac{\sigma^2}{2\theta}(1-e^{-2\theta t}) \to \frac{\sigma^2}{2\theta}\)
7. CIR Bond Pricing - Affine Structure
Objective: Find zero-coupon bond price \(P(r,t,T)\) when \(dr = a(b-r)dt + \sigma\sqrt{r}dW\).
Ansatz: Assume affine form
Bond pricing PDE:
Substitute ansatz and simplify (lengthy algebra) to get ODEs:
Solutions (Ricatti equation for \(B\)):
where \(\gamma = \sqrt{a^2 + 2\sigma^2}\)
Note: Closed form exists, but complex. Numerical methods often used in practice.
8. GBM Exact Solution via Itô
Problem: Solve \(dS_t = \mu S_t dt + \sigma S_t dW_t\)
Method 1: Log transformation
\(f'(S) = 1/S, \quad f''(S) = -1/S^2\)
\(dX_t = \frac{1}{S_t}dS_t - \frac{1}{2S_t^2}(dS_t)^2\)
\(= \frac{1}{S_t}(\mu S_t dt + \sigma S_t dW_t) - \frac{1}{2S_t^2}\sigma^2 S_t^2 dt\)
\(= \mu dt + \sigma dW_t - \frac{\sigma^2}{2}dt\)
\(= (\mu - \frac{\sigma^2}{2})dt + \sigma dW_t\)
Integrate:
\(X_t - X_0 = (\mu - \frac{\sigma^2}{2})t + \sigma W_t\)
\(\ln S_t = \ln S_0 + (\mu - \frac{\sigma^2}{2})t + \sigma W_t\)
Exponentiate:
Verification: Apply Itô to \(S_t = S_0 e^{X_t}\):
\(= S_t[(\mu-\frac{\sigma^2}{2})dt + \sigma dW_t] + \frac{1}{2}S_t\sigma^2 dt\)
\(= \mu S_t dt + \sigma S_t dW_t\) ✓
9. MGF and Moments of Log-Normal
For: \(S_t = S_0 e^{(\mu-\sigma^2/2)t + \sigma W_t}\), find \(\mathbb{E}[S_t^n]\)
Step 1: Rewrite
Step 2: Use MGF of normal
\(\mathbb{E}[e^{aW_t}] = e^{a^2t/2}\) (MGF of normal)
Step 3: Calculate expectation
\(= S_0^n e^{n(\mu-\sigma^2/2)t} e^{n^2\sigma^2 t/2}\)
\(= S_0^n e^{n\mu t + n(n-1)\sigma^2 t/2}\)
Special cases:
- \(n=1\): \(\mathbb{E}[S_t] = S_0 e^{\mu t}\)
- \(n=2\): \(\mathbb{E}[S_t^2] = S_0^2 e^{2\mu t + \sigma^2 t}\)
- \(\text{Var}(S_t) = \mathbb{E}[S_t^2] - (\mathbb{E}[S_t])^2 = S_0^2 e^{2\mu t}(e^{\sigma^2 t}-1)\)
10. Multivariate Itô's Lemma
For: \(dX_i = \mu_i dt + \sigma_i dW_i\) with \(\langle dW_i, dW_j \rangle = \rho_{ij}dt\)
Find \(dY\) where \(Y = f(X_1, X_2, ..., X_n, t)\)
Formula:
Key multiplication rules:
\(dX_i dX_j = \sigma_i \sigma_j dW_i dW_j = \sigma_i \sigma_j \rho_{ij}dt\)
Example: Basket option
Basket: \(V = w_1 S_1 + w_2 S_2\)
\(dV = w_1 dS_1 + w_2 dS_2\) (no second-order terms since V is linear)
Variance: \(\text{Var}(dV) = w_1^2\sigma_1^2S_1^2dt + w_2^2\sigma_2^2S_2^2dt + 2w_1w_2\rho\sigma_1\sigma_2S_1S_2dt\)
11. Change of Numéraire - Extended Theory
Concept: Price assets in units of different assets (not just cash)
The Big Idea - Why Change Numéraire?
Problem: Some derivatives are naturally denominated in other assets!
Examples: Options on forwards, quanto options, swaptions
Solution: Choose numéraire that simplifies the math
What is a Numéraire?
- A numéraire is a positive, traded asset used as unit of account
- Like choosing currency: price stocks in USD vs EUR vs gold
- Examples: cash (\(e^{rt}\)), zero-coupon bonds, portfolios
Setup: Let \(N_t\) be numéraire (positive tradable asset with dynamics)
Fundamental Theorem: Under \(N\)-forward measure \(\mathbb{Q}^N\):
Intuition: Relative prices (in numéraire units) have no drift!
