Pricing a European Call Option Using Girsanov's Theorem¶

We price a European call option on a stock whose price follows a geometric Brownian motion. We apply Girsanov's theorem to change from the real-world measure $\mathbb{P}$ to the risk-neutral measure $\mathbb{Q}$ under which the discounted asset price is a martingale.

Toy mechanism: reweighting a coin

Before the stochastic-calculus machinery, the idea of a measure change is already visible in one line of arithmetic. Consider a single coin under two probability measures: $\mathbb{P}(H) = p$ and $\mathbb{Q}(H) = q$. Any expectation under $\mathbb{Q}$ can be written as a $\mathbb{P}$-expectation reweighted by the likelihood ratio $L = \mathbb{Q}/\mathbb{P}$, i.e. $L(H) = q/p$, $L(T) = (1-q)/(1-p)$:

\[ \mathbb{E}^{\mathbb{Q}}[X] = p \cdot \frac{q}{p}\,X(H) + (1-p)\cdot \frac{1-q}{1-p}\,X(T) = \mathbb{E}^{\mathbb{P}}[L X]. \]

Girsanov's theorem is the continuous-time analogue: the Radon–Nikodym derivative $Z_T = d\mathbb{Q}/d\mathbb{P}$ plays the role of $L$, and the only nontrivial content is computing it explicitly for a Brownian filtration. The whole derivation below is just "pick $L$ so that the discounted stock has zero drift, then compute the call's $\mathbb{Q}$-expectation."

Where this fits

Roadmap row(s): Replication, PDE, Measure change.
Builds on: The Black-Scholes formula (the result to be derived).
Feeds into: Probabilistic interpretation (the Girsanov drift shift $d_1 = d_2 + \sigma\sqrt{T}$) and Properties and bounds (PDE structure underlying the Greeks).

Prerequisites

This section assumes familiarity with Itô's formula, exponential (Doléans-Dade) martingales, equivalent measure changes, and the role of filtrations. Readers without this background may treat the derivation as a structural map and revisit after the stochastic-calculus chapter.

1 Model Setup and Measure Change¶

Recall (see § Risk-Neutral Measure and § GBM SDE Solution): starting from $dS_t = \mu S_t\,dt + \sigma S_t\,dW_t$ under $\mathbb{P}$, Girsanov with market price of risk $\theta = (\mu - r)/\sigma$ and Radon–Nikodym derivative $Z_T = \exp(-\theta W_T - \tfrac{1}{2}\theta^2 T)$ produces $\mathbb{Q}$ under which $\tilde W_t = W_t + \theta t$ is Brownian, so

\[ dS_t = rS_t\,dt + \sigma S_t\,d\tilde W_t, \qquad S_T = S_0\exp\!\left((r - \tfrac{1}{2}\sigma^2)T + \sigma\tilde W_T\right), \qquad \log S_T \sim \mathcal N\!\left(\log S_0 + (r-\tfrac{1}{2}\sigma^2)T,\,\sigma^2 T\right). \]

The Novikov condition $\mathbb{E}^{\mathbb{P}}[\exp(\tfrac{1}{2}\theta^2 T)] < \infty$ is automatic here since $\theta$ is constant (Exercise 7). The goal is to evaluate

\[ C_0 = \mathbb{E}^{\mathbb{Q}}\!\left[e^{-rT}(S_T - K)^+\right]. \]

2 Step: Compute the Call Option Price¶

Section goal: evaluating $\mathbb{E}^{\mathbb{Q}}[(S_T - K)^+]$ in closed form by Gaussian integration.

Define:

\[ d_1 := \frac{\log(S_0/K) + (r + \frac{1}{2} \sigma^2)T}{\sigma \sqrt{T}}, \quad d_2 := d_1 - \sigma \sqrt{T} \]

Then the Black–Scholes formula gives:

\[ C_0 = S_0 \mathcal{N}(d_1) - K e^{-rT} \mathcal{N}(d_2) \]

where $\mathcal{N}(\cdot)$ is the standard normal cumulative distribution function.

Interpretation of the Terms¶

$\mathcal{N}(d_2) = \mathbb{Q}(S_T > K)$: the risk-neutral probability that the option expires in the money.
$\mathcal{N}(d_1) = \mathbb{Q}^S(S_T > K)$: the probability under the stock numéraire measure that the option expires in the money.

