From Bank-Account Pricing to a New Numeraire
A numeraire is a strictly positive traded asset used as the unit in which other values are measured. The bank account is the usual numeraire, but another asset can be more convenient when the payoff is naturally expressed relative to it. Changing numeraire changes the probability measure used for conditional expectations; it does not change the economic value of the claim.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Let \(r\) be a constant interest rate and \(A_t=e^{rt}\) the bank-account value, with \(A_0=1\). Let \(\mathbb{Q}\) be the risk-neutral measure associated with \(A_t\). For this tutorial, assume the bank-account pricing formula is valid for the terminal payoff \(H\): \(H/A_T\) is integrable under \(\mathbb{Q}\), and the time-\(t\) value is \(V_t=A_t\mathbb{E}^{\mathbb{Q}}[H/A_T\mid\mathcal{F}_t]\).
Let \(N_t\) be another traded numeraire, strictly positive at all times on the horizon \([0,T]\), with \(N_0>0\). Assume its bank-account-discounted value \(N_t/A_t\) is a true \(\mathbb{Q}\)-martingale, not merely a local martingale. This ensures that the process below is a probability density process. In applications, these assumptions and the integrability of the relevant payoff ratios must be checked rather than inferred from positivity alone.
The New-Numeraire Pricing Formula
Proof. Since \(Z_t>0\), conditional Bayes' formula for equivalent measures gives $$ \mathbb{E}^{\mathbb{Q}^N}[Y\mid\mathcal{F}_t] =\frac{\mathbb{E}^{\mathbb{Q}}[Z_TY\mid\mathcal{F}_t]}{Z_t} $$ for an integrable \(Y\). Apply this with \(Y=H/N_T\). The density ratio satisfies $$ \frac{Z_T}{Z_t} =\frac{N_T/A_T}{N_t/A_t}. $$ Therefore $$ \mathbb{E}^{\mathbb{Q}^N}\left[\frac{H}{N_T}\,\middle|\,\mathcal{F}_t\right] =\mathbb{E}^{\mathbb{Q}}\left[ \frac{Z_T}{Z_t}\frac{H}{N_T}\,\middle|\,\mathcal{F}_t\right] =\frac{A_t}{N_t}\mathbb{E}^{\mathbb{Q}}\left[\frac{H}{A_T}\,\middle|\,\mathcal{F}_t\right]. $$ Multiplying by \(N_t\) gives \(N_t\mathbb{E}^{\mathbb{Q}^N}[H/N_T\mid\mathcal{F}_t]=A_t\mathbb{E}^{\mathbb{Q}}[H/A_T\mid\mathcal{F}_t]=V_t\), which proves the claim. \(\square\)
The central operation is to divide the terminal payoff by the terminal numeraire, not by its current value. The density also includes its normalization by \(N_0/A_0\); omitting that factor generally fails to define a probability measure.
A Stock Numeraire in the Black–Scholes Model
Consider a stock \(S_t\) under \(\mathbb{Q}\) satisfying \(dS_t=rS_t\,dt+\sigma S_t\,dB_t\), where \(\sigma>0\). This is an SDE whose drift coefficient is \(rS_t\) and diffusion coefficient is \(\sigma S_t\). Its solution is \(S_t=S_0\exp((r-\tfrac12\sigma^2)t+\sigma B_t)\). The stochastic integrand \(\sigma S_t\) is progressively measurable and satisfies \(\mathbb{E}^{\mathbb{Q}}\int_0^T(\sigma S_s)^2\,ds<\infty\), since the lognormal second moment is finite on every finite time interval. The discounted stock \(S_t/A_t\) is a true \(\mathbb{Q}\)-martingale: its conditional expectation follows from the independent normal increment \(B_u-B_t\) and the normal exponential moment.
