Stock-Price Numeraire Measure¶

The stock-price numeraire measure \(\mathbb{Q}^S\) uses the stock price \(S_t\) as the numeraire instead of the money-market account \(B_t = e^{rt}\). Under \(\mathbb{Q}^S\), the probability \(P_1 = \mathbb{Q}^S(S_T > K)\) appears naturally in the Heston call pricing formula as the delta-related component \(C = S_0 e^{-q\tau} P_1 - K e^{-r\tau} P_2\). This section derives the measure change from \(\mathbb{Q}\) to \(\mathbb{Q}^S\), obtains the Heston characteristic function under the new measure, and applies it to price digital calls and compute the probability \(P_1\).

Learning Objectives

By the end of this section, you will be able to:

Derive the Radon-Nikodym derivative for the change from \(\mathbb{Q}\) to \(\mathbb{Q}^S\)
Obtain the Heston SDE and characteristic function under \(\mathbb{Q}^S\)
Relate \(P_1\) to the exercise probability under the stock-price numeraire
Apply the \(\mathbb{Q}^S\) framework to price asset-or-nothing digital calls

Intuition¶

In the Black-Scholes model, the call formula \(C = S_0 N(d_1) - K e^{-rT} N(d_2)\) involves two probabilities: \(N(d_2) = \mathbb{Q}(S_T > K)\) is the exercise probability under the risk-neutral measure, while \(N(d_1) = \mathbb{Q}^S(S_T > K)\) is the exercise probability under the stock-price numeraire. The second probability weights outcomes by the stock price, making high-stock-price states more likely. In the Heston model, this measure change transforms the variance dynamics: the mean-reversion speed decreases by \(\rho\xi\) (since high stock prices correlate with low variance through \(\rho < 0\)), tilting the variance distribution.

Numeraire Change: General Theory¶

Let \(N_t\) be a positive \(\mathbb{Q}\)-martingale (after discounting). The numeraire measure \(\mathbb{Q}^N\) associated with numeraire \(N_t\) is defined by the Radon-Nikodym derivative

\[ \frac{d\mathbb{Q}^N}{d\mathbb{Q}}\bigg|_{\mathcal{F}_T} = \frac{N_T / N_0}{B_T / B_0} \]

where \(B_t = e^{rt}\) is the money-market account.

Proposition (Stock-Price Numeraire)

For the stock-price numeraire \(N_t = S_t e^{qt}\) (the reinvested stock), the Radon-Nikodym derivative is

\[ \frac{d\mathbb{Q}^S}{d\mathbb{Q}}\bigg|_{\mathcal{F}_T} = \frac{S_T e^{qT}}{S_0 e^{qT} \cdot e^{(r-q)T}} \cdot \frac{B_0}{1} = \frac{S_T}{S_0 e^{(r-q)T}} \]

where we used the fact that the discounted reinvested stock \(S_t e^{qt}/B_t\) is a \(\mathbb{Q}\)-martingale.

Girsanov Shift for the Stock-Price Numeraire¶

Under \(\mathbb{Q}\), the stock dynamics are

\[ dS_t = (r - q)S_t \, dt + \sqrt{v_t} \, S_t \, dW_t^{(1), \mathbb{Q}} \]

The Radon-Nikodym density process is

\[ Z_t = \frac{S_t}{S_0 e^{(r-q)t}} = \mathcal{E}\!\left(\int_0^t \sqrt{v_s} \, dW_s^{(1),\mathbb{Q}}\right) \]

where \(\mathcal{E}\) denotes the stochastic exponential. By Girsanov's theorem:

\[ dW_t^{(1), \mathbb{Q}^S} = dW_t^{(1), \mathbb{Q}} - \sqrt{v_t} \, dt \]

\[ dW_t^{(2), \mathbb{Q}^S} = dW_t^{(2), \mathbb{Q}} - \rho\sqrt{v_t} \, dt \]

The second equation follows because \(W^{(2),\mathbb{Q}}\) has correlation \(\rho\) with \(W^{(1),\mathbb{Q}}\), so the Girsanov drift for \(W^{(2)}\) picks up a factor of \(\rho\).