If stock/gold ratio drifts, you could arbitrage by trading stock vs gold.
No-arbitrage ⇒ relative prices must be martingales.
Step 1: Radon-Nikodym Derivative for Numéraire Change
\(\frac{d\mathbb{Q}^N}{d\mathbb{Q}}\Big|_{\mathcal{F}_T} = \frac{N_T/N_0}{B_T/B_0}\)
where \(B_t = e^{rt}\) is money market account
Simplified: \(\frac{d\mathbb{Q}^N}{d\mathbb{Q}} = \frac{N_T e^{-rT}}{N_0}\) (discounted numéraire)
Why this form? Under \(\mathbb{Q}\), \(e^{-rt}N_t\) is martingale, so this is a positive martingale with expectation 1.
Step 2: Pricing Formula Under New Measure
\(V_t = \mathbb{E}^{\mathbb{Q}}[e^{-r(T-t)}V_T|\mathcal{F}_t]\)
Under N-numéraire:
\(V_t = N_t \mathbb{E}^{\mathbb{Q}^N}\left[\frac{V_T}{N_T}\Big|\mathcal{F}_t\right]\)
Derivation of pricing formula:
\(\frac{V_t}{N_t} = \mathbb{E}^{\mathbb{Q}^N}\left[\frac{V_T}{N_T}\Big|\mathcal{F}_t\right]\)
Multiply both sides by \(N_t\):
\(V_t = N_t \mathbb{E}^{\mathbb{Q}^N}\left[\frac{V_T}{N_T}\Big|\mathcal{F}_t\right]\) ✓
Key Insight: No discounting needed! The numéraire \(N_t\) acts as the "discount factor"
Step 3: How to Change Brownian Motion
Under \(\mathbb{Q}^N\): \(dW_t^{\mathbb{Q}^N} = dW_t^{\mathbb{Q}} + \theta_t^N dt\)
where drift adjustment \(\theta_t^N\) depends on volatility of numéraire
Girsanov theorem for numéraire change:
\(\theta_t^N = -\sigma_N\) (negative of numéraire volatility)
Step 4: T-Forward Measure - The Most Important Case
Key property: \(P(T,T) = 1\) (pays $1 at maturity)
Why is this useful? Forward prices become martingales!
Proof that forward price is martingale under \(\mathbb{Q}^T\):
Under \(\mathbb{Q}^T\): Both \(S_t\) and \(P(t,T)\) are martingales when normalized by \(P(t,T)\)
\(\frac{S_t}{P(t,T)}\) is \(\mathbb{Q}^T\)-martingale
But \(\frac{S_t}{P(t,T)} = F(t,T)\) by definition!
Therefore: \(F(t,T)\) is \(\mathbb{Q}^T\)-martingale ✓
Implication: \(\mathbb{E}^{\mathbb{Q}^T}[F(T,T)|\mathcal{F}_t] = F(t,T)\)
Since \(F(T,T) = S_T\): \(\mathbb{E}^{\mathbb{Q}^T}[S_T|\mathcal{F}_t] = F(t,T)\)
Pricing formula for payoff depending on \(S_T\):
Payoff at \(T\): \(\max(F(T,T)-K, 0) = \max(S_T-K, 0)\)
Under \(\mathbb{Q}^T\), forward \(F(t,T)\) is martingale (no drift!)
If \(F(t,T)\) has vol \(\sigma_F\), then \(F(T,T) \sim \text{Lognormal}\)
Price: \(C_t = P(t,T)[F(t,T)N(d_1) - KN(d_2)]\)
where \(d_1 = \frac{\ln(F/K) + \sigma_F^2(T-t)/2}{\sigma_F\sqrt{T-t}}\)
This is Black's formula!
A caplet on LIBOR \(L(T,T+\delta)\) with strike \(K\):
Payoff (at \(T+\delta\)): \(\delta \max(L(T,T+\delta)-K, 0)\)
Natural numéraire: \(P(t,T+\delta)\)
Under \(\mathbb{Q}^{T+\delta}\), forward LIBOR is martingale!
Price: \(Caplet_t = \delta P(t,T+\delta)[F_L N(d_1) - KN(d_2)]\)
where \(F_L = \mathbb{E}^{\mathbb{Q}^{T+\delta}}[L(T,T+\delta)]\) (forward LIBOR)
Step 5: Annuity Measure for Swaptions
Swap rate: \(S(t) = \frac{P(t,T_0)-P(t,T_n)}{\sum_{i=1}^n \delta_i P(t,T_i)}\)
Annuity: \(A(t) = \sum_{i=1}^n \delta_i P(t,T_i)\) (present value of basis point)
Choose numéraire \(N_t = A(t)\):
Swaption pricing: Option to enter swap at rate \(K\)
Payoff: \(A(T_0)\max(S(T_0)-K, 0)\)
Price: \(V_t = A(t)[S(t)N(d_1) - KN(d_2)]\)
Again Black's formula, but for swap rate!
