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Examples: Constructing Risk-Neutral Measures

This page is a catalog of how risk-neutral measures look in concrete models. Each example follows the same template from § Construction of the Risk-Neutral Measure:

  1. identify \(\mathbb{P}\)-dynamics,
  2. solve \(\boldsymbol{\mu} - r\mathbf{1} = \Sigma\boldsymbol{\theta}\) for the market price of risk,
  3. apply Girsanov,
  4. read off the risk-neutral dynamics.

Completeness vs incompleteness is determined by the rank of \(\Sigma\) — see § Market Price of Risk, multi-asset case for the taxonomy.


Example 1: Black-Scholes Model (Single Stock)

Under \(\mathbb{P}\): \(dS_t = \mu S_t\,dt + \sigma S_t\,dW_t^{\mathbb{P}}\) with constant \(\mu\), \(\sigma\), \(r\).

Market price of risk: \(\theta = (\mu - r)/\sigma\).

Under \(\mathbb{Q}\):

\[ dS_t = rS_t\,dt + \sigma S_t\,dW_t^{\mathbb{Q}} \]

Example 2: Stock with Dividends

Under \(\mathbb{P}\): \(dS_t = (\mu - q)S_t\,dt + \sigma S_t\,dW_t^{\mathbb{P}}\) with continuous dividend yield \(q\).

Market price of risk: \(\theta = (\mu - r)/\sigma\) (dividends cancel).

Under \(\mathbb{Q}\):

\[ dS_t = (r - q)S_t\,dt + \sigma S_t\,dW_t^{\mathbb{Q}} \]

Example 3: Two Correlated Stocks

Two stocks driven by Brownian motions \(W^1, W^2\) with correlation \(\rho\):

\[ \begin{cases} dS_t^1 = \mu_1 S_t^1\,dt + \sigma_1 S_t^1\,dW_t^{1,\mathbb{P}} \\ dS_t^2 = \mu_2 S_t^2\,dt + \sigma_2 S_t^2\,(\rho\,dW_t^{1,\mathbb{P}} + \sqrt{1-\rho^2}\,dW_t^{2,\mathbb{P}}) \end{cases} \]

Recall (see § Market Price of Risk) the vector equation \(\boldsymbol{\mu} - r\mathbf{1} = \Sigma\boldsymbol{\theta}\). Here \(\Sigma\) is the \(2 \times 2\) lower-triangular matrix with rows \((\sigma_1, 0)\) and \((\sigma_2\rho,\,\sigma_2\sqrt{1-\rho^2})\), giving \(\theta_1 = (\mu_1 - r)/\sigma_1\) and \(\theta_2 = [(\mu_2 - r) - \sigma_2\rho\theta_1]/(\sigma_2\sqrt{1-\rho^2})\).

With \(n = d = 2\) and \(\Sigma\) invertible, \(\boldsymbol{\theta}\) is unique and the market is complete.


Example 4: Incomplete Market (Stochastic Volatility)

The Heston model has two sources of randomness but only one traded asset:

\[ \begin{cases} dS_t = \mu S_t\,dt + \sqrt{V_t}S_t\,dW_t^{1,\mathbb{P}} \\ dV_t = \kappa(\bar{V} - V_t)\,dt + \xi\sqrt{V_t}\,dW_t^{2,\mathbb{P}} \end{cases} \]

with \(\text{Corr}(dW^1, dW^2) = \rho\).

No-arbitrage determines \(\theta_1 = (\mu - r)/\sqrt{V_t}\), but \(\theta_2\) (the volatility risk premium) remains free — any choice satisfying integrability gives a valid \(\mathbb{Q}\). Common specifications include \(\theta_2 = 0\) and \(\theta_2 = \lambda\sqrt{V_t}\). Different choices yield different option prices: the market is incomplete.


Example 5: Foreign Exchange

Under \(\mathbb{P}\): \(dX_t = \mu_X X_t\,dt + \sigma_X X_t\,dW_t^{\mathbb{P}}\) where \(X_t\) is the exchange rate (domestic per foreign).

No-arbitrage (covered interest parity): \(\mu_X = r_d - r_f\), so \(\theta = 0\).

Under \(\mathbb{Q}\):

\[ dX_t = (r_d - r_f)X_t\,dt + \sigma_X X_t\,dW_t^{\mathbb{Q}} \]

Example 6: Vasicek Interest Rate Model

Under \(\mathbb{P}\): \(dr_t = \kappa(\bar{r} - r_t)\,dt + \sigma\,dW_t^{\mathbb{P}}\).

