From a Numeraire to a Probability Measure
A change of numeraire is also a change of probability measure. The basic idea is to weight outcomes according to how large the new numeraire is relative to the bank account at the horizon. To make this weighting a probability measure, the relative value must have the correct expectation; to obtain the useful time-dependent density, it must also be a martingale.
There is a notation issue worth fixing at the outset: \(B_t\) denotes Brownian motion, while the bank-account value is written \(B_t^{\mathrm{bank}}\). In formulas that use \(B_t\) for the bank account, \(B_t^{\mathrm{bank}}\) is that same bank-account process. Fix a horizon \(T>0\) and a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{0\le t\le T},\mathbb{P})\). Let \(\mathbb{Q}^B\) be the bank-account risk-neutral measure on \(\mathcal{F}_T\), equivalent to \(\mathbb{P}\), and suppose \(B_t\) is Brownian motion under \(\mathbb{Q}^B\). Assume the bank account is strictly positive, with deterministic initial value \(B_0^{\mathrm{bank}}>0\). A new numeraire \(N_t\) is assumed adapted, finite and strictly positive, with deterministic \(N_0>0\).
Under the bank-account risk-neutral measure, discounted gains of traded assets are typically local martingales, and under appropriate integrability conditions may be true martingales. For a numeraire change, the condition needed below is specifically that the discounted value of the new numeraire is a true martingale. For a non-dividend-paying asset this is its discounted price; for a dividend-paying asset one must use its discounted cum-dividend gains, not simply its ex-dividend price.
This is stronger than requiring \(X\) to be a local martingale. A nonnegative local martingale is a supermartingale, and its expectation can decrease; in that case normalizing \(X_T\) by \(X_0\) need not give a density with expectation one.
The Numeraire-Change Density
Proof. Set \(Z_T=X_T/X_0\). Since \(X_T\) is integrable and \(X_0\) is a deterministic positive constant, \(Z_T\) is integrable. The martingale assumption gives \(\mathbb{E}^{\mathbb{Q}^B}[Z_T]=\mathbb{E}^{\mathbb{Q}^B}[X_T]/X_0=1\), and \(Z_T>0\) almost surely. Thus \(A\mapsto\mathbb{E}^{\mathbb{Q}^B}[\mathbf{1}_A Z_T]\) defines a probability measure on \(\mathcal{F}_T\). If \(\mathbb{Q}^B(A)=0\), this expectation is zero, so \(\mathbb{Q}^N(A)=0\). Conversely, if \(\mathbb{Q}^N(A)=0\), then \(\mathbb{E}^{\mathbb{Q}^B}[\mathbf{1}_A Z_T]=0\); strict positivity of \(Z_T\) implies \(\mathbb{Q}^B(A)=0\). Hence the measures are equivalent. For \(t\le T\), the defining property of conditional expectation and the martingale property of \(X\) give $$ \mathbb{E}^{\mathbb{Q}^B}[Z_T\mid\mathcal{F}_t] =\frac{\mathbb{E}^{\mathbb{Q}^B}[X_T\mid\mathcal{F}_t]}{X_0} =\frac{X_t}{X_0}. $$ The conditional expectation on the left is the density of the restriction of \(\mathbb{Q}^N\) to \(\mathcal{F}_t\). Also, \(X_t/X_0\) is integrable, nonnegative, and has expectation one at every \(t\), so the displayed density process is a true martingale. \(\square\)
The terminal formula therefore follows from a precise requirement, not just from positivity of the numeraire: the normalized terminal relative value must have expectation one. The true-martingale hypothesis supplies that normalization and ensures the stated intermediate-time densities.
Worked Example: Choosing the Bank Account Again
Let \(N_t=B_t^{\mathrm{bank}}\) for all \(t\). Compute the density and the resulting measure.
Here \(X_t=N_t/B_t^{\mathrm{bank}}=1\), so \(X_0=X_T=1\). The terminal density is \(X_T/X_0=1/1=1\), and the density process at every time is also \(X_t/X_0=1/1=1\). Thus \(\mathbb{Q}^N(A)=\mathbb{E}^{\mathbb{Q}^B}[\mathbf{1}_A]=\mathbb{Q}^B(A)\) for every \(A\in\mathcal{F}_T\).
Check: the terminal density is positive and its expectation is \(\mathbb{E}^{\mathbb{Q}^B}[1]=1\), as required. Choosing the original numeraire produces no change of measure.