Step 6: Complete Example - Foreign Exchange
Domestic rate \(r_d\), foreign rate \(r_f\)
Standard (domestic cash numéraire):
\(dX_t = (r_d - r_f)X_t dt + \sigma X_t dW_t^{\mathbb{Q}^d}\)
(Interest rate parity gives drift)
Change to foreign cash numéraire:
In foreign terms, price is \(1/X_t\)
Under \(\mathbb{Q}^f\): \(d(1/X_t)\) has drift \((r_f-r_d)/X_t\)
Equivalently: \(dX_t = (r_f - r_d)X_t dt + \sigma X_t dW_t^{\mathbb{Q}^f}\)
(Drift reverses!)
FX Option pricing:
Domestic view: \(C = e^{-r_d T}[F_X N(d_1) - KN(d_2)]\)
where \(F_X = X_0 e^{(r_d-r_f)T}\) (FX forward)
Foreign view: Put on \(1/X\)
Same price by put-call parity across currencies!
Step 7: Stock Numéraire - "Quantos"
Example: \(\max(S_T^{USD}-K, 0)\) paid in EUR at fixed rate
Issue: Stock and FX are correlated!
If \(dS = \mu_S S dt + \sigma_S S dW_S\) and \(dX = \mu_X X dt + \sigma_X X dW_X\)
with \(\langle dW_S, dW_X \rangle = \rho dt\)
Quanto adjustment: Drift changes by \(-\rho\sigma_S\sigma_X\)
Under quanto measure: \(dS = (r - \rho\sigma_S\sigma_X)S dt + \sigma_S S dW\)
Japanese investor wants S&P 500 call settled in JPY at fixed rate.
\(\sigma_S = 20\%\) (S&P vol), \(\sigma_X = 10\%\) (USD/JPY vol), \(\rho = -0.5\)
Standard drift: \(r_{USD} = 5\%\)
Quanto drift: \(r_{quanto} = 0.05 - (-0.5)(0.20)(0.10) = 0.05 + 0.01 = 6\%\)
Effect: Use 6% instead of 5% in Black-Scholes!
Why? Negative correlation means when S&P rises, JPY typically strengthens,
partially hedging the payoff. Must adjust drift upward to compensate.
General Numéraire Change Formula
\(\frac{d\mathbb{Q}^{N_2}}{d\mathbb{Q}^{N_1}} = \frac{N_2(T)/N_2(0)}{N_1(T)/N_1(0)}\)
Drift adjustment for asset \(S\):
\(\mu^{N_2} = \mu^{N_1} - \rho_{S,N_2}\sigma_S\sigma_{N_2} + \rho_{S,N_1}\sigma_S\sigma_{N_1}\)
where \(\rho_{S,N}\) = correlation between \(S\) and numéraire \(N\)
Practical Algorithm for Pricing
Step 2: Choose numéraire that makes key quantity martingale
Step 3: Write payoff in terms of that numéraire
Step 4: Take expectation (now simple because martingale!)
Step 5: Multiply by numéraire value at \(t\)
1. Forgetting that \(N_T\) appears in both numerator and denominator
2. Using wrong volatility after numéraire change
3. Confusing forward price with futures price (different for rates!)
4. Not accounting for correlation in quanto adjustments
Why Numéraire Changes Matter in Practice
- Simplicity: Black's formula instead of complex PDE
- Market convention: Caps/floors quoted in terms of forward LIBOR
- Numerical stability: Monte Carlo with forward measure converges faster
- Hedging: Natural hedge ratios in the "right" units
| Numéraire | Measure | What's Martingale | Use Case |
|---|---|---|---|
| Cash \(e^{rt}\) | \(\mathbb{Q}\) | Discounted assets | Standard pricing |
| Zero-coupon \(P(t,T)\) | \(\mathbb{Q}^T\) | Forward prices | Options on forwards |
| Annuity \(\sum \delta_i P(t,T_i)\) | \(\mathbb{Q}^A\) | Swap rates | Swaptions |
| Stock \(S_t\) | \(\mathbb{Q}^S\) | Reciprocal stock | Exchange options |
| Foreign bond | \(\mathbb{Q}^f\) | FX rate | Currency options |
12. Quadratic Variation & Volatility Estimation
Definition: Quadratic variation of process \(X_t\) over \([0,T]\):
Key results:
- Brownian motion: \([W,W]_T = T\) (not random!)
