Change of Numeraire: Alternative Derivation of Black-Scholes¶

Everything in this subsubsection follows from one sentence:

A change of numeraire is a reweighting of probabilitieson sample paths.

The § Heat Equation and § Feynman–Kac subsubsections derived the Black–Scholes formula

\[ C = S\mathcal{N}(d_1) - Ke^{-r\tau}\mathcal{N}(d_2) \]

but did not answer a basic question: why exactly two terms, each a normal CDF? The change-of-numeraire viewpoint answers it directly: the two CDFs are two probabilities of the same event $\{S_T > K\}$ under two different probability measures, each natural to a different "unit of account" (numeraire).

\[ \mathcal{N}(d_1) = \mathbb{Q}^S(S_T>K),\quad \mathcal{N}(d_2) = \mathbb{Q}(S_T>K) \]

We build the picture in the order a reader naturally builds intuition: a one-step binomial toy first, where the measure change is just finite reweighting of a finite probability space, then the continuous-time Girsanov machinery that lifts the toy to Black–Scholes.

Terminology: stock measure vs. $T$-forward measure

The measure $\mathbb{Q}^S$ obtained by taking the stock $S_t$ as numeraire is called the stock measure (or share measure). In some texts it is loosely called a "forward measure", but this terminology should be avoided: in interest-rate theory the $T$-forward measure $\mathbb{Q}^T$ is a different object, defined by taking the zero-coupon bond $P(t,T)$ as numeraire. The two measures coincide only when interest rates are deterministic and the underlying is itself a bond. Throughout this subsubsection, "stock measure" means $\mathbb{Q}^S$ with numeraire $S_t$; the $T$-forward measure $\mathbb{Q}^T$ appears only in the summary table and in Exercise 6.

1. Why Numeraires? A One-Step Binomial Toy¶

Before any Girsanov machinery, work the simplest possible measure change on a finite probability space.

1.1 Setup: a One-Step Binomial Market¶

At time $1$ the stock takes value $S_u$ (up) or $S_d$ (down), with $S_u > S_0 e^r > S_d$ (no-arbitrage). Under the risk-neutral measure $\mathbb{Q}$, the probability of the up move is fixed by the martingale condition $e^{-r}\, \mathbb{E}^{\mathbb{Q}}[S_1] = S_0$:

\[ q := \mathbb{Q}(\text{up}) = \frac{S_0\, e^r - S_d}{S_u - S_d}, \qquad 1 - q = \mathbb{Q}(\text{down}) = \frac{S_u - S_0\, e^r}{S_u - S_d} \]

The Radon–Nikodym derivative against the physical measure $\mathbb{P}$ is simply the ratio of probabilities on each state — a vector of two numbers. The measure change is just a finite reweighting.

1.2 Pricing a Call in Two Numeraires¶

Price a European call $C_1 = (S_1 - K)^+$ with $S_d < K < S_u$. Under $\mathbb{Q}$ with the money-market numeraire $B_t = e^{rt}$:

\[ C_0 = e^{-r}\, \mathbb{E}^{\mathbb{Q}}[(S_1 - K)^+] = e^{-r}\, q\, (S_u - K) \]

Split:

\[ C_0 = \underbrace{e^{-r}\, q\, S_u}_{\text{"$S$-piece"}} - \underbrace{K\, e^{-r}\, q}_{\text{"$\$$-piece"}} \]

The $\$$-piece reads cleanly as $K$ times the discounted risk-neutral probability of exercise $e^{-r}\, \mathbb{Q}(S_1 > K)$. The $S$-piece is the expected discounted in-the-money stock value $e^{-r}\, \mathbb{E}^{\mathbb{Q}}[S_1\, \mathbb{1}_{S_1 > K}]$ — a product, not yet "$S_0$ times a probability." The question of the subsubsection is whether it can be made to look like the $\$$-piece, by changing measures.

1.3 The Stock as Numeraire¶

Yes — reweight $\mathbb{Q}$ once more. Define the stock measure $\mathbb{Q}^S$ by the Radon–Nikodym derivative

\[ \frac{d\mathbb{Q}^S}{d\mathbb{Q}} := \frac{S_1}{S_0\, e^r} \]

— the ratio of the stock's terminal value to its $\mathbb{Q}$-expectation. By the martingale condition $\mathbb{E}^{\mathbb{Q}}[d\mathbb{Q}^S / d\mathbb{Q}] = 1$, so $\mathbb{Q}^S$ is a valid probability measure. State-by-state:

\[ \mathbb{Q}^S(\text{up}) = q\, \frac{S_u}{S_0\, e^r}, \qquad \mathbb{Q}^S(\text{down}) = (1 - q)\, \frac{S_d}{S_0\, e^r} \]

The up state is upweighted by the factor $S_u / (S_0 e^r) > 1$; the down state is downweighted by $S_d / (S_0 e^r) < 1$. Probability mass flows toward the in-the-money state. This is the entire content of switching numeraires.