Heston Dynamics Under the Stock-Price Numeraire¶

Substituting the Girsanov shifts into the \(\mathbb{Q}\)-dynamics:

Stock price:

\[ dS_t = (r - q)S_t \, dt + \sqrt{v_t} S_t \left(dW_t^{(1),\mathbb{Q}^S} + \sqrt{v_t} \, dt\right) \]

\[ = (r - q + v_t) S_t \, dt + \sqrt{v_t} S_t \, dW_t^{(1), \mathbb{Q}^S} \]

Variance:

\[ dv_t = \kappa(\theta - v_t) \, dt + \xi\sqrt{v_t}\left(dW_t^{(2),\mathbb{Q}^S} + \rho\sqrt{v_t}\,dt\right) \]

\[ = \left[\kappa\theta - (\kappa - \rho\xi)v_t\right] dt + \xi\sqrt{v_t} \, dW_t^{(2),\mathbb{Q}^S} \]

Theorem (Heston Under Stock-Price Numeraire)

Under \(\mathbb{Q}^S\), the Heston dynamics are

\[ dS_t = (r - q + v_t) S_t \, dt + \sqrt{v_t} S_t \, dW_t^{(1), \mathbb{Q}^S} \]

\[ dv_t = \kappa^*(\theta^* - v_t) \, dt + \xi\sqrt{v_t} \, dW_t^{(2), \mathbb{Q}^S} \]

with modified mean-reversion parameters:

\[ \kappa^* = \kappa - \rho\xi, \qquad \theta^* = \frac{\kappa\theta}{\kappa - \rho\xi} \]

The vol-of-vol \(\xi\) and correlation \(\rho\) are unchanged.

Effect of Negative Correlation

For the empirically relevant case \(\rho < 0\): \(\kappa^* = \kappa + |\rho|\xi > \kappa\) (faster mean reversion) and \(\theta^* = \kappa\theta/(\kappa + |\rho|\xi) < \theta\) (lower long-run variance). This makes intuitive sense: the \(\mathbb{Q}^S\) measure weights high-stock-price scenarios more heavily, and negative correlation implies these scenarios have lower variance.

Characteristic Function Under the Stock-Price Numeraire¶

The CF of \(\log S_T\) under \(\mathbb{Q}^S\) has the same affine structure but with modified parameters.

Proposition (CF Under Stock-Price Numeraire)

The characteristic function under \(\mathbb{Q}^S\) is

\[ \varphi_1(u) = \exp\bigl(C_1(\tau, u) + D_1(\tau, u) v_0 + iu\log S_0\bigr) \]

where \(C_1\) and \(D_1\) solve the same Riccati system as the \(\mathbb{Q}\)-CF but with \(\kappa\) replaced by \(\kappa^* = \kappa - \rho\xi\) and \(\theta\) replaced by \(\theta^* = \kappa\theta/\kappa^*\).

Alternatively, \(\varphi_1\) can be obtained from the \(\mathbb{Q}\)-CF by the shift formula:

\[ \varphi_1(u) = \frac{\varphi(u - i)}{S_0 e^{(r-q)\tau}} \]

where \(\varphi\) is the standard Heston CF under \(\mathbb{Q}\).

Proof of the shift formula. By definition:

\[ \varphi_1(u) = \mathbb{E}^{\mathbb{Q}^S}[e^{iu\log S_T}] = \mathbb{E}^{\mathbb{Q}}\!\left[\frac{S_T}{S_0 e^{(r-q)\tau}} e^{iu\log S_T}\right] = \frac{\mathbb{E}^{\mathbb{Q}}[e^{(iu+1)\log S_T}]}{S_0 e^{(r-q)\tau}} \]

\[ = \frac{\varphi(u - i)}{S_0 e^{(r-q)\tau}} \]

where we used \(e^{\log S_T} e^{iu\log S_T} = e^{i(u-i)\log S_T}\) and \(\varphi(u-i) = \mathbb{E}^{\mathbb{Q}}[e^{i(u-i)\log S_T}]\). \(\square\)

Application: Exercise Probability P1¶

The European call price decomposes as

\[ C = S_0 e^{-q\tau} P_1 - K e^{-r\tau} P_2 \]

where \(P_2 = \mathbb{Q}(S_T > K)\) and \(P_1 = \mathbb{Q}^S(S_T > K)\).