- Itô process \(dX = \mu dt + \sigma dW\): \([X,X]_T = \int_0^T \sigma^2_s ds\)
- GBM: \([S,S]_T = \int_0^T \sigma^2 S_t^2 dt\)
Realized volatility estimator:
\(\hat{\sigma}^2 = \frac{1}{T}\sum_{i=1}^n (\ln S_{t_i} - \ln S_{t_{i-1}})^2\)
This converges to true \(\sigma^2\) as \(n \to \infty\)
Warning: Market microstructure noise biases this at ultra-high frequencies!
13. Lévy's Characterization of BM
Theorem: A continuous process \(M_t\) is a Brownian motion if and only if:
- \(M_0 = 0\)
- \(M_t\) is a martingale
- \([M,M]_t = t\) (quadratic variation)
Application: Proving time-changed BM
Define \(A_t = \int_0^t \sigma_s^2 ds\)
Then \(B_t = X_{A^{-1}_t}\) is a standard Brownian motion!
Implication: Any continuous martingale with right quadratic variation IS a Brownian motion (possibly time-changed)
14. Tanaka's Formula & Local Time
For: Brownian motion \(W_t\), Tanaka's formula gives:
where \(L_t^0\) is local time at 0 (time spent at zero)
Key insight: \(|W_t|\) is NOT an Itô process (not differentiable at 0)
Occupation density formula:
where \(L_t^x\) = local time at level \(x\)
Application: Pricing barrier options, path-dependent payoffs
15. Fokker-Planck Equation Derivation
For SDE: \(dX_t = \mu(X_t,t)dt + \sigma(X_t,t)dW_t\)
Find evolution of probability density \(p(x,t)\)
Method: Infinitesimal generator
Kolmogorov forward: \(\frac{\partial p}{\partial t} = \mathcal{L}^* p\)
where \(\mathcal{L}^*\) is adjoint operator
Fokker-Planck PDE:
Example: O-U process stationary distribution
\(0 = -\frac{d}{dx}[\theta(\mu-x)p] + \frac{\sigma^2}{2}\frac{d^2p}{dx^2}\)
Solution: \(p(x) = \mathcal{N}(\mu, \sigma^2/(2\theta))\) ✓
Classic Interview Problems
Problem 1: Path-Dependent Claim
Q: Under GBM with \(\mu=0.08, \sigma=0.3, S_0=100\), find \(\mathbb{E}[S_T^2]\) for \(T=1\).
Solution:
\(S_T = S_0 e^{(\mu-\sigma^2/2)T + \sigma W_T}\)
\(S_T^2 = S_0^2 e^{2(\mu-\sigma^2/2)T + 2\sigma W_T}\)
\(\mathbb{E}[S_T^2] = S_0^2 e^{2(\mu-\sigma^2/2)T}\mathbb{E}[e^{2\sigma W_T}]\)
Since \(W_T \sim N(0,T)\), \(\mathbb{E}[e^{aW_T}] = e^{a^2T/2}\)
\(\mathbb{E}[S_T^2] = 100^2 e^{2(0.08-0.045)(1)}e^{2\sigma^2 T} = 10000 e^{0.07}e^{0.18}\)
\(= 10000 e^{0.25} = \boxed{12,840}\)
Alternative: \(\mathbb{E}[S_T^2] = S_0^2 e^{2\mu T + \sigma^2 T} = 10000 e^{0.16+0.09} = 10000(1.284)\)
Problem 2: First Passage Time
Q: Stock at $100, volatility 20%. What's probability it hits $120 before $90?
Solution: Use Gambler's Ruin formula for BM with drift.
Let \(X = \ln S\), \(X_0 = \ln 100\), barriers at \(a=\ln 90, b=\ln 120\)
Under risk-neutral measure (\(\mu=r\)), assume \(r \approx 0\) for simplicity (symmetric)
For symmetric BM: \(P(\text{hit } b \text{ first}) = \frac{X_0 - a}{b - a}\)
\(= \frac{\ln(100/90)}{\ln(120/90)} = \frac{0.1054}{0.2877} = \boxed{0.366}\)
With drift \(\mu\): \(P = \frac{e^{2\mu a/\sigma^2} - e^{2\mu X_0/\sigma^2}}{e^{2\mu a/\sigma^2} - e^{2\mu b/\sigma^2}}\)
Problem 3: Variance of Integrated Process
Q: For O-U process \(dX = \theta(\mu-X)dt + \sigma dW\), find \(\text{Var}\left(\int_0^T X_t dt\right)\).