Now compute the $S$-piece directly under $\mathbb{Q}^S$:

\[ e^{-r}\, q\, S_u = S_0\, \cdot\, \frac{q\, S_u}{S_0\, e^r} = S_0\, \mathbb{Q}^S(\text{up}) = S_0\, \mathbb{Q}^S(S_1 > K) \]

The product becomes exactly "$S_0$ times a probability," with the probability computed under the new measure.

1.4 The Two-Numeraire Decomposition¶

Putting both pieces together:

\[ C_0 = S_0\, \mathbb{Q}^S(S_1 > K) - K\, e^{-r}\, \mathbb{Q}(S_1 > K) \]

Each term is unit $\times$ probability of exercise, with the unit and the measure matched consistently. The $\$$-piece uses the bond numeraire and the risk-neutral measure; the $S$-piece uses the stock numeraire and the stock measure. The two probabilities are of the same event, computed under two different reweightings of the underlying probability space.

1.5 The Continuous-Time Lift¶

This is the binomial analogue of the Black–Scholes formula

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

with the identification

\[ \mathcal{N}(d_2) = \mathbb{Q}(S_T > K), \qquad \mathcal{N}(d_1) = \mathbb{Q}^S(S_T > K) \]

The two CDFs in the Black–Scholes formula are two probabilities of the same event under two different measures. The discrete reweighting $S_1 / (S_0 e^r)$ becomes, in the continuum limit, the Doléans exponential

\[ \frac{d\mathbb{Q}^S}{d\mathbb{Q}}\bigg|_{\mathcal{F}_T} = \frac{S_T}{S_0\, e^{rT}} = \exp\!\left(\sigma\, W_T^{\mathbb{Q}} - \tfrac{1}{2}\sigma^2\, T\right) \]

— the Girsanov density. Under $\mathbb{Q}^S$, Brownian motion picks up a drift $\sigma\, t$, and $W_t^{\mathbb{Q}^S} := W_t^{\mathbb{Q}} - \sigma\, t$ is a standard Brownian motion. The rest of this subsubsection is the rigorous continuous-time form of the two-state calculation above — Doléans replaces "vector of two numbers," Girsanov replaces "finite reweighting."

Core principle

A change of numeraire is a reweighting of probabilities. Each numeraire $N_t$ comes with its own measure $\mathbb{Q}^N$, defined by the requirement that prices expressed in units of $N_t$ are $\mathbb{Q}^N$-martingales. The Radon–Nikodym derivative $d\mathbb{Q}^N / d\mathbb{Q}$ encodes the relative reweighting: a finite ratio per state in discrete time, a Doléans exponential in continuous time.

This is the mechanism. Black–Scholes is the application: $\mathcal{N}(d_1)$ and $\mathcal{N}(d_2)$ are probabilities of the same event $\{S_T > K\}$ under the stock measure $\mathbb{Q}^S$ and the risk-neutral measure $\mathbb{Q}$ respectively.

2. Numeraire and Pricing Measures¶

1. Definition of Numeraire¶

A numeraire $N_t$ is a strictly positive traded asset used as a unit of account. All asset prices are expressed relative to the numeraire:

\[ \text{Relative price} = \frac{S_t}{N_t} \]

Key property: In an arbitrage-free market, there exists a probability measure (called the numeraire measure or $N$-measure) under which all asset prices relative to $N_t$ are martingales.

2. Standard Risk-Neutral Measure¶

In the Black-Scholes framework, the standard numeraire is the money market account:

\[ B_t = e^{rt} \]

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

The relative price $S_t/B_t = e^{-rt}S_t$ is a martingale
This gives the standard BS dynamics: $dS_t = rS_t dt + \sigma S_t dW_t^{\mathbb{Q}}$

Option pricing formula:

\[ V_0 = \mathbb{E}^{\mathbb{Q}}\left[\frac{V_T}{B_T}\right] = e^{-rT}\mathbb{E}^{\mathbb{Q}}[V_T] \]

3. General Numeraire Change¶

1. Fundamental Theorem¶

Numeraire Change Theorem: Let $N_t$ be any numeraire (strictly positive traded asset). There exists a unique equivalent probability measure $\mathbb{Q}^N$ under which all asset prices relative to $N_t$ are martingales.

Specifically, for any traded asset $S_t$:

\[ \frac{S_t}{N_t} = \mathbb{E}^{\mathbb{Q}^N}\left[\frac{S_T}{N_T} \Big| \mathcal{F}_t\right] \]

2. Radon–Nikodym Derivative¶

The measure $\mathbb{Q}^N$ is related to the standard risk-neutral measure $\mathbb{Q}$ via:

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

Intuition: We reweight paths according to the terminal value of the numeraire (discounted).