The Gil-Pelaez formula gives

\[ P_1 = \frac{1}{2} + \frac{1}{\pi}\int_0^{\infty} \text{Re}\!\left[\frac{e^{-iu\log K}\varphi_1(u)}{iu}\right] du \]

Since \(\varphi_1(u) = \varphi(u-i)/(S_0 e^{(r-q)\tau})\), this can be evaluated using the same Heston CF code with a shifted argument.

Implementation Note

To compute \(P_1\) from \(\varphi\), evaluate \(\varphi(u - i)\) for real \(u\). This requires the CF at a complex argument with imaginary part \(-1\). The Albrecher stable formulation handles this without modification, since the stability condition \(|g| < 1\) extends to complex arguments satisfying \(\text{Im}(u) \geq -1\) (which corresponds to the existence of the first moment \(\mathbb{E}^{\mathbb{Q}}[S_T] < \infty\)).

Application: Asset-or-Nothing Digital Call¶

An asset-or-nothing call pays \(S_T\) if \(S_T > K\) and zero otherwise. Its price is

\[ V_{\text{AoN}} = e^{-r\tau} \mathbb{E}^{\mathbb{Q}}[S_T \mathbf{1}_{S_T > K}] = S_0 e^{-q\tau} \mathbb{Q}^S(S_T > K) = S_0 e^{-q\tau} P_1 \]

This shows that the \(\mathbb{Q}^S\) measure is the natural pricing measure for asset-contingent payoffs.

Numerical Example¶

Using \(S_0 = 100\), \(K = 100\), \(r = 0.05\), \(q = 0\), \(v_0 = 0.04\), \(\kappa = 1.5\), \(\theta = 0.04\), \(\xi = 0.3\), \(\rho = -0.7\), \(\tau = 1\).

Risk-neutral parameters (\(\mathbb{Q}\)): \(\kappa = 1.5\), \(\theta = 0.04\).

Stock-numeraire parameters (\(\mathbb{Q}^S\)):

\[ \kappa^* = 1.5 - (-0.7)(0.3) = 1.5 + 0.21 = 1.71 \]

\[ \theta^* = \frac{1.5 \times 0.04}{1.71} = 0.0351 \]

Probabilities:

\[ P_2 = \mathbb{Q}(S_T > K) = 0.5134 \]

\[ P_1 = \mathbb{Q}^S(S_T > K) = 0.5748 \]

Note that \(P_1 > P_2\) always holds for \(r > q\), because the \(\mathbb{Q}^S\) measure weights high-stock-price outcomes more heavily.

Option prices:

\[ C = 100(0.5748) - 100 e^{-0.05}(0.5134) = 57.48 - 48.85 = 8.63 \]

\[ \text{Asset-or-nothing call} = S_0 P_1 = 100 \times 0.5748 = 57.48 \]

\[ \text{Cash-or-nothing call} = e^{-r\tau} P_2 = 0.9512 \times 0.5134 = 0.4884 \]

Delta Interpretation

The delta of the European call is

\[ \Delta = e^{-q\tau} P_1 = 0.5748 \]

This is the expected value under \(\mathbb{Q}^S\) of the exercise indicator. The delta is always between 0 and 1 for calls (and between \(-1\) and 0 for puts), and it equals \(P_1\) when \(q = 0\).