Solution: This is complex. Key steps:
1. Use \(\text{Var}(I) = \int_0^T\int_0^T \text{Cov}(X_s, X_t)ds\,dt\)
2. O-U covariance: \(\text{Cov}(X_s,X_t) = \frac{\sigma^2}{2\theta}e^{-\theta|t-s|}(1-e^{-2\theta\min(s,t)})\)
3. Double integral gives (for large \(T\)):
\(\text{Var}(I) \approx \frac{\sigma^2 T}{2\theta^2}\)
Intuition: Integrated variance grows linearly in \(T\), unlike BM which gives \(T^3/3\).
Problem 4: Risk-Neutral vs Physical Measure
Q: Stock has \(\mu=12\%, \sigma=25\%, r=5\%\). Find 1-year forward price and compare expected values under both measures.
Solution:
Forward price (no-arbitrage): \(F = S_0 e^{rT} = 100e^{0.05} = 105.13\)
Under \(\mathbb{P}\): \(\mathbb{E}^{\mathbb{P}}[S_T] = S_0 e^{\mu T} = 100e^{0.12} = 112.75\)
Under \(\mathbb{Q}\): \(\mathbb{E}^{\mathbb{Q}}[S_T] = S_0 e^{rT} = 105.13\)
Key insight: Forward price equals risk-neutral expectation, NOT physical expectation!
Market price of risk: \(\lambda = \frac{\mu-r}{\sigma} = \frac{0.07}{0.25} = 0.28\)
Problem 5: Time-Changed Brownian Motion
Q: If \(X_t = W_{A_t}\) where \(dA_t = v_t dt\) (random time change), what process does \(X_t\) follow?
Solution: By time-change theorem (Dubins-Schwarz):
\(dX_t = \sqrt{v_t}dW_t\)
This is local volatility structure!
If \(v_t\) follows another SDE (e.g., CIR), we get stochastic volatility.
Application: Heston model can be viewed as time-changed BM where clock runs at random speed determined by variance process.
Problem 6: Reflection Principle
Q: BM starts at 0. Find \(P(\max_{0 \leq t \leq T} W_t > a)\).
Solution: Use reflection principle:
\(P(M_T > a) = P(W_T > a) + P(W_T < -a) = 2P(W_T > a)\)
where \(M_T = \max_{0 \leq t \leq T} W_t\)
\(= 2\Phi\left(-\frac{a}{\sqrt{T}}\right) = 2N\left(-\frac{a}{\sqrt{T}}\right)\)
Example: \(T=1, a=1.5\)
\(P(M_1 > 1.5) = 2N(-1.5) = 2(0.0668) = \boxed{0.1336}\)
Application: Pricing barrier options!
Problem 7: Digital Option Pricing
Q: Price a digital call paying $100 if \(S_T > K\), else $0. Given: \(S_0=50, K=55, r=0.04, \sigma=0.25, T=0.5\).
Solution: Digital call = \(Qe^{-rT}N(d_2)\) where \(Q=100\)
\(d_2 = \frac{\ln(50/55)+(0.04-0.25^2/2)(0.5)}{0.25\sqrt{0.5}}\)
\(= \frac{-0.0953 + 0.0044}{0.1768} = -0.514\)
\(N(-0.514) = 0.3036\)
Price = \(100 e^{-0.04(0.5)}(0.3036) = 100(0.9802)(0.3036) = \boxed{29.76}\)
Check: Regular call would be worth more since it pays \(S_T-K\) not just fixed amount.
Problem 8: Down-and-Out Call
Q: Price a down-and-out call: payoff = \(\max(S_T-K,0)\) if \(S_t > H\) for all \(t\), else 0. Given: \(S_0=100, K=100, H=90, r=0.05, \sigma=0.3, T=1\).
Solution: Use closed-form barrier formula:
\(C_{DO} = C_{vanilla} - (H/S_0)^{2\lambda}C(H^2/S_0, K, ...)\)
where \(\lambda = (r+\sigma^2/2)/\sigma^2 = (0.05+0.045)/0.09 = 1.056\)
Step 1: Regular call with \(S_0=100, K=100\):
\(d_1=0.316, d_2=0.016, C = 100(0.624)-100e^{-0.05}(0.506) = \boxed{14.26}\)
Step 2: Adjustment term with \(S_0=H^2/S_0=81, K=100\):
\(d_1=-0.464, d_2=-0.764, C = 81(0.321)-100e^{-0.05}(0.222) = 4.90\)
Step 3: \(C_{DO} = 14.26 - (0.9)^{2(1.056)}(4.90) = 14.26 - 3.73 = \boxed{10.53}\)
Barrier option worth ~26% less than vanilla!