3. Girsanov Connection¶

The change of numeraire induces a change of Brownian motion via Girsanov's theorem. If under $\mathbb{Q}$:

\[ dN_t = \mu_N N_t dt + \sigma_N N_t dW_t^{\mathbb{Q}} \]

then under $\mathbb{Q}^N$:

\[ dW_t^{\mathbb{Q}^N} = dW_t^{\mathbb{Q}} - \sigma_N dt \]

Equivalently, $dW_t^{\mathbb{Q}} = dW_t^{\mathbb{Q}^N} + \sigma_N dt$. The drift adjustment reflects the covariance between the asset and the numeraire.

4. Stock Numeraire and the Stock Measure¶

1. Setup: Stock as Numeraire¶

Choose $N_t = S_t$ (the underlying stock itself).

The associated measure $\mathbb{Q}^S$ is called the stock measure (or share measure). As noted in the terminology warning above, it should not be conflated with the $T$-forward measure $\mathbb{Q}^T$ of interest-rate theory, which uses a zero-coupon bond as numeraire.

2. Relative Prices Under Q^S¶

Under the stock measure, all assets relative to $S_t$ are martingales.

Money market account relative to stock:

\[ \frac{B_t}{S_t} = \mathbb{E}^{\mathbb{Q}^S}\left[\frac{B_T}{S_T} \Big| \mathcal{F}_t\right] \]

Strike relative to stock:

\[ \frac{K}{S_t} = \mathbb{E}^{\mathbb{Q}^S}\left[\frac{K}{S_T} \Big| \mathcal{F}_t\right] \]

3. Radon–Nikodym Derivative¶

The stock measure $\mathbb{Q}^S$ is related to $\mathbb{Q}$ by the density process

\[ Z_t := \frac{d\mathbb{Q}^S}{d\mathbb{Q}}\bigg|_{\mathcal{F}_t} = \frac{S_t / B_t}{S_0 / B_0} = \frac{S_t e^{-rt}}{S_0} \]

evaluated at $t = T$:

\[ \frac{d\mathbb{Q}^S}{d\mathbb{Q}}\Big|_{\mathcal{F}_T} = \frac{S_T e^{-rT}}{S_0} \]

Since $S_t e^{-rt} / S_0 = \exp(-\tfrac{1}{2}\sigma^2 t + \sigma W_t^{\mathbb{Q}})$ is a Doléans-Dade exponential with bounded volatility $\sigma$, Novikov's condition $\mathbb{E}^{\mathbb{Q}}[\exp(\tfrac{1}{2}\sigma^2 T)] < \infty$ is trivially satisfied, so $Z_t$ is a true $\mathbb{Q}$-martingale (not merely a local martingale) with $\mathbb{E}^{\mathbb{Q}}[Z_T] = 1$. This is the precise integrability condition that makes the change of measure rigorous; under it, $\mathbb{Q}^S$ is a well-defined probability measure equivalent to $\mathbb{Q}$ on $\mathcal{F}_T$.

4. Brownian Motion Under Q^S¶

If $dS_t = rS_t dt + \sigma S_t dW_t^{\mathbb{Q}}$ under $\mathbb{Q}$, then by Girsanov:

\[ dW_t^{\mathbb{Q}^S} = dW_t^{\mathbb{Q}} - \sigma dt \]

Substituting $dW_t^{\mathbb{Q}} = dW_t^{\mathbb{Q}^S} + \sigma dt$ into the stock dynamics:

\[ dS_t = rS_t dt + \sigma S_t(dW_t^{\mathbb{Q}^S} + \sigma dt) = (r + \sigma^2)S_t dt + \sigma S_t dW_t^{\mathbb{Q}^S} \]

Key property: The ratio $e^{rt}/S_t$ is a martingale under $\mathbb{Q}^S$, since its drift vanishes.

5. Black-Scholes via Stock Numeraire¶

1. Call Option Valuation¶

We want to price a European call with payoff $(S_T - K)^+$.

Step 1: Express payoff in numeraire units

Divide by $S_T$:

\[ \frac{(S_T - K)^+}{S_T} = \left(1 - \frac{K}{S_T}\right)^+ = \left(1 - \frac{K}{S_T}\right)\mathbf{1}_{\{S_T > K\}} \]

Step 2: Change to stock measure

By the numeraire change theorem:

\[ \frac{C_0}{S_0} = \mathbb{E}^{\mathbb{Q}^S}\left[\frac{C_T}{S_T}\right] = \mathbb{E}^{\mathbb{Q}^S}\left[\left(1 - \frac{K}{S_T}\right)\mathbf{1}_{\{S_T > K\}}\right] \]