Derivation Sketch¶

Split the discounted call payoff into a stock-receipt term and a strike-payment term:

\[ C_0 = e^{-rT}\mathbb{E}^{\mathbb{Q}}\!\left[(S_T - K)^+\right] = \underbrace{e^{-rT}\mathbb{E}^{\mathbb{Q}}\!\left[S_T\,\mathbf{1}_{\{S_T > K\}}\right]}_{\text{stock term}} - \underbrace{Ke^{-rT}\,\mathbb{Q}(S_T > K)}_{\text{strike term}} \]

The strike term is immediate from the lognormal distribution of $S_T$ under $\mathbb{Q}$:

\[ Ke^{-rT}\,\mathbb{Q}(S_T > K) = Ke^{-rT}\,\mathcal{N}(d_2) \]

For the stock term, change numéraire from the money-market account to the stock itself. Define the stock numéraire measure $\mathbb{Q}^S$ by

\[ \frac{d\mathbb{Q}^S}{d\mathbb{Q}} = \frac{e^{-rT}S_T}{S_0} \]

(the discounted terminal stock value, normalised so that $\mathbb{E}^{\mathbb{Q}}[d\mathbb{Q}^S/d\mathbb{Q}] = 1$ by the martingale property of the discounted stock). Pulling this weight into the expectation,

\[ e^{-rT}\mathbb{E}^{\mathbb{Q}}\!\left[S_T\,\mathbf{1}_{\{S_T > K\}}\right] = S_0\,\mathbb{E}^{\mathbb{Q}^S}\!\left[\mathbf{1}_{\{S_T > K\}}\right] = S_0\,\mathbb{Q}^S(S_T > K) = S_0\,\mathcal{N}(d_1) \]

The final identity $\mathbb{Q}^S(S_T > K) = \mathcal{N}(d_1)$ holds because Girsanov shifts the stock drift from $r$ to $r + \sigma^2$ under $\mathbb{Q}^S$, which replaces $d_2 = \frac{\log(S_0/K) + (r - \frac{1}{2}\sigma^2)T}{\sigma\sqrt{T}}$ by $d_1 = \frac{\log(S_0/K) + (r + \frac{1}{2}\sigma^2)T}{\sigma\sqrt{T}}$ in the standardisation (the full drift-shift calculation is in Probabilistic interpretation). Combining,

\[ C_0 = S_0\,\mathcal{N}(d_1) - Ke^{-rT}\,\mathcal{N}(d_2) \]

Each term pairs with its natural measure: $\mathcal{N}(d_2)$ is the exercise probability under the money-market numéraire $\mathbb{Q}$, and $\mathcal{N}(d_1)$ is the exercise probability under the stock numéraire $\mathbb{Q}^S$. Forcing both integrals to stay under $\mathbb{Q}$ also works — it just routes through a Gaussian-integral completing-the-square computation that produces $\mathcal{N}(d_1)$ algebraically rather than measure-theoretically; that route is carried out in Exercise 8.

Key insight. The real-world drift $\mu$ does not appear: Girsanov absorbs $\mu - r$ into the change of measure, so option prices depend only on $r, \sigma, S_0, K, T$. The Feynman–Kac bridge $V(t,S) = e^{-r(T-t)}\mathbb{E}^{\mathbb{Q}}[\Phi(S_T)\mid S_t = S]$ that links this risk-neutral expectation to the BS PDE is developed in § Risk-Neutral Measure and § Feynman-Kac.

Exercises¶

Exercise 1. Suppose the real-world drift is $\mu = 12\%$, the risk-free rate is $r = 3\%$, and the volatility is $\sigma = 20\%$. Compute the market price of risk $\theta = \frac{\mu - r}{\sigma}$ and write down the Radon–Nikodym derivative $Z_T$ explicitly for $T = 1$ year.