Problem 9: Asian Option via MC
Q: Price arithmetic Asian call: payoff = \(\max(\bar{S}-K, 0)\) where \(\bar{S} = \frac{1}{n}\sum_{i=1}^n S_{t_i}\). Given: \(S_0=100, K=100, r=0.05, \sigma=0.2, T=1, n=12\) (monthly).
Solution: No closed form! Use Monte Carlo:
Step 1: Generate path: \(S_{t_i} = S_{t_{i-1}}e^{(r-\sigma^2/2)\Delta t + \sigma\sqrt{\Delta t}Z_i}\)
Step 2: Calculate average: \(\bar{S} = \frac{1}{12}\sum_{i=1}^{12} S_{i/12}\)
Step 3: Payoff: \(\max(\bar{S}-100, 0)\)
Step 4: Repeat 10,000 times, average and discount
Typical result: \(C_{Asian} \approx 5.2\) vs regular call \(\approx 8.0\)
Why cheaper? Averaging reduces volatility: \(\sigma_{avg} \approx \sigma/\sqrt{n}\)
Problem 10: Stochastic Vol vs Jump
Q: Market shows implied vol smile: 90% strike has 25% vol, ATM 20% vol, 110% strike 18% vol. Is this stochastic vol or jumps?
Solution - Analyze pattern:
Observed: Downward sloping (skew) - OTM puts expensive
Stochastic vol (Heston): Symmetric smile due to \(\rho<0\) creates skew. ✓ Matches!
Jump-diffusion: Negative jumps create left skew. ✓ Also matches!
Distinguishing test: Look at term structure:
- Stoch vol: Smile flattens with maturity (mean reversion)
- Jumps: Smile persists or steepens (jumps compound)
Answer: Need more maturities! But equity markets typically show BOTH effects, with stoch vol dominant for short-term, jumps for long-term.
Problem 11: Correlation from Options
Q: Two stocks: \(\sigma_1=0.3, \sigma_2=0.25\). Basket option on \(S_1+S_2\) has implied vol 0.24. What's correlation?
Solution: For equal weights \(w_1=w_2=0.5\):
\(\sigma_{basket}^2 = w_1^2\sigma_1^2 + w_2^2\sigma_2^2 + 2w_1w_2\rho\sigma_1\sigma_2\)
\(0.24^2 = 0.25(0.09) + 0.25(0.0625) + 0.5\rho(0.3)(0.25)\)
\(0.0576 = 0.0225 + 0.0156 + 0.0375\rho\)
\(0.0195 = 0.0375\rho\)
\(\rho = 0.0195/0.0375 = \boxed{0.52}\)
Interpretation: Positive correlation reduces diversification benefit.
Check: If \(\rho=1\): \(\sigma_{basket}=0.275\). If \(\rho=0\): \(\sigma_{basket}=0.196\). Our 0.24 is between, so \(\rho=0.52\) makes sense!
Problem 12: American Put via Binomial
Q: Price 6-month American put, \(S_0=50, K=52, r=0.06, \sigma=0.4\). Use 2-step tree.
Solution: \(\Delta t = 0.25\)
Step 1: Calculate parameters
\(u = e^{\sigma\sqrt{\Delta t}} = e^{0.4\sqrt{0.25}} = e^{0.2} = 1.221\)
\(d = 1/u = 0.819\)
\(q = \frac{e^{r\Delta t}-d}{u-d} = \frac{e^{0.015}-0.819}{0.402} = \frac{0.196}{0.402} = 0.488\)
Step 2: Build tree
Node values: \(S_{uu}=74.5, S_{ud}=50, S_{dd}=33.5\)
Terminal payoffs: \(P_{uu}=0, P_{ud}=2, P_{dd}=18.5\)
Step 3: Backward induction at \(t=0.25\)
Up node (\(S_u=61.05\)): Hold = \(e^{-0.015}[0.488(0)+0.512(2)] = 1.00\)
Exercise = \(\max(52-61.05, 0) = 0\). Choose hold: \(P_u=1.00\)
Down node (\(S_d=40.95\)): Hold = \(e^{-0.015}[0.488(2)+0.512(18.5)] = 10.35\)
Exercise = \(52-40.95 = 11.05\). Early exercise optimal! \(P_d=11.05\)
Step 4: Initial value
\(P_0 = e^{-0.015}[0.488(1.00)+0.512(11.05)] = \boxed{6.06}\)
Compare: European put ≈ 5.80, so early exercise premium ≈ $0.26
Problem 13: Volatility Surface Arbitrage
Q: Implied vol surface shows: 95 strike: 22%, 100 strike: 20%, 105 strike: 23%. Is this arbitrage-free?
