Girsanov's Theorem and the Market Price of Risk

Worked Solution

How to Think About It: Under the physical measure, the stock drifts at rate $\mu$, which reflects investors' required return for bearing risk. But for pricing derivatives, we do not care about $\mu$ -- we care about what drift would make the discounted stock a martingale. The answer is $r$, the risk-free rate. Girsanov's theorem is the machine that swaps one drift for the other by redefining what "standard Brownian motion" means. The quantity that gets absorbed in the switch is the market price of risk $\theta = (\mu - r)/\sigma$, which is just the Sharpe ratio of the stock.

Quick Sanity Checks: Under $\mathbb{Q}$, the drift of $S_t$ must become $r$ (otherwise discounted prices would not be martingales). The volatility $\sigma$ should stay the same -- Girsanov changes the drift, not the diffusion coefficient. If $\mu = r$ (no risk premium), the two measures should coincide and $\theta = 0$.

Derivation:

Start from the $\mathbb{P}$-dynamics:

$dS_t = \mu S_t \, dt + \sigma S_t \, dW_t^{\mathbb{P}}$

Step 1 -- Define the market price of risk. For the discounted stock $e^{-rt} S_t$ to be a $\mathbb{Q}$-martingale, we need to remove the excess drift $\mu - r$. The market price of risk is:

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

This is the amount of extra drift per unit of volatility -- i.e., the Sharpe ratio.

Step 2 -- Apply Girsanov's theorem. Define a new process:

$W_t^{\mathbb{Q}} = W_t^{\mathbb{P}} + \theta t$

Girsanov's theorem guarantees that $W_t^{\mathbb{Q}}$ is a standard Brownian motion under the measure $\mathbb{Q}$ defined by the Radon-Nikodym derivative:

$\frac{d\mathbb{Q}}{d\mathbb{P}} \bigg|_{\mathcal{F}_T} = \exp\left(-\theta W_T^{\mathbb{P}} - \frac{\theta^2 T}{2}\right)$

This is the exponential martingale (sometimes called the Doleans-Dade exponential) that tilts the probability measure.

Step 3 -- Rewrite the SDE under $\mathbb{Q}$. Substitute $dW_t^{\mathbb{P}} = dW_t^{\mathbb{Q}} - \theta \, dt$ into the original SDE:

$dS_t = \mu S_t \, dt + \sigma S_t \left(dW_t^{\mathbb{Q}} - \theta \, dt\right)$

$= \mu S_t \, dt - \sigma \theta S_t \, dt + \sigma S_t \, dW_t^{\mathbb{Q}}$

$= (\mu - \sigma \theta) S_t \, dt + \sigma S_t \, dW_t^{\mathbb{Q}}$

Since $\sigma \theta = \mu - r$, this simplifies to:

$dS_t = r S_t \, dt + \sigma S_t \, dW_t^{\mathbb{Q}}$

Step 4 -- Verify the martingale property. The discounted stock price $\tilde{S}_t = e^{-rt} S_t$ satisfies:

$d\tilde{S}_t = \sigma \tilde{S}_t \, dW_t^{\mathbb{Q}}$

which is a driftless SDE -- confirming $\tilde{S}_t$ is a $\mathbb{Q}$-martingale, as required.

Practical Interpretation: The market price of risk $\theta = (\mu - r)/\sigma$ is the Sharpe ratio of the stock. It quantifies how much excess return (above the risk-free rate) investors demand per unit of volatility. When you switch to the risk-neutral measure, you are stripping out this risk premium. This is why $\mu$ never appears in Black-Scholes -- the option price depends on $\sigma$ and $r$, not on $\mu$. A trader's key takeaway: two stocks with the same volatility but different drifts produce the same option prices. The drift is "priced in" by the measure change.

Note that the volatility $\sigma$ is invariant under the change of measure. This is a fundamental property: Girsanov only tilts the drift, not the diffusion. Economically, this means the magnitude of random shocks is the same regardless of how you assign probabilities to outcomes -- it is the expected direction of those shocks that differs.

Answer:

• Market price of risk: $\theta = (\mu - r)/\sigma$
• $\mathbb{Q}$-dynamics: $dS_t = r S_t \, dt + \sigma S_t \, dW_t^{\mathbb{Q}}$
• Radon-Nikodym derivative: $\frac{d\mathbb{Q}}{d\mathbb{P}}\big|_{\mathcal{F}_T} = \exp\left(-\theta W_T^{\mathbb{P}} - \frac{\theta^2 T}{2}\right)$
• The drift changes from $\mu$ to $r$; the volatility $\sigma$ is unchanged.