Solution to Exercise 1

The market price of risk is:

\[ \theta = \frac{\mu - r}{\sigma} = \frac{0.12 - 0.03}{0.20} = \frac{0.09}{0.20} = 0.45 \]

The Radon–Nikodym derivative for $T = 1$ is:

\[ Z_T = \exp\left(-\theta W_T - \frac{1}{2}\theta^2 T\right) = \exp\left(-0.45\, W_1 - \frac{1}{2}(0.45)^2 \cdot 1\right) \]

\[ = \exp\left(-0.45\, W_1 - 0.10125\right) \]

where $W_1 \sim \mathcal{N}(0, 1)$ under $\mathbb{P}$. This random variable converts $\mathbb{P}$-expectations to $\mathbb{Q}$-expectations: $\mathbb{E}^{\mathbb{Q}}[X] = \mathbb{E}^{\mathbb{P}}[Z_T X]$ for any $\mathcal{F}_T$-measurable $X$.

Note that $\theta = 0.45$ represents the excess return per unit of risk. The Sharpe ratio of the stock is $0.45$, meaning investors earn $0.45$ units of excess return for each unit of volatility risk they bear.

Exercise 2. Starting from the real-world SDE $dS_t = \mu S_t \, dt + \sigma S_t \, dW_t$ and the Girsanov change $\tilde{W}_t = W_t + \theta t$, substitute $dW_t = d\tilde{W}_t - \theta \, dt$ to verify that the risk-neutral SDE becomes $dS_t = r S_t \, dt + \sigma S_t \, d\tilde{W}_t$.

Solution to Exercise 2

Starting with the real-world SDE: $dS_t = \mu S_t\, dt + \sigma S_t\, dW_t$.

The Girsanov change defines $\tilde{W}_t = W_t + \theta t$ where $\theta = \frac{\mu - r}{\sigma}$. Therefore:

\[ dW_t = d\tilde{W}_t - \theta\, dt \]

Substituting into the SDE:

\[ dS_t = \mu S_t\, dt + \sigma S_t(d\tilde{W}_t - \theta\, dt) \]

\[ = \mu S_t\, dt + \sigma S_t\, d\tilde{W}_t - \sigma\theta S_t\, dt \]

\[ = (\mu - \sigma\theta) S_t\, dt + \sigma S_t\, d\tilde{W}_t \]

Substituting $\theta = \frac{\mu - r}{\sigma}$:

\[ \mu - \sigma\theta = \mu - \sigma \cdot \frac{\mu - r}{\sigma} = \mu - (\mu - r) = r \]

Therefore:

\[ dS_t = r S_t\, dt + \sigma S_t\, d\tilde{W}_t \]

This is the risk-neutral SDE, confirming that under $\mathbb{Q}$, the stock grows at the risk-free rate $r$ instead of the real-world drift $\mu$.

Exercise 3. Under the risk-neutral measure $\mathbb{Q}$, write the explicit solution for $S_T$ and verify that $\log S_T$ is normally distributed. State the mean and variance of $\log S_T$ under $\mathbb{Q}$ and confirm that the expected value $\mathbb{E}^{\mathbb{Q}}[S_T] = S_0 e^{rT}$.

Solution to Exercise 3

Under $\mathbb{Q}$, the SDE $dS_t = rS_t\, dt + \sigma S_t\, d\tilde{W}_t$ has the explicit solution:

\[ S_T = S_0 \exp\left[\left(r - \frac{1}{2}\sigma^2\right)T + \sigma\tilde{W}_T\right] \]

Taking logarithms:

\[ \log S_T = \log S_0 + \left(r - \frac{1}{2}\sigma^2\right)T + \sigma\tilde{W}_T \]

Since $\tilde{W}_T \sim \mathcal{N}(0, T)$ under $\mathbb{Q}$:

\[ \log S_T \sim \mathcal{N}\left(\log S_0 + \left(r - \frac{1}{2}\sigma^2\right)T,\; \sigma^2 T\right) \]

Mean: $\mathbb{E}^{\mathbb{Q}}[\log S_T] = \log S_0 + (r - \frac{1}{2}\sigma^2)T$.

Variance: $\text{Var}^{\mathbb{Q}}(\log S_T) = \sigma^2 T$.

Expected value of $S_T$: Since $S_T$ is log-normal, $\mathbb{E}^{\mathbb{Q}}[S_T] = \exp(\text{mean} + \frac{1}{2}\text{variance})$:

\[ \mathbb{E}^{\mathbb{Q}}[S_T] = \exp\left[\log S_0 + \left(r - \frac{1}{2}\sigma^2\right)T + \frac{1}{2}\sigma^2 T\right] = S_0 e^{rT} \]

This confirms that under the risk-neutral measure, the stock's expected return is the risk-free rate $r$, as required by the martingale property of the discounted asset price.

