Numéraire Techniques¶

Numéraire techniques provide a unified framework for changing probability measures and simplifying derivative pricing by choosing an appropriate reference asset.

What is a numéraire?¶

A numéraire is a strictly positive tradable asset \(N_t\) used to measure value. Prices expressed in units of \(N_t\) are

\[ \tilde S_t = \frac{S_t}{N_t}. \]

Fundamental theorem of numéraire change¶

For any admissible numéraire \(N_t\), there exists a probability measure \(\mathbb{Q}^N\) such that

\[ \frac{S_t}{N_t} \text{ is a martingale under } \mathbb{Q}^N. \]

This generalizes the risk-neutral measure concept.

Examples of numeraires¶

Common choices include: - money-market account \(B_t\) → risk-neutral measure, - zero-coupon bond \(P(t,T)\) → T-forward measure, - swap annuity → swap measure.

Each choice simplifies pricing of specific products.

Pricing with numéraires¶

If payoff \(V_T\) is measurable at \(T\),

\[ V_t = N_t\,\mathbb{E}^{\mathbb{Q}^N}\left[ \frac{V_T}{N_T} \middle| \mathcal{F}_t \right]. \]

Choosing \(N_t\) wisely can remove discounting or complex drifts.

Key takeaways¶

Numéraire choice determines the pricing measure.
Forward measures are special cases of numéraire techniques.
Proper numéraire selection simplifies valuation and dynamics.

Exercises¶

Exercise 1. Let \(N_t^{(1)} = B_t\) (money-market account) and \(N_t^{(2)} = P(t, T)\) (zero-coupon bond). Write down the Radon--Nikodym derivative \(d\mathbb{Q}^{(2)}/d\mathbb{Q}^{(1)}|_{\mathcal{F}_t}\) and verify that it defines a valid density process (i.e., it is a positive martingale under \(\mathbb{Q}^{(1)}\) with initial value 1).

Solution to Exercise 1

Radon--Nikodym derivative. Let \(N_t^{(1)} = B_t\) and \(N_t^{(2)} = P(t, T)\). The general numéraire change formula gives

\[ \frac{d\mathbb{Q}^{(2)}}{d\mathbb{Q}^{(1)}}\bigg|_{\mathcal{F}_t} = \frac{N_t^{(2)} / N_0^{(2)}}{N_t^{(1)} / N_0^{(1)}} = \frac{P(t, T) / P(0, T)}{B_t / 1} = \frac{P(t, T)}{P(0, T) \, B_t} \]

Verification of positivity. Since \(P(t, T) > 0\) (bond prices are strictly positive for \(t < T\)), \(P(0, T) > 0\), and \(B_t > 0\), the density is strictly positive.

Verification of initial value. At \(t = 0\):

\[ \frac{P(0, T)}{P(0, T) \cdot B_0} = \frac{P(0, T)}{P(0, T) \cdot 1} = 1 \]

Verification of the martingale property under \(\mathbb{Q}^{(1)}\). We need to show that \(L_t = P(t, T)/(P(0, T) B_t)\) is a \(\mathbb{Q}^{(1)}\)-martingale. Note that \(L_t = \frac{1}{P(0,T)} \cdot \frac{P(t,T)}{B_t}\).

Under \(\mathbb{Q}^{(1)} = \mathbb{Q}\) (the risk-neutral measure with numéraire \(B_t\)), the discounted price of any tradable asset is a martingale. Since \(P(t, T)\) is the price of a tradable asset (the zero-coupon bond), \(P(t, T)/B_t\) is a \(\mathbb{Q}\)-martingale. Multiplying by the positive constant \(1/P(0, T)\) preserves the martingale property.

Therefore \(L_t\) is a positive \(\mathbb{Q}^{(1)}\)-martingale with \(L_0 = 1\), confirming it is a valid Radon--Nikodym density process.

Exercise 2. A derivative pays \(V_T = S_T^2\) at time \(T\), where \(S_t\) follows geometric Brownian motion \(dS_t = rS_t\,dt + \sigma S_t\,dW_t^{\mathbb{Q}}\) under the risk-neutral measure with numéraire \(B_t\). Compute \(\mathbb{E}^{\mathbb{Q}}[e^{-rT}S_T^2]\) directly. Then choose \(N_t = S_t\) as the numéraire and compute the same price as \(S_0\,\mathbb{E}^{\mathbb{Q}^S}[S_T]\), verifying that the two approaches agree. What are the dynamics of \(S_t\) under the measure \(\mathbb{Q}^S\)?