Market price of risk: specified exogenously (e.g., \(\theta_t = \lambda\)) because the short rate is not a traded asset.

Under \(\mathbb{Q}\) (with \(\theta = \lambda\)):

\[ dr_t = \kappa^*(\bar{r}^* - r_t)\,dt + \sigma\,dW_t^{\mathbb{Q}} \]

where \(\kappa^* = \kappa\) and \(\bar{r}^* = \bar{r} - \sigma\lambda/\kappa\).


Summary Table

Model # Assets # BMs Complete? \(\boldsymbol{\theta}\) unique?
Black-Scholes 1 1 Yes Yes
Multi-stock \(n\) \(n\) Yes Yes
Stochastic volatility 1 2 No No
FX 1 1 Yes Yes
Interest rate Bonds 1 Often incomplete No

The unifying structural insight, visible in every row of the table above:

\[ \boxed{\;\text{Only the drift changes under } \mathbb{Q};\quad \text{the volatility structure } \Sigma \text{ is preserved.}\;} \]

See § Market Price of Risk, Exercise 2 for the Girsanov derivation.


Exercises

Exercise 1. In the Black-Scholes model with \(\mu = 0.15\), \(r = 0.03\), and \(\sigma = 0.30\), compute the market price of risk \(\theta\) and the Radon–Nikodym derivative \(Z_1\) for a specific path where \(W_1^{\mathbb{P}} = -0.5\). Is this path upweighted or downweighted under \(\mathbb{Q}\)?

Solution to Exercise 1

With \(\mu = 0.15\), \(r = 0.03\), and \(\sigma = 0.30\):

\[ \theta = \frac{\mu - r}{\sigma} = \frac{0.15 - 0.03}{0.30} = 0.40 \]

For the specific path with \(W_1^{\mathbb{P}} = -0.5\) and \(T = 1\):

\[ Z_1 = \exp\!\left(-\theta W_1^{\mathbb{P}} - \frac{1}{2}\theta^2 \cdot 1\right) = \exp\!\left(-0.40 \cdot (-0.5) - \frac{1}{2}(0.16)\right) \]
\[ = \exp(0.20 - 0.08) = \exp(0.12) \approx 1.1275 \]

Since \(Z_1 > 1\), this path is upweighted under \(\mathbb{Q}\). Intuitively, \(W_1^{\mathbb{P}} = -0.5\) corresponds to a below-average stock return. The risk-neutral measure overweights adverse outcomes (and underweights favorable ones), reflecting the risk adjustment embedded in \(\mathbb{Q}\).


Exercise 2. A stock pays a continuous dividend yield \(q = 0.02\) with \(\mu = 0.08\), \(\sigma = 0.20\), and \(r = 0.05\). Verify that the market price of risk is the same as in the no-dividend case. Write the risk-neutral dynamics and check that the discounted reinvested-dividend process is a \(\mathbb{Q}\)-martingale.

Solution to Exercise 2

With dividends, the stock dynamics are \(dS_t = (\mu - q)S_t\,dt + \sigma S_t\,dW_t^{\mathbb{P}}\). The discounted price for a dividend-paying stock must account for the reinvested dividends. Consider \(\tilde{S}_t = e^{-rt}e^{qt}S_t \cdot e^{-qt} = e^{-rt}S_t\). More carefully, the total return from holding the stock includes dividends, so the relevant discounted process is \(e^{-(r-q)t}S_t\) (or equivalently \(e^{-rt}\) times the dividend-reinvested portfolio).

The market price of risk is computed from the excess return of the total return process. Since the stock pays dividends at rate \(q\), the total instantaneous return is \(\mu\,dt + \sigma\,dW_t^{\mathbb{P}}\). The excess over \(r\) divided by \(\sigma\) gives:

\[ \theta = \frac{\mu - r}{\sigma} = \frac{0.08 - 0.05}{0.20} = 0.15 \]

This is the same as without dividends (where \(\mu\) would be the total return). Under \(\mathbb{Q}\):

\[ dS_t = (r - q)S_t\,dt + \sigma S_t\,dW_t^{\mathbb{Q}} \]

The discounted reinvested-dividend process \(\tilde{S}_t = e^{-(r-q)t}S_t\) satisfies:

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

The drift vanishes, confirming \(\tilde{S}_t\) is a \(\mathbb{Q}\)-martingale.


Exercise 3. For the two correlated stocks model, take \(\mu_1 = 0.12\), \(\mu_2 = 0.08\), \(r = 0.03\), \(\sigma_1 = 0.25\), \(\sigma_2 = 0.30\), and \(\rho = 0.4\). Solve for \(\theta_1\) and \(\theta_2\). Verify your answer by checking that both excess returns \(\mu_i - r\) are reproduced by \(\Sigma\boldsymbol{\theta}\).