Exercise 4. Using the Feynman-Kac connection, show that if $V(t, S)$ satisfies the Black-Scholes PDE with terminal condition $V(T, S_T) = (S_T - K)^+$, then the discounted process $\tilde{V}(t) = e^{-rt} V(t, S_t)$ is a martingale under $\mathbb{Q}$. Verify by applying Ito's lemma to $\tilde{V}(t)$ and showing the drift vanishes.

Solution to Exercise 4

Let $V(t, S)$ satisfy the Black-Scholes PDE:

\[ V_t + rSV_S + \frac{1}{2}\sigma^2 S^2 V_{SS} = rV \]

Define $\tilde{V}(t) = e^{-rt}V(t, S_t)$. By Ito's lemma applied to $e^{-rt}V(t, S_t)$:

\[ d\tilde{V} = -re^{-rt}V\,dt + e^{-rt}dV \]

Now apply Ito's lemma to $V(t, S_t)$ under $\mathbb{Q}$ where $dS_t = rS_t\,dt + \sigma S_t\,d\tilde{W}_t$:

\[ dV = V_t\,dt + V_S\,dS + \frac{1}{2}V_{SS}(dS)^2 \]

\[ = V_t\,dt + V_S(rS\,dt + \sigma S\,d\tilde{W}) + \frac{1}{2}V_{SS}\sigma^2 S^2\,dt \]

\[ = \left(V_t + rSV_S + \frac{1}{2}\sigma^2 S^2 V_{SS}\right)dt + \sigma S V_S\,d\tilde{W} \]

Using the PDE constraint $V_t + rSV_S + \frac{1}{2}\sigma^2 S^2 V_{SS} = rV$:

\[ dV = rV\,dt + \sigma S V_S\,d\tilde{W} \]

Substituting back:

\[ d\tilde{V} = -re^{-rt}V\,dt + e^{-rt}(rV\,dt + \sigma S V_S\,d\tilde{W}) \]

\[ = e^{-rt}\sigma S V_S\,d\tilde{W} \]

The drift term vanishes, leaving only the stochastic integral with respect to $\tilde{W}$. Therefore $\tilde{V}(t) = e^{-rt}V(t, S_t)$ is a (local) martingale under $\mathbb{Q}$.

Exercise 5. Consider two traders who agree on all market parameters except the real-world drift: Trader A believes $\mu = 8\%$ while Trader B believes $\mu = 15\%$. Show that both traders arrive at the same Black-Scholes option price, and explain why the drift $\mu$ does not appear in the pricing formula despite appearing in the Radon–Nikodym derivative.

Solution to Exercise 5

Trader A ($\mu = 8\%$) computes:

\[ \theta_A = \frac{0.08 - r}{\sigma} \]

Trader B ($\mu = 15\%$) computes:

\[ \theta_B = \frac{0.15 - r}{\sigma} \]

Each trader defines a different Radon–Nikodym derivative:

\[ Z_T^A = \exp\left(-\theta_A W_T - \frac{1}{2}\theta_A^2 T\right), \quad Z_T^B = \exp\left(-\theta_B W_T - \frac{1}{2}\theta_B^2 T\right) \]

However, both arrive at the same risk-neutral measure $\mathbb{Q}$ under which $dS_t = rS_t\,dt + \sigma S_t\,d\tilde{W}_t$. This is because the Girsanov transformation absorbs the entire drift difference:

Trader A's $\mathbb{P}_A$-Brownian motion $W_t^A$ satisfies $\tilde{W}_t = W_t^A + \theta_A t$
Trader B's $\mathbb{P}_B$-Brownian motion $W_t^B$ satisfies $\tilde{W}_t = W_t^B + \theta_B t$

Both lead to the same $\tilde{W}_t$ under $\mathbb{Q}$, and hence the same risk-neutral distribution for $S_T$. The option price is:

\[ C_0 = e^{-rT}\mathbb{E}^{\mathbb{Q}}[(S_T - K)^+] = S_0\mathcal{N}(d_1) - Ke^{-rT}\mathcal{N}(d_2) \]

Since $d_1$ and $d_2$ depend only on $S_0$, $K$, $r$, $\sigma$, and $T$ (not $\mu$), both traders get the same price.