Solution: Check butterfly spread condition:
Butterfly = \(C(95) - 2C(100) + C(105) \geq 0\)
Using BS with each vol (assume \(S=100, r=0.05, T=1\)):
\(C(95, \sigma=0.22) = 12.54\)
\(C(100, \sigma=0.20) = 10.45\)
\(C(105, \sigma=0.23) = 9.38\)
Butterfly = \(12.54 - 2(10.45) + 9.38 = 1.02 > 0\) ✓
BUT: Also check calendar spread and density conditions!
Density condition: \(\frac{\partial^2 C}{\partial K^2} \geq 0\)
Numerically: \(\frac{12.54-2(10.45)+9.38}{5^2} = 0.0408 > 0\) ✓
Conclusion: Surface passes basic arbitrage checks, but full check needs all strikes and maturities!
Problem 14: Delta Hedging P&L
Q: Sold 1 call @ $8 (BS value), delta=0.6. Stock moves $50→$52. Rehedge costs? Assume \(\Gamma=0.03, r=0\) for simplicity.
Solution:
Initial hedge: Buy 0.6 shares @ $50 = $30 outlay
After move: Option value change by Itô:
\(\Delta V = \Delta(S-S_0) + \frac{1}{2}\Gamma(S-S_0)^2 = 0.6(2) + 0.5(0.03)(4) = 1.26\)
New option value ≈ $9.26 (liability up $1.26)
Stock position value: \(0.6 \times 52 = 31.20\) (asset up $1.20)
P&L so far: \(1.20 - 1.26 = -0.06\) (small loss)
New delta: \(\Delta_{new} = 0.6 + \Gamma(S-S_0) = 0.6 + 0.03(2) = 0.66\)
Rehedge: Buy additional 0.06 shares @ $52 = $3.12
Total P&L: \(-0.06 - 3.12 + 3.12 = \boxed{-0.06}\)
Source of loss: Gamma! The \(\frac{1}{2}\Gamma(dS)^2\) term. This is theta decay in discrete time.
Problem 15: Vol Arbitrage Trade
Q: GARCH(1,1) forecasts \(\sigma_{tomorrow}=25\%\) (annualized). Market ATM option (30 days) trades at 30% implied vol. Design a trade.
Solution - Realized vs Implied:
View: Realized vol will be ~25% but market prices 30% → market overpricing!
Trade: SELL straddle (sell call + sell put at ATM)
Hedge: Delta-hedge daily
P&L source:
- Collect premium based on 30% vol
- Pay hedging costs based on realized 25% vol
- Profit ≈ \(\frac{1}{2}S^2\Gamma(\sigma_{implied}^2-\sigma_{realized}^2)T\)
Risk: If realized vol spikes above 30%, lose money!
Size: For \(S=100, T=30/365\), theoretical edge ≈ $0.80 per straddle
Reality check: Transaction costs, gamma risk, vol clustering → actual edge smaller
Problem 16: Quanto Options
Q: Japanese investor wants payoff \(\max(S_T^{USD}-K,0)\) settled in JPY at fixed rate 110 JPY/USD. Stock vol 25%, FX vol 10%, correlation -0.5. Adjust drift?
Solution - Quanto adjustment:
Standard: \(dS = rS dt + \sigma_S S dW_S\)
Quanto: Need to adjust for correlation between asset and FX
Adjustment: Replace drift \(r\) with \(r - \rho\sigma_S\sigma_{FX}\)
\(r_{quanto} = r - (-0.5)(0.25)(0.10) = r + 0.0125\)
Effect: Positive correlation adjustment! (negative \(\rho\) increases effective drift)
Intuition: When stock goes up (good for call), typically JPY strengthens (bad for conversion). Negative correlation provides natural hedge, making option less valuable → need higher drift to compensate.
Example: If \(r_{USD}=5\%\), use \(r_{quanto}=6.25\%\) in BS formula
Problem 17: VIX Futures Pricing
Q: VIX spot = 20. 1-month VIX future = 22. Is this contango or backwardation? What does it imply?
Solution:
Structure: Future > Spot → Contango
Typical regime: VIX mean-reverts to ~20. Current low vol → expect increase
Cost of carry: VIX futures don't follow standard cost-of-carry!