Solution to Exercise 2

Direct computation under \(\mathbb{Q}\). Under \(\mathbb{Q}\), \(S_t = S_0 \exp((r - \sigma^2/2)t + \sigma W_t^{\mathbb{Q}})\), so

\[ S_T^2 = S_0^2 \exp\!\left((2r - \sigma^2)T + 2\sigma W_T^{\mathbb{Q}}\right) \]

\[ \mathbb{E}^{\mathbb{Q}}[e^{-rT}S_T^2] = S_0^2 e^{-rT} \exp\!\left((2r - \sigma^2)T\right) \mathbb{E}^{\mathbb{Q}}\!\left[\exp(2\sigma W_T^{\mathbb{Q}})\right] \]

Since \(W_T^{\mathbb{Q}} \sim N(0, T)\), \(\mathbb{E}[\exp(2\sigma W_T^{\mathbb{Q}})] = \exp(2\sigma^2 T)\). Therefore:

\[ \mathbb{E}^{\mathbb{Q}}[e^{-rT}S_T^2] = S_0^2 \exp\!\left(-rT + 2rT - \sigma^2 T + 2\sigma^2 T\right) = S_0^2 \exp\!\left((r + \sigma^2)T\right) \]

Computation using \(N_t = S_t\) as numéraire. The pricing formula gives

\[ V_0 = S_0 \, \mathbb{E}^{\mathbb{Q}^S}\!\left[\frac{S_T^2}{S_T}\right] = S_0 \, \mathbb{E}^{\mathbb{Q}^S}[S_T] \]

Dynamics under \(\mathbb{Q}^S\). The Radon--Nikodym derivative is

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

Since \(S_t/B_t\) has volatility \(\sigma\), Girsanov's theorem gives \(dW_t^S = dW_t^{\mathbb{Q}} - \sigma\,dt\), where \(W_t^S\) is a Brownian motion under \(\mathbb{Q}^S\).

Under \(\mathbb{Q}^S\), substituting \(dW_t^{\mathbb{Q}} = dW_t^S + \sigma\,dt\):

\[ dS_t = rS_t\,dt + \sigma S_t(dW_t^S + \sigma\,dt) = (r + \sigma^2)S_t\,dt + \sigma S_t\,dW_t^S \]

Therefore \(S_T = S_0 \exp((r + \sigma^2/2)T + \sigma W_T^S)\) and

\[ \mathbb{E}^{\mathbb{Q}^S}[S_T] = S_0 \exp\!\left((r + \sigma^2/2)T\right) \exp(\sigma^2 T/2) = S_0 \exp\!\left((r + \sigma^2)T\right) \]

Final price:

\[ V_0 = S_0 \cdot S_0 \exp\!\left((r + \sigma^2)T\right) = S_0^2 \exp\!\left((r + \sigma^2)T\right) \]

Both methods agree.

Exercise 3. Suppose there are two tradable assets \(N_t^{(1)}\) and \(N_t^{(2)}\) with associated measures \(\mathbb{Q}^{(1)}\) and \(\mathbb{Q}^{(2)}\). Show that the Radon--Nikodym derivative for the composite change \(\mathbb{Q}^{(1)} \to \mathbb{Q}^{(2)}\) is

\[ \frac{d\mathbb{Q}^{(2)}}{d\mathbb{Q}^{(1)}}\bigg|_{\mathcal{F}_t} = \frac{N_t^{(2)} / N_0^{(2)}}{N_t^{(1)} / N_0^{(1)}} \]

and verify that this is consistent with the individual changes from a common reference measure \(\mathbb{Q}\).

Solution to Exercise 3

From a common reference measure. Let \(\mathbb{Q}\) be any reference measure. The changes from \(\mathbb{Q}\) to \(\mathbb{Q}^{(k)}\) (for \(k = 1, 2\)) are given by

\[ \frac{d\mathbb{Q}^{(k)}}{d\mathbb{Q}}\bigg|_{\mathcal{F}_t} = \frac{N_t^{(k)} / N_0^{(k)}}{B_t} \]

where \(B_t\) is the numéraire for \(\mathbb{Q}\).