Solution to Exercise 3

Given \(\mu_1 = 0.12\), \(\mu_2 = 0.08\), \(r = 0.03\), \(\sigma_1 = 0.25\), \(\sigma_2 = 0.30\), \(\rho = 0.4\):

From \(\theta_1 = (\mu_1 - r)/\sigma_1\):

\[ \theta_1 = \frac{0.12 - 0.03}{0.25} = 0.36 \]

From \(\theta_2 = [(\mu_2 - r) - \sigma_2\rho\theta_1]/(\sigma_2\sqrt{1 - \rho^2})\):

\[ \theta_2 = \frac{(0.08 - 0.03) - 0.30 \cdot 0.4 \cdot 0.36}{0.30\sqrt{1 - 0.16}} = \frac{0.05 - 0.0432}{0.30 \cdot \sqrt{0.84}} \]
\[ = \frac{0.0068}{0.30 \cdot 0.9165} = \frac{0.0068}{0.2750} \approx 0.0247 \]

Verification: The volatility matrix is \(\Sigma = \begin{pmatrix} 0.25 & 0 \\ 0.30 \cdot 0.4 & 0.30\sqrt{0.84} \end{pmatrix} = \begin{pmatrix} 0.25 & 0 \\ 0.12 & 0.2750 \end{pmatrix}\).

\[ \Sigma\boldsymbol{\theta} = \begin{pmatrix} 0.25 \cdot 0.36 + 0 \cdot 0.0247 \\ 0.12 \cdot 0.36 + 0.2750 \cdot 0.0247 \end{pmatrix} = \begin{pmatrix} 0.09 \\ 0.0432 + 0.0068 \end{pmatrix} = \begin{pmatrix} 0.09 \\ 0.05 \end{pmatrix} \]

This matches \(\boldsymbol{\mu} - r\mathbf{1} = (0.09, 0.05)^{\top}\).


Exercise 4. In the Heston stochastic volatility model, explain why the volatility risk premium \(\theta_2\) cannot be determined by no-arbitrage. If one practitioner sets \(\theta_2 = 0\) and another sets \(\theta_2 = -0.5\sqrt{V_t}\), write the risk-neutral variance dynamics under each choice. Which choice produces a lower long-run mean of variance under \(\mathbb{Q}\)?

Solution to Exercise 4

In the Heston model, \(dS_t = \mu S_t\,dt + \sqrt{V_t}S_t\,dW_t^{1,\mathbb{P}}\) and \(dV_t = \kappa(\bar{V} - V_t)\,dt + \xi\sqrt{V_t}\,dW_t^{2,\mathbb{P}}\). There are two Brownian motions but only one traded risky asset \(S\). The no-arbitrage condition for \(S\) determines \(\theta_1 = (\mu - r)/\sqrt{V_t}\), but \(\theta_2\) (associated with \(W^2\)) is unconstrained because no traded asset's return depends solely on \(W^2\) in a way that pins down \(\theta_2\). This is the hallmark of an incomplete market.

Under \(\mathbb{Q}\) with a general \(\theta_2\), the variance dynamics become:

\[ dV_t = \kappa(\bar{V} - V_t)\,dt + \xi\sqrt{V_t}(dW_t^{2,\mathbb{Q}} - \theta_2\,dt) \]

Choice 1: \(\theta_2 = 0\):

\[ dV_t = \kappa(\bar{V} - V_t)\,dt + \xi\sqrt{V_t}\,dW_t^{2,\mathbb{Q}} \]

Long-run mean: \(\bar{V}\).

Choice 2: \(\theta_2 = -0.5\sqrt{V_t}\):

\[ dV_t = [\kappa(\bar{V} - V_t) + 0.5\xi V_t]\,dt + \xi\sqrt{V_t}\,dW_t^{2,\mathbb{Q}} \]
\[ = [(\kappa\bar{V}) - (\kappa - 0.5\xi)V_t]\,dt + \xi\sqrt{V_t}\,dW_t^{2,\mathbb{Q}} \]

This has the form \(\kappa^*(\bar{V}^* - V_t)\,dt\) with \(\kappa^* = \kappa - 0.5\xi\) and \(\bar{V}^* = \kappa\bar{V}/(\kappa - 0.5\xi)\).

