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.

1 Model Setup¶

Let the stock price $S_t$ evolve under the real-world measure $\mathbb{P}$ as:

\[ dS_t = \mu S_t\, dt + \sigma S_t\, dW_t, \quad S_0 > 0 \]

Let:

$K > 0$ be the strike price,
$T > 0$ be the maturity,
$r > 0$ be the risk-free interest rate,
$\sigma > 0$ the volatility,
$\mu$ the drift (real-world expected return).

The European call option payoff at time $T$ is:

\[ \max(S_T - K, 0) = (S_T - K)^+ \]

Our goal is to compute:

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

2 Step 1: Change of Measure Using Girsanov¶

We define the market price of risk:

\[ \theta := \frac{\mu - r}{\sigma} \]

Girsanov's theorem tells us that under the new measure $\mathbb{Q}$ defined by:

\[ \left. \frac{d\mathbb{Q}}{d\mathbb{P}} \right|_{\mathcal{F}_T} = Z_T = \exp\left( -\theta W_T - \frac{1}{2} \theta^2 T \right) \]

the process:

\[ \tilde{W}_t := W_t + \theta t \]

is a Brownian motion under $\mathbb{Q}$, and under $\mathbb{Q}$ the stock price satisfies:

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

This is the risk-neutral dynamic. The drift has been shifted from $\mu$ to $r$, while the volatility $\sigma$ remains unchanged.

3 Step 2: Explicit Distribution of S_T under Q¶

Solving the SDE under $\mathbb{Q}$:

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

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

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

4 Step 3: Compute the Call Option Price¶

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:

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

where $\Phi(\cdot)$ is the standard normal cumulative distribution function.

Interpretation of the Terms¶

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

Derivation Sketch¶

From risk-neutral pricing:

\[ \begin{aligned} C_0 &= e^{-rT} \mathbb{E}^{\mathbb{Q}}[(S_T - K)^+] \\ &= e^{-rT} \int_K^\infty (s - K)\, f_{S_T}^{\mathbb{Q}}(s)\, ds \\ &= S_0 \Phi(d_1) - K e^{-rT} \Phi(d_2) \end{aligned} \]

The integral is evaluated by substituting $S_T = S_0 e^{(r - \sigma^2/2)T + \sigma\sqrt{T}\, Z}$ with $Z \sim N(0,1)$, completing the square, and recognizing the resulting expressions as normal CDF evaluations.

5 Final Result¶

The European call price under Girsanov (risk-neutral pricing) is:

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

where:

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

6 Summary of the Logic¶

Start under $\mathbb{P}$: $dS_t = \mu S_t\, dt + \sigma S_t\, dW_t$
Use Girsanov to define $\mathbb{Q}$ via the exponential martingale:

\[ Z_T = \exp\left( -\theta W_T - \frac{1}{2} \theta^2 T \right), \quad \theta = \frac{\mu - r}{\sigma} \]

Under $\mathbb{Q}$: the drift becomes $r$ instead of $\mu$, giving the risk-neutral SDE.
Price the option via risk-neutral expectation: $C_0 = e^{-rT}\,\mathbb{E}^{\mathbb{Q}}[(S_T - K)^+]$.
Evaluate the expectation using the lognormal distribution of $S_T$ under $\mathbb{Q}$ to obtain the Black–Scholes formula.

Key Insight

The real-world drift $\mu$ does not appear in the final pricing formula. Girsanov's theorem absorbs the drift difference $\mu - r$ into the change of measure, so that option prices depend only on $r$, $\sigma$, $S_0$, $K$, and $T$.

QuantPie Derivation via Feynman-Kac¶

Connection to the Black-Scholes PDE¶

The Black-Scholes PDE and risk-neutral pricing are connected through the Feynman-Kac formula. For a European option with payoff $\Phi(S_T)$:

\[ \begin{array}{lcc} \text{Black-Scholes Equation} & & \displaystyle \frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2 S^2\frac{\partial^2 V}{\partial S^2} + rS\frac{\partial V}{\partial S} - rV = 0\\ \\ \text{Terminal Condition} & & \displaystyle V(T, S_T) = \Phi(S_T)\\ \\ \text{Risk-Neutral SDE} & & \displaystyle dS = rS\,dt + \sigma S\,dB^{\mathbb{Q}}\\ \\ \text{Feynman-Kac Result} & & \displaystyle V(t, S) = e^{-r(T-t)}\mathbb{E}^{\mathbb{Q}}[\Phi(S_T) | S_t = S] \end{array} \]