Step 3: Decompose expectation

\[ \frac{C_0}{S_0} = \mathbb{E}^{\mathbb{Q}^S}[\mathbf{1}_{\{S_T > K\}}] - K\mathbb{E}^{\mathbb{Q}^S}\left[\frac{1}{S_T}\mathbf{1}_{\{S_T > K\}}\right] \]

2. First Term: $\mathbb{Q}^S(S_T > K)$¶

Under $\mathbb{Q}^S$, the shifted drift $(r + \sigma^2)$ derived in §4 above gives a GBM with log-normal terminal law

\[ \ln S_T \sim \mathcal{N}\!\left(\ln S_0 + (r + \tfrac{1}{2}\sigma^2)T,\ \sigma^2 T\right) \quad \text{under } \mathbb{Q}^S. \]

Recall (Gaussian tail computation)

Reducing $\{S_T > K\}$ to a standard-normal tail and evaluating it is the same completion-of-square computation carried out canonically in § Feynman–Kac (and equivalently in § Heat Equation via the heat-kernel picture). We do not repeat it.

The result is

\[ \mathbb{Q}^S(S_T > K) = \mathcal{N}(d_1), \qquad d_1 = \frac{\ln(S_0/K) + (r + \tfrac{1}{2}\sigma^2)T}{\sigma\sqrt{T}}. \]

The probabilistic content — and the reason this subsubsection exists — is that $\mathcal{N}(d_1)$ is the exercise probability under the stock measure, not under $\mathbb{Q}$.

3. Second Term: Change Back to $\mathbb{Q}$¶

For the second term, the abstract Bayes / Radon–Nikodym formula

\[ \mathbb{E}^{\mathbb{Q}^S}[X] = \mathbb{E}^{\mathbb{Q}}\left[\frac{d\mathbb{Q}^S}{d\mathbb{Q}}\bigg|_{\mathcal{F}_T} \cdot X\right] \]

with $X = S_T^{-1}\mathbf{1}_{\{S_T > K\}}$ and the explicit density $\frac{d\mathbb{Q}^S}{d\mathbb{Q}}\big|_{\mathcal{F}_T} = S_T e^{-rT}/S_0$ from §3 gives

\[ \mathbb{E}^{\mathbb{Q}^S}\left[\frac{1}{S_T}\mathbf{1}_{\{S_T > K\}}\right] = \mathbb{E}^{\mathbb{Q}}\left[\frac{S_T e^{-rT}}{S_0} \cdot \frac{1}{S_T}\mathbf{1}_{\{S_T > K\}}\right] = \frac{e^{-rT}}{S_0}\,\mathbb{Q}(S_T > K). \]

The two $S_T$ factors cancel cleanly — the structural reason the change of numeraire works. By the canonical computation in § Feynman–Kac,

\[ \mathbb{Q}(S_T > K) = \mathcal{N}(d_2), \qquad d_2 = d_1 - \sigma\sqrt{T}. \]

4. Final Result¶

Combining the two terms,

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

— the same formula obtained in § Heat Equation and § Feynman–Kac, but now reinterpreted: each $\mathcal{N}(d_i)$ is a probability of $\{S_T > K\}$, measured under a different numeraire. $\mathcal{N}(d_1)$ is the stock-measure probability; $\mathcal{N}(d_2)$ is the risk-neutral one. That asymmetry is the subsubsection's signature insight.

6. Foreign Exchange Options¶

1. Setup: Quanto Options¶

Consider an option on foreign exchange rate $X_t$ (domestic per foreign).

Two numeraires:

Domestic money market: $B_t^d = e^{r_d t}$
Foreign money market: $B_t^f = e^{r_f t}$

Exchange rate dynamics under domestic risk-neutral measure $\mathbb{Q}^d$:

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

2. Foreign Numeraire Measure¶

Choose numeraire $N_t = X_t B_t^f = X_t e^{r_f t}$ (foreign money market converted to domestic).

The Radon–Nikodym derivative is the density process $N_t / B_t^d$ normalised to start at $1$:

\[ \frac{d\mathbb{Q}^f}{d\mathbb{Q}^d}\bigg|_{\mathcal{F}_t} = \frac{N_t / B_t^d}{N_0 / B_0^d} = \frac{X_t e^{(r_f - r_d) t}}{X_0}, \qquad \frac{d\mathbb{Q}^f}{d\mathbb{Q}^d}\bigg|_{\mathcal{F}_T} = \frac{X_T e^{(r_f - r_d)T}}{X_0} \]

This is a true $\mathbb{Q}^d$-martingale (Novikov is satisfied since $\sigma$ is bounded), so the change of measure is well defined.