Summary¶

The stock-price numeraire measure \(\mathbb{Q}^S\) is obtained from the risk-neutral measure \(\mathbb{Q}\) by the Radon-Nikodym derivative \(d\mathbb{Q}^S/d\mathbb{Q} = S_T / (S_0 e^{(r-q)\tau})\). Under \(\mathbb{Q}^S\), the Heston variance process has modified mean-reversion parameters \(\kappa^* = \kappa - \rho\xi\) and \(\theta^* = \kappa\theta/\kappa^*\), while \(\xi\) and \(\rho\) are unchanged. The characteristic function under \(\mathbb{Q}^S\) is obtained either by solving the Riccati system with modified parameters or by the shift formula \(\varphi_1(u) = \varphi(u-i)/(S_0 e^{(r-q)\tau})\). The probability \(P_1 = \mathbb{Q}^S(S_T > K)\) is the delta of the call (when \(q = 0\)) and prices asset-or-nothing digital options. The \(\mathbb{Q}^S\) framework completes the two-probability decomposition of the Heston call price.

Exercises¶

Exercise 1. The Radon-Nikodym derivative is \(d\mathbb{Q}^S/d\mathbb{Q} = S_T/(S_0 e^{(r-q)\tau})\). Verify that \(\mathbb{E}^{\mathbb{Q}}[d\mathbb{Q}^S/d\mathbb{Q}] = 1\) by using the fact that \(\mathbb{E}^{\mathbb{Q}}[S_T] = S_0 e^{(r-q)\tau}\) (the discounted stock is a \(\mathbb{Q}\)-martingale). Why is this expectation-one property necessary for \(\mathbb{Q}^S\) to be a valid probability measure?

Solution to Exercise 1

We compute

\[ \mathbb{E}^{\mathbb{Q}}\!\left[\frac{d\mathbb{Q}^S}{d\mathbb{Q}}\right] = \mathbb{E}^{\mathbb{Q}}\!\left[\frac{S_T}{S_0 e^{(r-q)\tau}}\right] = \frac{\mathbb{E}^{\mathbb{Q}}[S_T]}{S_0 e^{(r-q)\tau}} \]

Under \(\mathbb{Q}\), the discounted reinvested stock price \(e^{-(r-q)t}S_t\) is a martingale (since \(dS_t = (r-q)S_t\,dt + \sqrt{v_t}S_t\,dW_t^{(1),\mathbb{Q}}\)), so

\[ \mathbb{E}^{\mathbb{Q}}[S_T] = S_0 e^{(r-q)\tau} \]

Therefore

\[ \mathbb{E}^{\mathbb{Q}}\!\left[\frac{d\mathbb{Q}^S}{d\mathbb{Q}}\right] = \frac{S_0 e^{(r-q)\tau}}{S_0 e^{(r-q)\tau}} = 1 \]

Why this is necessary. For \(\mathbb{Q}^S\) to be a valid probability measure, it must assign total mass 1 to the sample space \(\Omega\). The Radon-Nikodym derivative \(Z = d\mathbb{Q}^S/d\mathbb{Q}\) defines \(\mathbb{Q}^S\) via \(\mathbb{Q}^S(A) = \mathbb{E}^{\mathbb{Q}}[Z\cdot\mathbf{1}_A]\) for any event \(A\). For the full space \(A = \Omega\):

\[ \mathbb{Q}^S(\Omega) = \mathbb{E}^{\mathbb{Q}}[Z] = 1 \]

If \(\mathbb{E}^{\mathbb{Q}}[Z] \neq 1\), then \(\mathbb{Q}^S\) would not be a probability measure. Additionally, \(Z\) must be strictly positive almost surely for \(\mathbb{Q}^S\) to be equivalent to \(\mathbb{Q}\) (not just absolutely continuous). Since \(S_T > 0\) almost surely in the Heston model (the geometric Brownian motion with stochastic volatility preserves positivity), \(Z > 0\) a.s., confirming equivalence.

Exercise 2. Under \(\mathbb{Q}^S\), the variance parameters change to \(\kappa^* = \kappa - \rho\xi\) and \(\theta^* = \kappa\theta/\kappa^*\). For \(\rho = -0.7\), \(\xi = 0.3\), \(\kappa = 1.5\), \(\theta = 0.04\), verify the numerical example values \(\kappa^* = 1.71\) and \(\theta^* = 0.0351\). Now consider \(\rho = +0.5\) (positive correlation). Compute \(\kappa^*\) and \(\theta^*\) and explain why the mean-reversion speed decreases and the long-run variance increases in this case.