Verification via Itô's Lemma¶

Starting from the PDE, assume $V$ satisfies:

\[ \frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2 S^2\frac{\partial^2 V}{\partial S^2} + rS\frac{\partial V}{\partial S} - rV = 0 \]

Apply Itô's Lemma to $V(t, S_t)$ under the risk-neutral dynamics:

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

\[ = V_t dt + V_S(rS\,dt + \sigma S\,dB^{\mathbb{Q}}) + \frac{1}{2}V_{SS}\sigma^2 S^2 dt \]

\[ = \left(V_t + rSV_S + \frac{1}{2}\sigma^2S^2V_{SS}\right)dt + \sigma SV_S dB^{\mathbb{Q}} \]

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

\[ dV = rV\,dt + \sigma SV_S dB^{\mathbb{Q}} \]

Discounted Value is a Martingale¶

Define the discounted option value:

\[ \tilde{V}(t) = \frac{V(t, S_t)}{e^{rt}} \]

Computing the differential:

\[ d\tilde{V} = \frac{dV}{e^{rt}} - r\frac{V}{e^{rt}}dt = \frac{rV\,dt + \sigma SV_S dB^{\mathbb{Q}}}{e^{rt}} - r\frac{V}{e^{rt}}dt \]

\[ = \frac{\sigma SV_S}{e^{rt}}dB^{\mathbb{Q}} = \sigma S\left(\frac{V_S}{e^{rt}}\right)dB^{\mathbb{Q}} \]

The drift term vanishes! Therefore $\tilde{V}(t)$ is a martingale under $\mathbb{Q}$.

Martingale Property Implies Risk-Neutral Pricing¶

Since $\tilde{V}(t)$ is a martingale:

\[ \tilde{V}(t) = \mathbb{E}^{\mathbb{Q}}[\tilde{V}(T) | \mathcal{F}_t] \]

\[ \frac{V(t, S_t)}{e^{rt}} = \mathbb{E}^{\mathbb{Q}}\left[\frac{V(T, S_T)}{e^{rT}} \,\Big|\, \mathcal{F}_t\right] \]

\[ \frac{V(t, S_t)}{e^{rt}} = \mathbb{E}^{\mathbb{Q}}\left[\frac{\Phi(S_T)}{e^{rT}} \,\Big|\, S_t = S\right] \]

\[ \boxed{V(t, S) = e^{-r(T-t)}\mathbb{E}^{\mathbb{Q}}[\Phi(S_T) | S_t = S]} \]

Why the Real-World Drift Doesn't Matter¶

Under the real-world measure $\mathbb{P}$:

\[ dS = \mu S\,dt + \sigma S\,dW \]

Under the risk-neutral measure $\mathbb{Q}$ defined by Girsanov:

\[ dS = rS\,dt + \sigma S\,d\tilde{W}^{\mathbb{Q}} \]

The change of measure via the Radon-Nikodym derivative:

\[ \frac{d\mathbb{Q}}{d\mathbb{P}}\Big|_{\mathcal{F}_T} = \exp\left(-\frac{\mu - r}{\sigma}W_T - \frac{1}{2}\left(\frac{\mu-r}{\sigma}\right)^2 T\right) \]

Key observation: The martingale property of $\tilde{V}(t)$ holds under $\mathbb{Q}$, not $\mathbb{P}$. The drift $\mu$ appears in the Radon-Nikodym derivative but not in the final formula because we compute the expectation under $\mathbb{Q}$.

This is why option prices are drift-neutral: they depend only on the volatility $\sigma$, the risk-free rate $r$, and the option structure, not on traders' expectations ($\mu$) about future stock returns.

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:

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

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 $\Phi(d_1)$ and $\Phi(d_2)$, and obtain the call price $C_0 = S_0 \Phi(d_1) - K e^{-rT} \Phi(d_2)$.

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 $\Phi(d_1)$ and $\Phi(d_2)$

\[ \Phi(-0.0462) \approx 0.4816 \]

\[ \Phi(-0.2230) \approx 0.4118 \]

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.