Under $\mathbb{Q}^f$:

\[ dW_t^{\mathbb{Q}^f} = dW_t^{\mathbb{Q}^d} - \sigma dt \]

Exchange rate becomes:

\[ dX_t = (r_d - r_f + \sigma^2)X_t dt + \sigma X_t dW_t^{\mathbb{Q}^f} \]

3. Call on FX Rate¶

Repeating the stock-numeraire argument of the previous subsubsection with $X_t e^{r_f t}$ in place of $S_t$ — the only change is that the discount rate is now $r_d$ and the foreign rate $r_f$ plays the role of a continuous dividend yield — gives the Garman–Kohlhagen formula

\[ C_0 = X_0 e^{-r_f T}\mathcal{N}(d_1) - K e^{-r_d T}\mathcal{N}(d_2), \qquad d_1 = \frac{\ln(X_0/K) + (r_d - r_f + \tfrac{1}{2}\sigma^2)T}{\sigma\sqrt{T}}, \quad d_2 = d_1 - \sigma\sqrt{T}. \]

The Gaussian tail evaluation is identical to the one in § Feynman–Kac; the new content here is the foreign–domestic symmetry: by reversing the roles of the two currencies one obtains the corresponding put formula and the put–call relation expressed in the foreign numeraire (Exercise 3).

7. When to Use Each Numeraire¶

Numeraire	Measure	Application	Advantage
Money market $B_t$	Risk-neutral $\mathbb{Q}$	Standard options	Familiar, drift = $r$
Stock $S_t$	Stock measure $\mathbb{Q}^S$ (a.k.a. share measure)	Equity options, forward contracts	$\mathcal{N}(d_1)$ as exercise probability
Zero-coupon bond $P(t,T)$	$T$-forward measure $\mathbb{Q}^T$ (distinct from $\mathbb{Q}^S$!)	Interest rate options	Forwards are $\mathbb{Q}^T$-martingales; simplifies swap pricing
Foreign money market	Foreign measure $\mathbb{Q}^f$	FX options	Symmetry between currencies

General principle: Choose the numeraire that eliminates drift from the payoff-relevant dynamics.

8. Summary¶

This subsubsection did not re-derive the Black–Scholes formula; it reinterpreted the formula already obtained in § Heat Equation and § Feynman–Kac. The signature insight: $\mathcal{N}(d_1)$ and $\mathcal{N}(d_2)$ are the same event $\{S_T > K\}$ measured under two different numeraires — the stock and the money market. The Radon–Nikodym density $S_T e^{-rT}/S_0$ links them, and Girsanov turns the density into a drift shift $\sigma$. The framework generalises immediately: replace $S$ by the foreign money account to get Garman–Kohlhagen, or by a zero-coupon bond to get the $T$-forward measure (Exercise 6).

In the operator language of § Introduction, the pricing semigroup $\mathcal{P}_\tau$ admits multiple expectation representations, one per numeraire — the Core Identity is invariant, only its decomposition into "probability × payoff factor" changes.

Exercises¶

Exercise 1. Let $N_t = e^{rt}$ be the money market account and $\mathbb{Q}$ the associated risk-neutral measure. Verify the Radon–Nikodym derivative formula by showing that $\frac{d\mathbb{Q}^S}{d\mathbb{Q}}\big|_{\mathcal{F}_T} = \frac{S_T e^{-rT}}{S_0}$ has expectation 1 under $\mathbb{Q}$.

Solution to Exercise 1

We need to show that $\mathbb{E}^{\mathbb{Q}}\left[\frac{S_T e^{-rT}}{S_0}\right] = 1$.

Under $\mathbb{Q}$, the discounted stock price $e^{-rt}S_t$ is a martingale, so:

\[ \mathbb{E}^{\mathbb{Q}}[e^{-rT}S_T] = e^{-r \cdot 0}S_0 = S_0 \]

Dividing both sides by $S_0$:

\[ \mathbb{E}^{\mathbb{Q}}\left[\frac{S_T e^{-rT}}{S_0}\right] = 1 \]

Alternatively, using the explicit formula $S_T = S_0 \exp\left((r - \frac{1}{2}\sigma^2)T + \sigma W_T^{\mathbb{Q}}\right)$:

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

This is the stochastic exponential $\mathcal{E}(\sigma W^{\mathbb{Q}})_T$. Using the moment generating function of the normal distribution with $W_T^{\mathbb{Q}} \sim \mathcal{N}(0, T)$:

\[ \mathbb{E}^{\mathbb{Q}}\left[\exp\left(-\frac{1}{2}\sigma^2 T + \sigma W_T^{\mathbb{Q}}\right)\right] = \exp\left(-\frac{1}{2}\sigma^2 T + \frac{1}{2}\sigma^2 T\right) = 1 \]

This confirms that the Radon–Nikodym derivative has unit expectation, as required for a valid change of measure.