Solution to Exercise 2

Verification for \(\rho = -0.7\):

\[ \kappa^* = \kappa - \rho\xi = 1.5 - (-0.7)(0.3) = 1.5 + 0.21 = 1.71 \]

\[ \theta^* = \frac{\kappa\theta}{\kappa^*} = \frac{1.5 \times 0.04}{1.71} = \frac{0.06}{1.71} = 0.0351 \]

Both match the numerical example.

For \(\rho = +0.5\):

\[ \kappa^* = 1.5 - (0.5)(0.3) = 1.5 - 0.15 = 1.35 \]

\[ \theta^* = \frac{1.5 \times 0.04}{1.35} = \frac{0.06}{1.35} = 0.0444 \]

The mean-reversion speed decreases (\(\kappa^* = 1.35 < \kappa = 1.5\)) and the long-run variance increases (\(\theta^* = 0.0444 > \theta = 0.04\)).

Explanation. The stock-price numeraire \(\mathbb{Q}^S\) weights high-stock-price outcomes more heavily. The sign of the effect on variance parameters depends on the correlation \(\rho\):

When \(\rho < 0\) (leverage effect): high stock prices are associated with low variance. Under \(\mathbb{Q}^S\), these low-variance states get more weight, so \(\theta^*\) decreases and \(\kappa^*\) increases (faster reversion to a lower level).
When \(\rho > 0\) (positive correlation): high stock prices are associated with high variance. Under \(\mathbb{Q}^S\), these high-variance states get more weight, so \(\theta^*\) increases and \(\kappa^*\) decreases (slower reversion to a higher level).

The formula \(\kappa^* = \kappa - \rho\xi\) makes this transparent: when \(\rho > 0\), the subtracted term is positive, reducing \(\kappa^*\).

Exercise 3. The shift formula states \(\varphi_1(u) = \varphi(u - i)/(S_0 e^{(r-q)\tau})\), which requires evaluating the Heston CF at \(u - i\) (a complex argument with imaginary part \(-1\)). Show that this evaluation is well-defined provided \(\mathbb{E}^{\mathbb{Q}}[S_T] < \infty\), which is equivalent to the first moment condition. What happens if you try to compute \(\varphi(u - 2i)\), corresponding to \(\mathbb{E}^{\mathbb{Q}}[S_T^2]\)? Under what Heston parameter condition does this second moment exist?

Solution to Exercise 3

The shift formula requires evaluating \(\varphi(u - i)\), which involves \(\mathbb{E}^{\mathbb{Q}}[e^{i(u-i)\log S_T}] = \mathbb{E}^{\mathbb{Q}}[e^{iu\log S_T + \log S_T}] = \mathbb{E}^{\mathbb{Q}}[S_T e^{iu\log S_T}]\).

For this to be well-defined, we need \(\mathbb{E}^{\mathbb{Q}}[S_T] < \infty\). In the Heston model, \(S_T\) has a lognormal-like distribution with stochastic variance, and \(\mathbb{E}^{\mathbb{Q}}[S_T] = S_0 e^{(r-q)\tau} < \infty\). This is the first moment condition, and it is always satisfied because \(S_t\) is a \(\mathbb{Q}\)-martingale after discounting.

For \(\varphi(u - 2i)\): This requires evaluating

\[ \varphi(u - 2i) = \mathbb{E}^{\mathbb{Q}}[e^{i(u-2i)\log S_T}] = \mathbb{E}^{\mathbb{Q}}[S_T^2 e^{iu\log S_T}] \]

At \(u = 0\), this gives \(\mathbb{E}^{\mathbb{Q}}[S_T^2]\), the second moment of the stock price. In the Heston model, this exists if and only if the Riccati ODE for \(D(\tau, u)\) does not explode when evaluated at \(u = -2i\) (i.e., with the substitution \(iu \to 2\)).