Worked Example: The Stock-Numeraire Density
Take \(N_t=S_t\). The normalized density at time \(T\) is $$ Z_T=\frac{S_T/A_T}{S_0/A_0} =\exp\left(\sigma B_T-\frac12\sigma^2T\right). $$ For a concrete calculation, set \(S_0=2\), \(r=\tfrac14\), \(\sigma=\tfrac12\), and \(T=2\). Then $$ Z_T=\exp\left(\frac12B_2-\frac14\right). $$ Indeed, if \(X\sim N(0,2)\), then \(\mathbb{E}[e^{uX}]=e^{u^2}\). Thus \(\mathbb{E}^{\mathbb{Q}}[Z_T]=e^{-1/4}e^{(1/2)^2}=1\). More generally, \(Z_t=\exp(\sigma B_t-\tfrac12\sigma^2t)\), and independent Brownian increments give \(\mathbb{E}^{\mathbb{Q}}[Z_u\mid\mathcal{F}_t]=Z_t\) for \(t\le u\le T\). It is therefore a true martingale density, not just a local-martingale candidate.
The exponential-martingale form also satisfies Novikov's condition: \(\mathbb{E}^{\mathbb{Q}}\exp(\tfrac12\int_0^T\sigma^2\,ds)=\exp(\tfrac12\sigma^2T)<\infty\). By the standard Girsanov theorem, \(B_t^N=B_t-\sigma t\) is Brownian motion under \(\mathbb{Q}^N\). Substituting \(dB_t=dB_t^N+\sigma\,dt\) into the stock SDE gives $$ dS_t=(r+\sigma^2)S_t\,dt+\sigma S_t\,dB_t^N. $$ The drift changes under the new measure; the diffusion coefficient remains \(\sigma S_t\).
Check: with the stated parameters, the density is \(\exp(\tfrac12B_2-\tfrac14)\), and its expectation is \(e^{-1/4}e^{1/4}=1\). The drift shift is \(\sigma^2=\tfrac14\), so the stock drift under \(\mathbb{Q}^N\) is \(r+\sigma^2=\tfrac14+\tfrac14=\tfrac12\).
The Girsanov theorem used here is a standard result: for a progressively measurable kernel satisfying Novikov's condition, the exponential density changes Brownian motion by subtracting the time-integrated kernel. Its general proof is not needed for the pricing identity above; the specific density and the condition that permits its use have been verified here.
Pricing Payoffs Relative to the Stock
Worked Example: A Deterministic Unit Payoff
Let \(H=1\), and use the stock as numeraire. Under \(\mathbb{Q}^N\), the explicit solution of the stock SDE just derived is $$ S_T=S_t\exp\left((r+\tfrac12\sigma^2)(T-t)+\sigma(B_T^N-B_t^N)\right). $$ Because \(B_T^N-B_t^N\) is independent of \(\mathcal{F}_t\) and is normal with mean zero and variance \(T-t\), its exponential moment gives $$ \mathbb{E}^{\mathbb{Q}^N}\left[\frac{1}{S_T}\,\middle|\,\mathcal{F}_t\right] =\frac{1}{S_t}\exp\left(-(r+\tfrac12\sigma^2)(T-t)+\tfrac12\sigma^2(T-t)\right) =\frac{e^{-r(T-t)}}{S_t}. $$ Thus the new-numeraire formula gives \(V_t=S_t\mathbb{E}^{\mathbb{Q}^N}[1/S_T\mid\mathcal{F}_t]=e^{-r(T-t)}\). This agrees with bank-account pricing: \(A_t\mathbb{E}^{\mathbb{Q}}[1/A_T\mid\mathcal{F}_t]=e^{rt}e^{-rT}=e^{-r(T-t)}\).
Worked Example: A Stock Payoff
For \(H=S_T\), its value relative to the stock numeraire at maturity is \(H/N_T=S_T/S_T=1\). Hence $$ V_t=S_t\mathbb{E}^{\mathbb{Q}^N}[1\mid\mathcal{F}_t]=S_t. $$ The conditional expectation is exactly one because the constant random variable \(1\) is integrable and its conditional expectation under any probability measure is \(1\). Independently, bank-account pricing gives $$ A_t\mathbb{E}^{\mathbb{Q}}\left[\frac{S_T}{A_T}\,\middle|\,\mathcal{F}_t\right] =A_t\frac{S_t}{A_t}=S_t, $$ using the true martingale property of the discounted stock.