Composite change. The chain rule for Radon--Nikodym derivatives gives

\[ \frac{d\mathbb{Q}^{(2)}}{d\mathbb{Q}^{(1)}}\bigg|_{\mathcal{F}_t} = \frac{d\mathbb{Q}^{(2)}/d\mathbb{Q}|_{\mathcal{F}_t}}{d\mathbb{Q}^{(1)}/d\mathbb{Q}|_{\mathcal{F}_t}} \]

Substituting:

\[ \frac{d\mathbb{Q}^{(2)}}{d\mathbb{Q}^{(1)}}\bigg|_{\mathcal{F}_t} = \frac{N_t^{(2)} / N_0^{(2)} / B_t}{N_t^{(1)} / N_0^{(1)} / B_t} = \frac{N_t^{(2)} / N_0^{(2)}}{N_t^{(1)} / N_0^{(1)}} \]

The common reference numéraire \(B_t\) cancels, confirming the formula is independent of the choice of \(\mathbb{Q}\).

Verification. At \(t = 0\): \(\frac{N_0^{(2)}/N_0^{(2)}}{N_0^{(1)}/N_0^{(1)}} = \frac{1}{1} = 1\). The density starts at 1.

The ratio \(\frac{N_t^{(2)}/N_0^{(2)}}{N_t^{(1)}/N_0^{(1)}} = \frac{1}{N_0^{(2)}/N_0^{(1)}} \cdot \frac{N_t^{(2)}}{N_t^{(1)}}\) is the ratio of two positive tradable assets (up to a constant), so it is positive. Under \(\mathbb{Q}^{(1)}\), the process \(N_t^{(2)}/N_t^{(1)}\) is a martingale (since \(\mathbb{Q}^{(1)}\) makes any tradable asset divided by \(N_t^{(1)}\) a martingale). Multiplying by a positive constant preserves the martingale property. Hence the density process is a valid positive \(\mathbb{Q}^{(1)}\)-martingale with initial value 1.

Exercise 4. Explain why the money-market account \(B_t = \exp(\int_0^t r_s\,ds)\) is a valid numéraire even when the short rate \(r_t\) is stochastic, whereas a zero-coupon bond \(P(t, T)\) is only useful as a numéraire for pricing payoffs at or before time \(T\). What goes wrong if you try to use \(P(t, T)\) to price a payoff at time \(T' > T\)?

Solution to Exercise 4

Why \(B_t\) is always a valid numéraire:

Strict positivity: \(B_t = \exp(\int_0^t r_s\,ds) > 0\) for all \(t \geq 0\), regardless of whether \(r_t\) is stochastic or even negative (the exponential of any real number is positive).
Tradability: \(B_t\) represents the value of continuously rolling over at the instantaneous short rate, which is a self-financing strategy.
No maturity constraint: \(B_t\) is defined for all \(t \geq 0\), so it can serve as a numéraire for payoffs at any time \(T\).

Why \(P(t, T)\) is limited to payoffs at or before \(T\):

Numéraire at payment date: For the pricing formula \(V_0 = N_0\,\mathbb{E}^{\mathbb{Q}^N}[V_{T'}/N_{T'}]\), the numéraire \(N_{T'}\) must be well-defined and strictly positive at the payment date \(T'\).
At maturity: \(P(T, T) = 1 > 0\), so \(P(t, T)\) works perfectly for payoffs at time \(T\).
Before maturity: For \(T' < T\), \(P(T', T) > 0\) is still well-defined, so \(P(t, T)\) also works.

What goes wrong for \(T' > T\):

For a payoff at \(T' > T\), the pricing formula requires

\[ V_0 = P(0, T)\,\mathbb{E}^{\mathbb{Q}^T}\!\left[\frac{V_{T'}}{P(T', T)}\right] \]

But \(P(T', T)\) for \(T' > T\) is not the price of the zero-coupon bond (which has already matured at \(T\)). After maturity, \(P(T, T) = 1\) and the bond ceases to exist as a tradable asset for \(t > T\). One would need to define how the proceeds are reinvested after \(T\), which brings us back to the money-market account. Formally, the process \(P(t, T)\) for \(t > T\) would need to be extended as \(P(t, T) = B_t / B_T\) (reinvesting at the short rate), but this is just the money-market account renormalized, defeating the purpose of using the bond as numéraire.

In summary, the bond numéraire naturally "expires" at \(T\), while the money-market account is perpetual.

Exercise 5. A foreign-exchange option pays \(\max(X_T - K, 0)\) in domestic currency at time \(T\), where \(X_t\) is the spot exchange rate. The domestic and foreign money-market accounts are \(B_t^d\) and \(B_t^f\). Show that \(X_t B_t^f / B_t^d\) is a martingale under the domestic risk-neutral measure, and use this to identify the natural numéraire for pricing this option. What is the drift of \(X_t\) under the domestic measure?