The drift $\mu$ does not appear because risk-neutral pricing is based on replication, not forecasting. The option can be perfectly hedged using the stock and bond, and the cost of this hedge is determined by $\sigma$ (which governs how much the stock moves) and $r$ (which governs the cost of financing), not by $\mu$ (which governs the direction of expected moves). Different beliefs about $\mu$ lead to different Radon–Nikodym derivatives but the same pricing measure.

Exercise 6. For the parameters $S_0 = 100$, $K = 105$, $r = 5\%$, $\sigma = 25\%$, $T = 0.5$, carry out the full Girsanov derivation: compute $d_1$, $d_2$, evaluate $\mathcal{N}(d_1)$ and $\mathcal{N}(d_2)$, and obtain the call price $C_0 = S_0 \mathcal{N}(d_1) - K e^{-rT} \mathcal{N}(d_2)$.

Solution to Exercise 6

Parameters: $S_0 = 100$, $K = 105$, $r = 0.05$, $\sigma = 0.25$, $T = 0.5$.

Step 1: Compute $d_1$

\[ d_1 = \frac{\ln(100/105) + (0.05 + 0.5 \times 0.0625) \times 0.5}{0.25\sqrt{0.5}} \]

\[ = \frac{-0.04879 + (0.05 + 0.03125) \times 0.5}{0.17678} = \frac{-0.04879 + 0.04063}{0.17678} = \frac{-0.00817}{0.17678} = -0.0462 \]

Step 2: Compute $d_2$

\[ d_2 = -0.0462 - 0.17678 = -0.2230 \]

Step 3: Evaluate $\mathcal{N}(d_1)$ and $\mathcal{N}(d_2)$

\[ \mathcal{N}(-0.0462) \approx 0.4816 \]

\[ \mathcal{N}(-0.2230) \approx 0.4118 \]

Step 4: Compute call price

\[ C_0 = 100 \times 0.4816 - 105 \times e^{-0.025} \times 0.4118 \]

\[ = 48.16 - 105 \times 0.9753 \times 0.4118 = 48.16 - 42.18 = 5.98 \]

The European call price is approximately $\$5.98$.

Exercise 7. The Novikov condition $\mathbb{E}^{\mathbb{P}}\left[\exp\left(\frac{1}{2}\theta^2 T\right)\right] < \infty$ guarantees that the Girsanov change of measure is well-defined. Show that this condition is automatically satisfied when $\theta$ is a constant. Discuss what could go wrong if $\theta$ were a stochastic process that grows too fast.

Solution to Exercise 7

When $\theta$ is a constant, the Novikov condition becomes:

\[ \mathbb{E}^{\mathbb{P}}\left[\exp\left(\frac{1}{2}\theta^2 T\right)\right] = \exp\left(\frac{1}{2}\theta^2 T\right) < \infty \]

Since $\theta$ and $T$ are finite constants, $\frac{1}{2}\theta^2 T$ is a finite number, and the exponential of a finite number is finite. Therefore the Novikov condition is automatically satisfied for any constant $\theta$. ✓

When $\theta_t$ is stochastic: The Novikov condition generalizes to:

\[ \mathbb{E}^{\mathbb{P}}\left[\exp\left(\frac{1}{2}\int_0^T \theta_t^2\, dt\right)\right] < \infty \]

If $\theta_t$ grows too fast (e.g., if $\theta_t$ itself depends on $W_t$ in a way that makes $\int_0^T \theta_t^2\, dt$ have heavy tails), this expectation can diverge.

What goes wrong: If the Novikov condition fails, the exponential martingale $Z_t = \exp(-\int_0^t \theta_s\, dW_s - \frac{1}{2}\int_0^t \theta_s^2\, ds)$ may fail to be a true martingale (it could be only a supermartingale with $\mathbb{E}[Z_T] < 1$). In this case, $\mathbb{Q}$ defined by $d\mathbb{Q}/d\mathbb{P} = Z_T$ would not be a valid probability measure (it would not integrate to $1$). The Girsanov change of drift would be invalid, and the resulting "risk-neutral" pricing could produce incorrect option prices or even allow arbitrage in the model. This is a genuine concern in stochastic volatility models where the market price of volatility risk can be unbounded.