\(F \neq S_0e^{rT}\) because VIX is not a traded asset
Instead: \(F = \mathbb{E}^{\mathbb{Q}}[VIX_T]\) where \(\mathbb{Q}\) includes variance risk premium
Trade implication:
- Long VIX futures in contango bleeds ~5-10% monthly (negative roll)
- Short VIX earns roll yield but has massive tail risk (VIX spikes)
Historical: VIX futures in contango ~70% of time. Backwardation during crises.
Feb 2018 "Volmageddon": VIX spiked 20→50, short vol products lost 80-95%!
Problem 18: Credit-Equity Hybrid
Q: Company has debt \(D=80M\), equity \(E=40M\), equity vol 40%. Estimate default probability in 1 year. Use Merton model.
Solution - Merton's approach:
Step 1: Firm value \(V = E + D = 120M\)
Step 2: Equity as call option
\(E = V N(d_1) - De^{-rT}N(d_2)\) and \(\sigma_E = \frac{VN(d_1)}{E}\sigma_V\)
This gives 2 equations, 2 unknowns: \(V, \sigma_V\)
Step 3: Solve iteratively (assume \(r=5\%\)):
Initial guess: \(\sigma_V = \sigma_E \times E/V = 0.4 \times 40/120 = 0.133\)
After iteration: \(V_0 \approx 124M, \sigma_V \approx 15\%\)
Step 4: Distance to default
\(DD = \frac{\ln(V/D) + (r-\sigma_V^2/2)T}{\sigma_V\sqrt{T}} = \frac{\ln(124/80)+0.039}{0.15} = 2.47\)
Step 5: Default probability
\(PD = N(-DD) = N(-2.47) = \boxed{0.68\%}\)
Reality check: BBB-rated bonds have ~1% 1-year PD, so this seems reasonable.
Problem 19: Interest Rate Parity Check
Q: USD rate 5%, EUR rate 3%, spot EUR/USD = 1.10, 1-year forward = 1.12. Arbitrage?
Solution - Covered Interest Parity:
Theory: \(F = S e^{(r_d - r_f)T}\)
\(F_{theory} = 1.10 \times e^{(0.05-0.03)(1)} = 1.10 \times 1.0202 = 1.1222\)
Market: \(F_{market} = 1.12 < 1.1222\)
Arbitrage exists! Forward is cheap relative to spot.
Trade:
- Borrow 1.10M USD at 5%
- Convert to 1M EUR at spot 1.10
- Invest 1M EUR at 3%
- Sell forward: lock in 1.12 rate to convert back
Payoffs at \(T=1\):
- Owe: 1.10M × 1.05 = 1.155M USD
- Have: 1M × 1.03 = 1.03M EUR → 1.03 × 1.12 = 1.1536M USD
- Profit = 1.1536 - 1.155 = -$1,400 LOSS!
Wait! Made error. Let me recalculate forward sale at 1.12:
Actually arbitrage goes OTHER way - forward overpriced!
Correct trade: Opposite direction for profit of ~$2,200
Problem 20: Path Integral for Transition
Q: For O-U process \(dX = \theta(\mu-X)dt + \sigma dW\), find transition density \(p(x,t|x_0,0)\).
Solution - Conditional distribution:
From earlier solution: \(X_t = x_0e^{-\theta t} + \mu(1-e^{-\theta t}) + \sigma e^{-\theta t}\int_0^t e^{\theta s}dW_s\)
Mean: \(m_t = x_0e^{-\theta t} + \mu(1-e^{-\theta t})\)
Variance: \(v_t = \frac{\sigma^2}{2\theta}(1-e^{-2\theta t})\)
Distribution: \(X_t | X_0=x_0 \sim \mathcal{N}(m_t, v_t)\)
Transition density:
Check limits:
- \(t \to 0\): \(v_t \to 0, m_t \to x_0\) → \(\delta(x-x_0)\) ✓
- \(t \to \infty\): \(m_t \to \mu, v_t \to \sigma^2/(2\theta)\) → stationary distribution ✓
Interview Success Strategies
1. Always check units! Vol daily vs annual, returns vs log-returns
2. Know your measures: Is this \(\mathbb{P}\) or \(\mathbb{Q}\)? Huge difference!
3. Itô's formula: Master this - it's 80% of quant interviews
4. Martingale test: Quick check if you've made mistakes
5. Feller condition: Always verify for CIR/Heston
6. Put-Call Parity: Use to check option pricing answers
7. Greeks sum: \(\Theta + rS\Delta + \frac{1}{2}\sigma^2S^2\Gamma = rV\)
8. Dimensional analysis: Does your answer have right units?
9. Limiting cases: Check \(t \to 0, T \to \infty, \sigma \to 0\), etc.
10. Numerical sanity: Does the number make economic sense?