Solution to Exercise 5

Martingale property of \(X_t B_t^f / B_t^d\). Under the domestic risk-neutral measure \(\mathbb{Q}^d\) (with numéraire \(B_t^d\)), every tradable asset denominated in domestic currency, discounted by \(B_t^d\), is a martingale.

Consider holding one unit of foreign currency. At time \(t\), its domestic value is \(X_t\) (the exchange rate). By investing this foreign currency in the foreign money-market account, the value at time \(t\) of an initial investment made at time \(0\) is \(X_t B_t^f\) in domestic currency. This is a tradable strategy (buy foreign currency, invest at foreign short rate). Therefore

\[ \frac{X_t B_t^f}{B_t^d} \]

must be a \(\mathbb{Q}^d\)-martingale.

Drift of \(X_t\) under \(\mathbb{Q}^d\). Let \(dX_t = \mu_X X_t\,dt + \sigma_X X_t\,dW_t^d\). The discounted foreign investment is

\[ \frac{X_t B_t^f}{B_t^d} = X_t \exp\!\left(\int_0^t (r_s^f - r_s^d)\,ds\right) \]

For this to be a martingale, applying Itô's formula (with constant rates for simplicity):

\[ d\!\left(\frac{X_t B_t^f}{B_t^d}\right) = \frac{B_t^f}{B_t^d}\left[(\mu_X + r^f - r^d)X_t\,dt + \sigma_X X_t\,dW_t^d\right] \]

Setting the drift to zero: \(\mu_X + r^f - r^d = 0\), so

\[ \mu_X = r^d - r^f \]

The exchange rate dynamics under \(\mathbb{Q}^d\) are

\[ dX_t = (r^d - r^f) X_t\,dt + \sigma_X X_t\,dW_t^d \]

This is the interest rate parity relation in continuous time: the drift of the exchange rate equals the interest rate differential.

Natural numéraire for the FX option. The payoff \(\max(X_T - K, 0)\) is in domestic currency at time \(T\). The natural numéraire is \(B_t^d\) (domestic money-market account) with the domestic risk-neutral measure \(\mathbb{Q}^d\):

\[ V_0 = e^{-r^d T}\,\mathbb{E}^{\mathbb{Q}^d}[\max(X_T - K, 0)] \]

Under \(\mathbb{Q}^d\), \(X_t\) is a GBM with drift \(r^d - r^f\), so \(X_T\) is lognormal and the Garman--Kohlhagen formula (the FX analogue of Black--Scholes) applies:

\[ V_0 = X_0 e^{-r^f T} N(d_1) - K e^{-r^d T} N(d_2) \]

where \(d_1 = \frac{\ln(X_0/K) + (r^d - r^f + \sigma_X^2/2)T}{\sigma_X\sqrt{T}}\) and \(d_2 = d_1 - \sigma_X\sqrt{T}\).

Exercise 6. Consider a stock \(S_t\) paying a continuous dividend yield \(q\). The reinvested stock price \(\hat{S}_t = S_t e^{qt}\) is a valid numéraire. Derive the Radon--Nikodym derivative from \(\mathbb{Q}\) (money-market numéraire) to \(\mathbb{Q}^{\hat{S}}\) (stock numéraire) and show that under \(\mathbb{Q}^{\hat{S}}\), the process \(B_t/\hat{S}_t\) is a martingale. Use this to price a European put option with payoff \(\max(K - S_T, 0)\).

Solution to Exercise 6

Setup. Under \(\mathbb{Q}\), the stock with continuous dividends satisfies \(dS_t = (r - q)S_t\,dt + \sigma S_t\,dW_t^{\mathbb{Q}}\). The reinvested stock price \(\hat{S}_t = S_t e^{qt}\) satisfies

\[ d\hat{S}_t = e^{qt}(dS_t + qS_t\,dt) = e^{qt}[(r - q)S_t + qS_t]\,dt + e^{qt}\sigma S_t\,dW_t^{\mathbb{Q}} \]

\[ = r\hat{S}_t\,dt + \sigma\hat{S}_t\,dW_t^{\mathbb{Q}} \]

So \(\hat{S}_t\) grows at rate \(r\) under \(\mathbb{Q}\), confirming it is a valid numéraire (\(\hat{S}_t/B_t\) is a \(\mathbb{Q}\)-martingale).