Exercise 8. Derive the Black–Scholes call price by direct Gaussian integration under $\mathbb{Q}$ alone (no change of numéraire). Substitute $S_T = S_0 e^{(r - \frac{1}{2}\sigma^2)T + \sigma\sqrt{T}\,Z}$ with $Z \sim \mathcal{N}(0,1)$ under $\mathbb{Q}$, show that $\{S_T > K\} = \{Z > -d_2\}$, split the resulting integral into a stock piece and a strike piece, and obtain the $\mathcal{N}(d_1)$ factor by completing the square in the exponent. Compare with the stock-numéraire derivation in the body: both must give the same answer.

Solution to Exercise 8

Under $\mathbb{Q}$, write $\tilde{W}_T = \sqrt{T}\,Z$ with $Z \sim \mathcal{N}(0,1)$, so

\[ S_T = S_0\,e^{(r - \frac{1}{2}\sigma^2)T + \sigma\sqrt{T}\,Z} \]

The exercise condition $S_T > K$ is equivalent to

\[ Z > \frac{\log(K/S_0) - (r - \frac{1}{2}\sigma^2)T}{\sigma\sqrt{T}} = -d_2 \]

Therefore

\[ C_0 = e^{-rT}\int_{-d_2}^{\infty}\!\left(S_0\,e^{(r - \frac{1}{2}\sigma^2)T + \sigma\sqrt{T}\,z} - K\right)\frac{1}{\sqrt{2\pi}}e^{-z^2/2}\,dz \]

Split into a stock integral $A$ and a strike integral $B$:

\[ A = e^{-rT}S_0\int_{-d_2}^{\infty} e^{(r - \frac{1}{2}\sigma^2)T + \sigma\sqrt{T}\,z}\,\frac{1}{\sqrt{2\pi}}e^{-z^2/2}\,dz \]

\[ B = Ke^{-rT}\int_{-d_2}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-z^2/2}\,dz \]

The strike integral is the standard normal tail probability $\mathbb{P}(Z > -d_2) = \mathcal{N}(d_2)$:

\[ B = Ke^{-rT}\,\mathcal{N}(d_2) \]

In the stock integral, the prefactor $e^{-rT}\cdot e^{rT} = 1$ cancels, leaving

\[ A = S_0\int_{-d_2}^{\infty}\frac{1}{\sqrt{2\pi}}\exp\!\left(-\tfrac{1}{2}\sigma^2 T + \sigma\sqrt{T}\,z - \tfrac{1}{2}z^2\right)dz \]

Complete the square in the exponent:

\[ -\tfrac{1}{2}\sigma^2 T + \sigma\sqrt{T}\,z - \tfrac{1}{2}z^2 = -\tfrac{1}{2}(z - \sigma\sqrt{T})^2 \]

Substitute $u = z - \sigma\sqrt{T}$, $du = dz$. The lower limit shifts to $-d_2 - \sigma\sqrt{T} = -d_1$:

\[ A = S_0\int_{-d_1}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-u^2/2}\,du = S_0\,\mathcal{N}(d_1) \]

Combining,

\[ C_0 = A - B = S_0\,\mathcal{N}(d_1) - Ke^{-rT}\,\mathcal{N}(d_2) \]

Comparison with the body derivation: the stock-numéraire route produces $\mathcal{N}(d_1)$ as the probability $\mathbb{Q}^S(S_T > K)$ — a measure-theoretic statement. The route above produces the same $\mathcal{N}(d_1)$ as the result of completing the square in a Gaussian integral — an algebraic statement. The two routes are not just numerically equal: the completing-the-square step $-\tfrac{1}{2}(z - \sigma\sqrt{T})^2$ is exactly the Radon–Nikodym density $d\mathbb{Q}^S/d\mathbb{Q} = \exp(\sigma\tilde{W}_T - \tfrac{1}{2}\sigma^2 T)$ multiplied against the $\mathbb{Q}$-density, which is why the integration variable shifts by $\sigma\sqrt{T}$. The numéraire change makes structurally explicit what the integral computes by mechanical substitution.