The relevant condition involves the discriminant of the Riccati equation. The characteristic exponent \(D(\tau, u)\) solves

\[ \frac{dD}{d\tau} = \frac{1}{2}(iu)^2 - \frac{1}{2}(iu) + [\rho\xi(iu) - \kappa]D + \frac{1}{2}\xi^2 D^2 \]

Setting \(w = iu\), for the second moment we need \(w = 2\). The discriminant is

\[ d = (\kappa - \rho\xi w)^2 - \xi^2(w^2 - w) = (\kappa - 2\rho\xi)^2 - \xi^2(4 - 2) = (\kappa - 2\rho\xi)^2 - 2\xi^2 \]

The second moment exists for all \(\tau > 0\) provided \(D(\tau)\) remains finite, which requires \(\kappa - 2\rho\xi > \xi\sqrt{2}\) (or more precisely, that the Riccati ODE does not blow up on \([0, \tau]\)). For the numerical parameters \(\kappa = 1.5\), \(\rho = -0.7\), \(\xi = 0.3\):

\[ \kappa - 2\rho\xi = 1.5 - 2(-0.7)(0.3) = 1.5 + 0.42 = 1.92 \]

\[ \xi\sqrt{2} = 0.3\sqrt{2} = 0.424 \]

Since \(1.92 > 0.424\), the second moment exists. In general, the condition for the \(n\)-th moment \(\mathbb{E}^{\mathbb{Q}}[S_T^n] < \infty\) becomes more restrictive as \(n\) increases, eventually failing for sufficiently large \(n\) (the moment explosion phenomenon studied by Andersen and Piterbarg, 2007).

Exercise 4. For a European put, \(\Delta_{\text{put}} = e^{-q\tau}(P_1 - 1)\) where \(P_1 = \mathbb{Q}^S(S_T > K)\). Using the numerical example (\(P_1 = 0.5748\), \(q = 0\)), compute \(\Delta_{\text{put}}\) and verify that \(\Delta_{\text{call}} - \Delta_{\text{put}} = e^{-q\tau}\) (put-call parity for deltas).

Solution to Exercise 4

Put delta:

\[ \Delta_{\text{put}} = e^{-q\tau}(P_1 - 1) = (1)(0.5748 - 1) = -0.4252 \]

Verification of put-call parity for deltas:

\[ \Delta_{\text{call}} - \Delta_{\text{put}} = e^{-q\tau}P_1 - e^{-q\tau}(P_1 - 1) = e^{-q\tau}P_1 - e^{-q\tau}P_1 + e^{-q\tau} = e^{-q\tau} \]

Numerically: \(0.5748 - (-0.4252) = 1.0000 = e^{-0\cdot 1} = e^{-q\tau}\). Confirmed.

This follows from put-call parity at the price level: \(C - P = S_0 e^{-q\tau} - Ke^{-r\tau}\). Differentiating with respect to \(S_0\):

\[ \frac{\partial C}{\partial S_0} - \frac{\partial P}{\partial S_0} = e^{-q\tau} \]

which gives \(\Delta_{\text{call}} - \Delta_{\text{put}} = e^{-q\tau}\). When \(q = 0\), the call and put deltas always sum to 1 in absolute value.

Exercise 5. The \(\mathbb{Q}^S\) measure weights high-stock-price outcomes more heavily, so \(P_1 > P_2\) when \(r > q\). Show this rigorously: \(P_1 - P_2 = \mathbb{E}^{\mathbb{Q}}[(S_T/(S_0 e^{(r-q)\tau}) - 1)\mathbf{1}_{S_T > K}] > 0\). Why is the inequality strict? What happens to \(P_1 - P_2\) as the strike \(K \to \infty\) (deep OTM call)?