Exercise 2. Under the stock measure $\mathbb{Q}^S$, the discounted bond price $B_t / S_t = e^{rt}/S_t$ is a martingale. Verify this explicitly by computing $d(e^{rt}/S_t)$ using Ito's lemma and the stock dynamics under $\mathbb{Q}^S$, and confirming that the drift vanishes.

Solution to Exercise 2

Under $\mathbb{Q}^S$, the stock dynamics are $dS_t = (r + \sigma^2)S_t \, dt + \sigma S_t \, dW_t^{\mathbb{Q}^S}$.

Define $Y_t = e^{rt}/S_t$. Apply Ito's lemma to $f(t, S) = e^{rt}/S$.

For $g(S) = 1/S$, we have $g'(S) = -1/S^2$ and $g''(S) = 2/S^3$. By Ito's lemma:

\[ d(1/S_t) = -\frac{1}{S_t^2} \, dS_t + \frac{1}{2} \cdot \frac{2}{S_t^3}(dS_t)^2 \]

Substituting $dS_t = (r + \sigma^2)S_t \, dt + \sigma S_t \, dW_t^{\mathbb{Q}^S}$ and $(dS_t)^2 = \sigma^2 S_t^2 \, dt$:

\[ d(1/S_t) = \frac{-(r + \sigma^2) + \sigma^2}{S_t} \, dt - \frac{\sigma}{S_t} \, dW_t^{\mathbb{Q}^S} = \frac{-r}{S_t} \, dt - \frac{\sigma}{S_t} \, dW_t^{\mathbb{Q}^S} \]

Therefore:

\[ dY_t = \frac{re^{rt}}{S_t} \, dt + e^{rt}\left(\frac{-r}{S_t} \, dt - \frac{\sigma}{S_t} \, dW_t^{\mathbb{Q}^S}\right) = -\sigma Y_t \, dW_t^{\mathbb{Q}^S} \]

The drift vanishes, confirming that $Y_t = e^{rt}/S_t$ is a martingale under $\mathbb{Q}^S$. $\square$

Exercise 3. Derive the Garman-Kohlhagen formula for a European call on a foreign exchange rate. Starting from the FX dynamics $dX_t = (r_d - r_f)X_t \, dt + \sigma X_t \, dW_t^{\mathbb{Q}^d}$, use the change of numeraire to the foreign money market and show that the call price is $C_0 = X_0 e^{-r_f T}\mathcal{N}(d_1) - Ke^{-r_d T}\mathcal{N}(d_2)$, where $d_1 = \frac{\ln(X_0/K) + (r_d - r_f + \frac{1}{2}\sigma^2)T}{\sigma\sqrt{T}}$.

Solution to Exercise 3

Under $\mathbb{Q}^d$, the FX rate satisfies $dX_t = (r_d - r_f)X_t \, dt + \sigma X_t \, dW_t^{\mathbb{Q}^d}$.

Step 1: Choose numeraire. Take the foreign money market account converted to domestic currency: $N_t = X_t e^{r_f t}$. The Radon–Nikodym derivative is:

\[ \frac{d\mathbb{Q}^f}{d\mathbb{Q}^d}\Big|_{\mathcal{F}_T} = \frac{X_T e^{(r_f - r_d)T}}{X_0} \]

Step 2: Girsanov shift. Under $\mathbb{Q}^f$: $dW_t^{\mathbb{Q}^f} = dW_t^{\mathbb{Q}^d} - \sigma \, dt$, so $dW_t^{\mathbb{Q}^d} = dW_t^{\mathbb{Q}^f} + \sigma \, dt$.

Under $\mathbb{Q}^f$, the FX dynamics become:

\[ dX_t = (r_d - r_f)X_t \, dt + \sigma X_t(dW_t^{\mathbb{Q}^f} + \sigma \, dt) = (r_d - r_f + \sigma^2)X_t \, dt + \sigma X_t \, dW_t^{\mathbb{Q}^f} \]

Step 3: Price the call. The call payoff is $(X_T - K)^+$. Using the numeraire $N_t = X_t e^{r_f t}$:

\[ \frac{C_0}{N_0} = \mathbb{E}^{\mathbb{Q}^f}\left[\frac{(X_T - K)^+}{N_T}\right] = \mathbb{E}^{\mathbb{Q}^f}\left[\frac{(X_T - K)^+}{X_T e^{r_f T}}\right] \]

\[ C_0 = X_0 \mathbb{E}^{\mathbb{Q}^f}\left[\frac{(X_T - K)^+}{X_T e^{r_f T}} \cdot e^{r_f T} \cdot \frac{X_T}{X_0}\right] \]

Alternatively, use the standard risk-neutral approach under $\mathbb{Q}^d$:

\[ C_0 = e^{-r_d T}\mathbb{E}^{\mathbb{Q}^d}[(X_T - K)^+] \]

Under $\mathbb{Q}^d$, $\ln X_T \sim \mathcal{N}(\ln X_0 + (r_d - r_f - \frac{1}{2}\sigma^2)T, \sigma^2 T)$. Following the standard Black-Scholes integral evaluation (splitting into two Gaussian integrals and completing the square):

\[ C_0 = X_0 e^{-r_f T}\mathcal{N}(d_1) - K e^{-r_d T}\mathcal{N}(d_2) \]

where:

\[ d_1 = \frac{\ln(X_0/K) + (r_d - r_f + \frac{1}{2}\sigma^2)T}{\sigma\sqrt{T}}, \quad d_2 = d_1 - \sigma\sqrt{T} \]

This is the Garman-Kohlhagen formula.