Radon--Nikodym derivative. From \(\mathbb{Q}\) to \(\mathbb{Q}^{\hat{S}}\):

\[ \frac{d\mathbb{Q}^{\hat{S}}}{d\mathbb{Q}}\bigg|_{\mathcal{F}_t} = \frac{\hat{S}_t / \hat{S}_0}{B_t} = \frac{S_t e^{qt}}{S_0 e^{rt}} \]

Since \(S_t = S_0\exp((r - q - \sigma^2/2)t + \sigma W_t^{\mathbb{Q}})\):

\[ \frac{d\mathbb{Q}^{\hat{S}}}{d\mathbb{Q}}\bigg|_{\mathcal{F}_t} = \exp\!\left(-\frac{\sigma^2}{2}t + \sigma W_t^{\mathbb{Q}}\right) \]

This is the standard exponential martingale.

Girsanov transformation. By Girsanov's theorem:

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

is a Brownian motion under \(\mathbb{Q}^{\hat{S}}\).

\(B_t/\hat{S}_t\) is a \(\mathbb{Q}^{\hat{S}}\)-martingale. Under \(\mathbb{Q}^{\hat{S}}\), any tradable asset divided by \(\hat{S}_t\) is a martingale. Since \(B_t\) is tradable, \(B_t/\hat{S}_t\) is a \(\mathbb{Q}^{\hat{S}}\)-martingale.

We can verify directly: \(B_t/\hat{S}_t = e^{rt}/(S_t e^{qt}) = e^{(r-q)t}/S_t\). Under \(\mathbb{Q}^{\hat{S}}\), using \(dW_t^{\mathbb{Q}} = dW_t^{\hat{S}} + \sigma\,dt\):

\[ dS_t = (r - q)S_t\,dt + \sigma S_t(dW_t^{\hat{S}} + \sigma\,dt) = (r - q + \sigma^2)S_t\,dt + \sigma S_t\,dW_t^{\hat{S}} \]

By Itô's formula for \(1/S_t\):

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

\[ = -(r - q)/S_t\,dt - \sigma/S_t\,dW_t^{\hat{S}} \]

Then \(d(e^{(r-q)t}/S_t) = e^{(r-q)t}[(r-q)/S_t\,dt + d(1/S_t)] = -\sigma e^{(r-q)t}/S_t\,dW_t^{\hat{S}}\), which is driftless, confirming the martingale property.

Pricing the European put. The put payoff is \(\max(K - S_T, 0) = K\max(1 - S_T/K, 0)\). Using numéraire \(\hat{S}_t\):

\[ P_0 = \hat{S}_0\,\mathbb{E}^{\mathbb{Q}^{\hat{S}}}\!\left[\frac{\max(K - S_T, 0)}{\hat{S}_T}\right] = S_0\,\mathbb{E}^{\mathbb{Q}^{\hat{S}}}\!\left[\frac{K\max(1 - S_T/K, 0)}{S_T e^{qT}}\right] \]

\[ = S_0 K e^{-qT}\,\mathbb{E}^{\mathbb{Q}^{\hat{S}}}\!\left[\frac{\max(1 - S_T/K, 0)}{S_T}\right] \]

Alternatively, using the standard approach, the put price under \(\mathbb{Q}\) is

\[ P_0 = e^{-rT}\mathbb{E}^{\mathbb{Q}}[\max(K - S_T, 0)] = Ke^{-rT}N(-d_2) - S_0 e^{-qT}N(-d_1) \]

where

\[ d_1 = \frac{\ln(S_0/K) + (r - q + \sigma^2/2)T}{\sigma\sqrt{T}}, \qquad d_2 = d_1 - \sigma\sqrt{T} \]

To use the stock numéraire more cleanly, write the put as \(\max(K - S_T, 0) = K\max(K^{-1}S_T^{-1}(K - S_T), 0) \cdot S_T/1\)... The most elegant numéraire approach for the put is to decompose:

\[ P_0 = K\,e^{-rT}\,\mathbb{Q}(S_T < K) - S_0\,e^{-qT}\,\mathbb{Q}^{\hat{S}}(S_T < K) \]

Under \(\mathbb{Q}\): \(\ln S_T \sim N(\ln S_0 + (r - q - \sigma^2/2)T,\;\sigma^2 T)\), so \(\mathbb{Q}(S_T < K) = N(-d_2)\).

Under \(\mathbb{Q}^{\hat{S}}\): \(\ln S_T \sim N(\ln S_0 + (r - q + \sigma^2/2)T,\;\sigma^2 T)\), so \(\mathbb{Q}^{\hat{S}}(S_T < K) = N(-d_1)\).

Therefore:

\[ P_0 = Ke^{-rT}N(-d_2) - S_0 e^{-qT}N(-d_1) \]

This is the Black--Scholes put formula with continuous dividends, derived via the numéraire change.