Solution to Exercise 5

We compute \(P_1 - P_2\) using the definition of \(\mathbb{Q}^S\):

\[ P_1 - P_2 = \mathbb{Q}^S(S_T > K) - \mathbb{Q}(S_T > K) \]

\[ = \mathbb{E}^{\mathbb{Q}}\!\left[\frac{S_T}{S_0 e^{(r-q)\tau}}\mathbf{1}_{S_T > K}\right] - \mathbb{E}^{\mathbb{Q}}[\mathbf{1}_{S_T > K}] \]

\[ = \mathbb{E}^{\mathbb{Q}}\!\left[\left(\frac{S_T}{S_0 e^{(r-q)\tau}} - 1\right)\mathbf{1}_{S_T > K}\right] \]

Why this is strictly positive when \(r > q\). On the event \(\{S_T > K\}\), the stock price is above the strike. The integrand \(\left(\frac{S_T}{S_0 e^{(r-q)\tau}} - 1\right)\mathbf{1}_{S_T > K}\) can be positive or negative depending on whether \(S_T\) is above or below \(S_0 e^{(r-q)\tau}\). However, the key insight is that conditional on \(S_T > K\), the values of \(S_T\) above \(S_0 e^{(r-q)\tau}\) are weighted by \((S_T/(S_0 e^{(r-q)\tau}) - 1) > 0\), and these large-\(S_T\) outcomes contribute more in absolute terms than the deficit from outcomes where \(K < S_T < S_0 e^{(r-q)\tau}\).

More rigorously, define \(F = S_0 e^{(r-q)\tau}\) (the forward price). Then

\[ P_1 - P_2 = \frac{1}{F}\mathbb{E}^{\mathbb{Q}}[(S_T - F)\mathbf{1}_{S_T > K}] \]

Since \(\mathbb{E}^{\mathbb{Q}}[S_T - F] = 0\) (martingale property), we have \(\mathbb{E}^{\mathbb{Q}}[(S_T - F)\mathbf{1}_{S_T \leq K}] = -\mathbb{E}^{\mathbb{Q}}[(S_T - F)\mathbf{1}_{S_T > K}]\). On \(\{S_T \leq K\}\), \(S_T - F\) is mostly negative (for reasonable \(K\)), so \(\mathbb{E}^{\mathbb{Q}}[(S_T - F)\mathbf{1}_{S_T \leq K}] < 0\), which means \(\mathbb{E}^{\mathbb{Q}}[(S_T - F)\mathbf{1}_{S_T > K}] > 0\). Hence \(P_1 - P_2 > 0\).

The inequality is strict because \(S_T\) has a continuous distribution (no atoms) and \(\mathbb{Q}(S_T > K) > 0\) for any finite \(K\).

Behavior as \(K \to \infty\). Both \(P_1\) and \(P_2\) converge to 0 (the call goes deep OTM and is almost never exercised). The difference \(P_1 - P_2\) also converges to 0, but \(P_1/P_2\) can diverge because \(P_1\) decays more slowly than \(P_2\). Intuitively, the \(\mathbb{Q}^S\) measure overweights the extreme right tail of \(S_T\), so the probability of exercising a deep OTM call is relatively higher under \(\mathbb{Q}^S\) than under \(\mathbb{Q}\).

Exercise 6. Consider an asset-or-nothing put, which pays \(S_T\) if \(S_T < K\). Show that its price is \(S_0 e^{-q\tau}(1 - P_1)\) and verify this against the numerical example: \(\text{AoN put} = 100(1 - 0.5748) = 42.52\). Use the decomposition \(S_0 e^{-q\tau} = \text{AoN call} + \text{AoN put}\) to confirm consistency.

Solution to Exercise 6

Asset-or-nothing put price. The asset-or-nothing put pays \(S_T\mathbf{1}_{S_T < K}\). Its price is

\[ V_{\text{AoN put}} = e^{-r\tau}\mathbb{E}^{\mathbb{Q}}[S_T\mathbf{1}_{S_T < K}] \]

We use the identity \(\mathbf{1}_{S_T < K} = 1 - \mathbf{1}_{S_T > K}\) (ignoring the zero-probability event \(S_T = K\)):

\[ V_{\text{AoN put}} = e^{-r\tau}\mathbb{E}^{\mathbb{Q}}[S_T] - e^{-r\tau}\mathbb{E}^{\mathbb{Q}}[S_T\mathbf{1}_{S_T > K}] \]

\[ = e^{-r\tau}S_0 e^{(r-q)\tau} - S_0 e^{-q\tau}P_1 = S_0 e^{-q\tau}(1 - P_1) \]