Exercise 4. Consider the exchange option (Margrabe's formula) with payoff $(S_T^{(1)} - S_T^{(2)})^+$ where both assets follow GBM with correlation $\rho$. Using $S_t^{(2)}$ as numeraire, show that the option price is $V_0 = S_0^{(1)}\mathcal{N}(d_1) - S_0^{(2)}\mathcal{N}(d_2)$ and determine the effective volatility $\hat{\sigma}$ that appears in $d_1$ and $d_2$.

Solution to Exercise 4

Let $S_t^{(1)}$ and $S_t^{(2)}$ follow correlated GBMs under $\mathbb{Q}$:

\[ dS_t^{(i)} = rS_t^{(i)} \, dt + \sigma_i S_t^{(i)} \, dW_t^{(i)}, \quad d\langle W^{(1)}, W^{(2)}\rangle_t = \rho \, dt \]

Step 1: Use $S_t^{(2)}$ as numeraire. The Radon–Nikodym derivative is:

\[ \frac{d\mathbb{Q}^{S^{(2)}}}{d\mathbb{Q}}\Big|_{\mathcal{F}_T} = \frac{S_T^{(2)} e^{-rT}}{S_0^{(2)}} \]

Step 2: Pricing formula. The exchange option price is:

\[ V_0 = e^{-rT}\mathbb{E}^{\mathbb{Q}}[(S_T^{(1)} - S_T^{(2)})^+] \]

Using the numeraire $S_t^{(2)}$:

\[ \frac{V_0}{S_0^{(2)}} = \mathbb{E}^{\mathbb{Q}^{S^{(2)}}}\left[\left(\frac{S_T^{(1)}}{S_T^{(2)}} - 1\right)^+\right] \]

Step 3: Dynamics of the ratio. Define $R_t = S_t^{(1)}/S_t^{(2)}$. By Ito's lemma:

\[ \frac{dR_t}{R_t} = (\sigma_1 \, dW_t^{(1)} - \sigma_2 \, dW_t^{(2)}) + (\sigma_2^2 - \rho\sigma_1\sigma_2) \, dt \]

Under $\mathbb{Q}^{S^{(2)}}$, the Girsanov shift removes the drift from $R_t/1$ (since $R_t = S_t^{(1)}/S_t^{(2)}$ must be a martingale under this measure). The volatility of $R_t$ is:

\[ \hat{\sigma} = \sqrt{\sigma_1^2 - 2\rho\sigma_1\sigma_2 + \sigma_2^2} \]

Under $\mathbb{Q}^{S^{(2)}}$, $R_t$ is a driftless geometric Brownian motion with volatility $\hat{\sigma}$.

Step 4: Apply Black-Scholes to the ratio. The problem reduces to pricing a call on $R_T$ with strike 1 in a world with zero interest rate:

\[ \frac{V_0}{S_0^{(2)}} = R_0 \mathcal{N}(d_1) - \mathcal{N}(d_2) \]

where $R_0 = S_0^{(1)}/S_0^{(2)}$, and:

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

Multiplying by $S_0^{(2)}$:

\[ V_0 = S_0^{(1)}\mathcal{N}(d_1) - S_0^{(2)}\mathcal{N}(d_2) \]

This is Margrabe's formula. The effective volatility $\hat{\sigma} = \sqrt{\sigma_1^2 - 2\rho\sigma_1\sigma_2 + \sigma_2^2}$ is the volatility of the log-ratio $\ln(S^{(1)}/S^{(2)})$.

Exercise 5. Explain why the term $\mathcal{N}(d_1)$ in the Black-Scholes call formula is both the delta of the option and the probability of exercise under the stock measure. Is this a coincidence, or does the change-of-numeraire framework make this relationship transparent? Justify your answer.

Solution to Exercise 5

The relationship between $\mathcal{N}(d_1)$ being both the delta and the stock-measure probability of exercise is not a coincidence -- it is a direct consequence of the change-of-numeraire framework.