Check: the relative payoff is identically one at \(T\), so its conditional expectation is one at every earlier time. Multiplying by the current numeraire returns \(S_t\), as required for a claim that delivers one share.
Worked Example: A Squared-Stock Payoff
Let \(H=S_T^2\), with \(\tau=T-t\). The stock-numeraire formula requires the conditional expectation of \(S_T^2/S_T=S_T\), not of \(S_T^2\). Under \(\mathbb{Q}^N\), the solution above and the normal exponential moment yield $$ \mathbb{E}^{\mathbb{Q}^N}[S_T\mid\mathcal{F}_t] =S_t\exp\left((r+\tfrac12\sigma^2)\tau+\tfrac12\sigma^2\tau\right) =S_t e^{(r+\sigma^2)\tau}. $$ It follows that \(V_t=S_t^2e^{(r+\sigma^2)\tau}\).
Check by bank-account pricing. Under \(\mathbb{Q}\), the stock solution gives $$ \mathbb{E}^{\mathbb{Q}}[S_T^2\mid\mathcal{F}_t] =S_t^2\exp\left(2r\tau+\sigma^2\tau\right), $$ because the square of the random exponential contributes the normal moment \(e^{(2\sigma)^2\tau/2}=e^{2\sigma^2\tau}\), while the squared solution's deterministic exponent contributes \(e^{(2r-\sigma^2)\tau}\). Discounting over \(\tau\) gives \(e^{-r\tau}\mathbb{E}^{\mathbb{Q}}[S_T^2\mid\mathcal{F}_t]=S_t^2e^{(r+\sigma^2)\tau}\), matching the new-numeraire result.
For \(r=\tfrac14\), \(\sigma=\tfrac12\), and \(\tau=2\), the exponent is \((\tfrac14+\tfrac14)2=1\), so the value is \(S_t^2e\). The direct bank-account calculation has exponent \((r+\sigma^2)\tau=1\) after discounting as well.
What the Change of Numeraire Does—and Does Not Do
The formula is useful when \(H/N_T\) has a simple distribution under \(\mathbb{Q}^N\), even if \(H/A_T\) is awkward under \(\mathbb{Q}\). It is a change of units combined with a change of measure, not a new pricing principle. The risk-neutral measure \(\mathbb{Q}\) is associated with the bank account; \(\mathbb{Q}^N\) is associated with the chosen numeraire. Neither is the original probability measure \(\mathbb{P}\) unless the model happens to make them equal.
A common mistake is to omit the density normalization. In the stock example, writing \(Z_T=S_T/A_T\) instead of \(Z_T=(S_T/A_T)/(S_0/A_0)\) gives \(\mathbb{E}^{\mathbb{Q}}[Z_T]=S_0/A_0\), not necessarily \(1\). For \(S_0=2\) and \(A_0=1\), this expectation is \(2\), so the proposed expression is not a probability density. The correction is to divide by \(S_0/A_0\), giving expectation \(1\). Another error is to use \(H/N_t\) inside the conditional expectation: the payoff is received at \(T\), so the numeraire in the ratio is \(N_T\); the current numeraire \(N_t\) multiplies the expectation outside it.
Finally, a positive discounted numeraire that is only a local martingale need not have expectation one. In that case its normalized terminal value may fail to define the probability measure required for this argument. The true-martingale and integrability hypotheses are therefore substantive conditions, not formal details.
Check Your Understanding
For each question, state any measure and integrability assumptions you use.
- Starting from \(Z_t=(N_t/A_t)/(N_0/A_0)\), derive the conditional Bayes identity that gives the new-numeraire pricing formula.
- Under the stock-numeraire measure, what is the Brownian motion in terms of \(B_t\), and what are the stock's drift and diffusion coefficients?
- Use the stock numeraire to price a deterministic payoff \(H=c\), where \(c\) is a constant, and compare with bank-account pricing.
- For \(H=S_T^3\), identify the payoff ratio to use under the stock-numeraire measure before computing any expectation.
- Explain why positivity of \(N_t/A_t\) alone does not guarantee that its normalized terminal value defines a probability measure.