Numerical verification. With \(S_0 = 100\), \(q = 0\), \(P_1 = 0.5748\):

\[ V_{\text{AoN put}} = 100(1 - 0.5748) = 100 \times 0.4252 = 42.52 \]

Consistency check. The decomposition requires

\[ S_0 e^{-q\tau} = V_{\text{AoN call}} + V_{\text{AoN put}} \]

since \(S_T\mathbf{1}_{S_T > K} + S_T\mathbf{1}_{S_T < K} = S_T\) (the stock pays either above or below \(K\)). Numerically:

\[ V_{\text{AoN call}} + V_{\text{AoN put}} = S_0 P_1 + S_0(1 - P_1) = 57.48 + 42.52 = 100.00 = S_0 e^{-q\tau} \]

Confirmed. This decomposition is the asset-or-nothing analogue of the forward price identity: the present value of receiving \(S_T\) unconditionally is \(S_0 e^{-q\tau}\), which splits into the AoN call (receive \(S_T\) if exercised) and AoN put (receive \(S_T\) if not exercised).

Exercise 7. The Feller condition under \(\mathbb{Q}^S\) requires \(2\kappa^*\theta^* \geq \xi^2\). Since \(\kappa^*\theta^* = \kappa\theta\) (verify this), the Feller condition under \(\mathbb{Q}^S\) is identical to that under \(\mathbb{Q}\). Prove this algebraically and discuss why measure changes within the affine family preserve the Feller condition.

Solution to Exercise 7

Algebraic proof. By definition:

\[ \kappa^*\theta^* = (\kappa - \rho\xi)\cdot\frac{\kappa\theta}{\kappa - \rho\xi} = \kappa\theta \]

The factor \((\kappa - \rho\xi)\) cancels exactly, giving \(\kappa^*\theta^* = \kappa\theta\) for any values of \(\rho\), \(\xi\), and \(\kappa\).

Feller condition. Under \(\mathbb{Q}\), the Feller condition is \(2\kappa\theta \geq \xi^2\). Under \(\mathbb{Q}^S\), it is \(2\kappa^*\theta^* \geq \xi^2\). Since \(\kappa^*\theta^* = \kappa\theta\) and \(\xi\) is unchanged:

\[ 2\kappa^*\theta^* \geq \xi^2 \iff 2\kappa\theta \geq \xi^2 \]

The conditions are identical.

Why affine measure changes preserve the Feller condition. The general pattern is as follows. The variance SDE under any measure in the affine family has the form

\[ dv_t = [\alpha - \beta v_t]\,dt + \xi\sqrt{v_t}\,dW_t \]

The Feller condition is \(2\alpha \geq \xi^2\). Both the \(\mathbb{P} \to \mathbb{Q}\) measure change (with \(\lambda_v = \lambda\sqrt{v_t}\)) and the \(\mathbb{Q} \to \mathbb{Q}^S\) measure change modify the drift by adding a term proportional to \(v_t\):

\(\mathbb{P} \to \mathbb{Q}\): adds \(-\xi\lambda v_t\,dt\), changing \(\beta\) but not \(\alpha\)
\(\mathbb{Q} \to \mathbb{Q}^S\): adds \(\rho\xi v_t\,dt\), changing \(\beta\) but not \(\alpha\)

In both cases, the constant term \(\alpha = \kappa\theta\) in the drift is invariant. Since the Feller condition depends only on \(\alpha\) and \(\xi\) (both unchanged), it is preserved.

This invariance holds because the Girsanov shift is proportional to the diffusion coefficient \(\xi\sqrt{v_t}\), so the drift adjustment is proportional to \(v_t\) (not to a constant or to \(\sqrt{v_t}\)). Any measure change that shifts the drift by a term proportional to \(v_t\) preserves the constant part of the drift and hence the Feller condition. This is the structural reason why the specification \(\lambda_v \propto \sqrt{v_t}\) (or the numeraire change induced by the stock) preserves both the affine form and the boundary behavior of the CIR process.