Delta as stock-measure probability. From the Black-Scholes formula $C = S\mathcal{N}(d_1) - Ke^{-rT}\mathcal{N}(d_2)$, differentiating with respect to $S$:

\[ \Delta = \frac{\partial C}{\partial S} = \mathcal{N}(d_1) + S\mathcal{N}'(d_1)\frac{\partial d_1}{\partial S} - Ke^{-rT}\mathcal{N}'(d_2)\frac{\partial d_2}{\partial S} \]

Since $\frac{\partial d_1}{\partial S} = \frac{\partial d_2}{\partial S} = \frac{1}{S\sigma\sqrt{T}}$ and $S\mathcal{N}'(d_1) = Ke^{-rT}\mathcal{N}'(d_2)$ (a standard identity), the last two terms cancel, giving $\Delta = \mathcal{N}(d_1)$.

Stock-measure probability. Under the stock measure $\mathbb{Q}^S$, the call price can be written as $C = S_0\mathbb{Q}^S(S_T > K) - Ke^{-rT}\mathbb{Q}(S_T > K)$. Comparing with the Black-Scholes formula:

\[ \mathbb{Q}^S(S_T > K) = \mathcal{N}(d_1) \]

Why this is transparent from the numeraire framework. Under the numeraire change to the stock, $C_0/S_0 = \mathbb{E}^{\mathbb{Q}^S}[(1 - K/S_T)^+]$. The derivative of $C_0$ with respect to $S_0$ equals the probability under $\mathbb{Q}^S$ that the option expires in the money, because $C_0 = S_0 \cdot \mathbb{Q}^S(S_T > K) - Ke^{-rT}\mathcal{N}(d_2)$ and the first term is linear in $S_0$ (through the dependence of $\mathbb{Q}^S(S_T > K)$ on $S_0$, with the derivative simplifying to $\mathcal{N}(d_1)$). The change-of-numeraire framework makes this relationship structurally transparent: the delta is the hedge ratio, which equals the probability of exercise under the measure where the stock is the numeraire.

Exercise 6. A zero-coupon bond maturing at time $T$ with price $P(t,T) = e^{-r(T-t)}$ can serve as a numeraire, giving rise to the $T$-forward measure $\mathbb{Q}^T$. Show that under $\mathbb{Q}^T$, the forward price $F(t,T) = S_t / P(t,T)$ is a martingale. Use this to re-derive the Black-Scholes call price starting from $C_0 = P(0,T)\mathbb{E}^{\mathbb{Q}^T}[(F(T,T) - K)^+]$.

Solution to Exercise 6

Step 1: Show $F(t,T)$ is a martingale under $\mathbb{Q}^T$. The forward price is:

\[ F(t,T) = \frac{S_t}{P(t,T)} = S_t e^{r(T-t)} \]

Under $\mathbb{Q}$, $dS_t = rS_t \, dt + \sigma S_t \, dW_t^{\mathbb{Q}}$. The numeraire is $P(t,T) = e^{-r(T-t)}$ with $dP = rP \, dt$ (deterministic). The Radon–Nikodym derivative:

\[ \frac{d\mathbb{Q}^T}{d\mathbb{Q}}\Big|_{\mathcal{F}_T} = \frac{P(T,T)/B_T}{P(0,T)/B_0} = \frac{1/e^{rT}}{e^{-rT}/1} = 1 \]

Since the bond is deterministic, $\mathbb{Q}^T = \mathbb{Q}$ and there is no Girsanov shift. Now compute $dF$:

\[ dF = d(S_t e^{r(T-t)}) = e^{r(T-t)} \, dS_t + S_t \, d(e^{r(T-t)}) \]

\[ = e^{r(T-t)}(rS_t \, dt + \sigma S_t \, dW_t) - rS_t e^{r(T-t)} \, dt = \sigma F_t \, dW_t^{\mathbb{Q}^T} \]

The drift vanishes, so $F(t,T)$ is a martingale under $\mathbb{Q}^T$.

Step 2: Re-derive the call price. Starting from:

\[ C_0 = P(0,T)\mathbb{E}^{\mathbb{Q}^T}[(F(T,T) - K)^+] \]

Note that $F(T,T) = S_T$ and $P(0,T) = e^{-rT}$. Under $\mathbb{Q}^T = \mathbb{Q}$, $F(t,T) = F(0,T)\exp(-\frac{1}{2}\sigma^2 t + \sigma W_t)$ with $F(0,T) = S_0 e^{rT}$.

So:

\[ F(T,T) = S_0 e^{rT} \exp\left(-\frac{1}{2}\sigma^2 T + \sigma W_T\right) \]

The problem is now a standard Black-Scholes pricing with the forward $F_0 = S_0 e^{rT}$ replacing $S_0$, zero interest rate (since discounting is already handled by $P(0,T)$), and the same volatility $\sigma$. By the Black formula:

\[ C_0 = e^{-rT}[F_0 \mathcal{N}(d_1) - K\mathcal{N}(d_2)] \]