Assuming \(\kappa > 0.5\xi\) (so mean reversion is maintained), \(\bar{V}^* = \kappa\bar{V}/(\kappa - 0.5\xi) > \bar{V}\). Therefore, Choice 2 produces a higher long-run mean of variance under \(\mathbb{Q}\), and Choice 1 produces the lower long-run mean \(\bar{V}\).


Exercise 5. In the FX example, the no-arbitrage condition implies \(\mu_X = r_d - r_f\). Derive this condition by requiring the domestic-currency value of a foreign money market investment to grow at rate \(r_d\) under \(\mathbb{Q}\). What happens to the market price of risk if \(\mu_X \neq r_d - r_f\)?

Solution to Exercise 5

The domestic-currency value of investing 1 unit of foreign currency in the foreign money market is \(V_t = X_t e^{r_f t}\). By Itô's formula:

\[ dV_t = e^{r_f t}dX_t + r_f X_t e^{r_f t}\,dt = V_t\!\left(\frac{dX_t}{X_t} + r_f\,dt\right) \]

Under \(\mathbb{Q}\), this discounted process \(\tilde{V}_t = e^{-r_d t}V_t\) must be a martingale:

\[ d\tilde{V}_t = \tilde{V}_t\!\left(\frac{dX_t}{X_t} + r_f\,dt - r_d\,dt\right) \]

For the drift of \(\tilde{V}_t\) to vanish, the drift of \(dX_t/X_t\) must equal \(r_d - r_f\):

\[ \mu_X = r_d - r_f \]

If \(\mu_X \neq r_d - r_f\), the market price of risk becomes

\[ \theta = \frac{\mu_X - (r_d - r_f)}{\sigma_X} \neq 0 \]

A Girsanov change of measure with this \(\theta\) is required to construct \(\mathbb{Q}\), and \(W_t^{\mathbb{Q}} = W_t^{\mathbb{P}} + \theta t\) absorbs the deviation. Under \(\mathbb{Q}\), the FX dynamics always become \(dX_t = (r_d - r_f)X_t\,dt + \sigma_X X_t\,dW_t^{\mathbb{Q}}\).


Exercise 6. In the Vasicek model with \(\kappa = 0.5\), \(\bar{r} = 0.04\), \(\sigma = 0.01\), and constant market price of risk \(\theta = 0.2\), compute \(\bar{r}^* = \bar{r} - \sigma\theta/\kappa\). Using the risk-neutral dynamics, compute the zero-coupon bond price \(P(0, T)\) for \(T = 5\) with \(r_0 = 0.03\).

Solution to Exercise 6

Given \(\kappa = 0.5\), \(\bar{r} = 0.04\), \(\sigma = 0.01\), \(\theta = 0.2\):

\[ \bar{r}^* = \bar{r} - \frac{\sigma\theta}{\kappa} = 0.04 - \frac{0.01 \cdot 0.2}{0.5} = 0.04 - 0.004 = 0.036 \]

Under \(\mathbb{Q}\), the dynamics are \(dr_t = \kappa(\bar{r}^* - r_t)\,dt + \sigma\,dW_t^{\mathbb{Q}}\) with \(\kappa^* = \kappa = 0.5\) and \(\bar{r}^* = 0.036\).

For the Vasicek model, the bond price is \(P(0,T) = A(0,T)e^{-B(0,T)r_0}\) where

\[ B(0,T) = \frac{1 - e^{-\kappa T}}{\kappa} = \frac{1 - e^{-0.5 \cdot 5}}{0.5} = \frac{1 - e^{-2.5}}{0.5} = \frac{1 - 0.08209}{0.5} = \frac{0.91791}{0.5} = 1.83583 \]
\[ A(0,T) = \exp\!\left[\left(\bar{r}^* - \frac{\sigma^2}{2\kappa^2}\right)(B(0,T) - T) - \frac{\sigma^2}{4\kappa}B(0,T)^2\right] \]

Computing: \(\sigma^2/(2\kappa^2) = 0.0001/0.5 = 0.0002\) and \(\sigma^2/(4\kappa) = 0.0001/2 = 0.00005\).

\[ A(0,5) = \exp\!\left[(0.036 - 0.0002)(1.83583 - 5) - 0.00005 \cdot (1.83583)^2\right] \]
\[ = \exp\!\left[0.0358 \cdot (-3.16417) - 0.00005 \cdot 3.37028\right] \]
\[ = \exp(-0.11328 - 0.00017) = \exp(-0.11345) \approx 0.89280 \]

Therefore:

\[ P(0,5) = 0.89280 \cdot e^{-1.83583 \cdot 0.03} = 0.89280 \cdot e^{-0.05507} = 0.89280 \cdot 0.94645 \approx 0.84498 \]