Continuous-Time Check: A Risky Numeraire
Consider a constant interest rate \(r\), so \(B_t^{\mathrm{bank}}=e^{rt}\), and let \(N_0=1\). Under \(\mathbb{Q}^B\), suppose the positive numeraire follows $$ dN_t=rN_t\,dt+N_t\,dB_t. $$ Here the stochastic integral means the Itô integral \(\int_0^t N_s\,dB_s\), with progressively measurable integrand \(N_s\) satisfying \(\mathbb{E}^{\mathbb{Q}^B}\int_0^T N_s^2\,ds<\infty\). The solution below has \(\mathbb{E}^{\mathbb{Q}^B}[N_s^2]=e^{(2r+1)s}\), so this integrability condition holds on every finite horizon.
Worked Example: The Density for a Geometric Brownian Numeraire
Find the density process and terminal density. Verify both the stochastic solution and the martingale normalization.
First define \(X_t=e^{-rt}N_t\). The candidate solution is \(X_t=e^{B_t-t/2}\), or equivalently \(N_t=e^{(r-1/2)t+B_t}\). To verify it using Itô’s formula, take \(f(t,x)=e^{x-t/2}\). Its derivatives are \(f_t=-\tfrac12f\), \(f_x=f\), and \(f_{xx}=f\). Since \(dB_t\,dB_t=dt\), Itô’s formula, including its second-derivative correction, gives $$ df(t,B_t)=f_t\,dt+f_x\,dB_t+\tfrac12 f_{xx}\,dt =-\tfrac12 f\,dt+f\,dB_t+\tfrac12 f\,dt =f\,dB_t. $$ Thus \(dX_t=X_t\,dB_t\), and \(N_t=e^{rt}X_t\) satisfies the stated equation by the product rule with the deterministic differentiable factor \(e^{rt}\).
The normalized relative value has \(X_0=1\). To check the martingale property directly, for \(0\le s\le t\), $$ \mathbb{E}^{\mathbb{Q}^B}[X_t\mid\mathcal{F}_s] =e^{B_s-t/2}\mathbb{E}^{\mathbb{Q}^B}[e^{B_t-B_s}\mid\mathcal{F}_s] =e^{B_s-t/2}e^{(t-s)/2} =e^{B_s-s/2} =X_s. $$ The middle equality uses that \(B_t-B_s\) is independent of \(\mathcal{F}_s\), is normally distributed with mean zero and variance \(t-s\), and has exponential moment \(e^{(t-s)/2}\). This also verifies integrability: \(\mathbb{E}^{\mathbb{Q}^B}[X_t]=e^{-t/2}e^{t/2}=1\).
Consequently the density process is \(Z_t=e^{B_t-t/2}\), and at horizon \(T\) the density is \(Z_T=e^{B_T-T/2}\). Independent check: the normal moment-generating function gives \(\mathbb{E}^{\mathbb{Q}^B}[e^{B_T-T/2}]=e^{-T/2}e^{T/2}=1\). Positivity is immediate, so the terminal density is normalized and gives an equivalent measure.
A Finite-State Check
Worked Example: A Two-State Numeraire Ratio
Take a one-period model with trivial \(\mathcal{F}_0\), terminal sigma-field \(\mathcal{F}_1\), and two outcomes \(\omega_1,\omega_2\). Suppose \(\mathbb{Q}^B(\{\omega_1\})=1/3\) and \(\mathbb{Q}^B(\{\omega_2\})=2/3\). Let the relative numeraire values be \(X_0=1\), \(X_1(\omega_1)=2\), and \(X_1(\omega_2)=1/2\). Compute the density and the probabilities under \(\mathbb{Q}^N\).
The conditional mean of the terminal ratio is $$ \mathbb{E}^{\mathbb{Q}^B}[X_1\mid\mathcal{F}_0] =\frac13(2)+\frac23\left(\frac12\right) =\frac23+\frac13=1=X_0. $$ Thus \(X\) is a martingale on this one-period filtration. The terminal density is \(X_1/X_0\), so its values are \(2\) on \(\omega_1\) and \(1/2\) on \(\omega_2\). The new probabilities are $$ \mathbb{Q}^N(\{\omega_1\})=\frac13(2)=\frac23,\qquad \mathbb{Q}^N(\{\omega_2\})=\frac23\left(\frac12\right)=\frac13. $$ They sum to \(2/3+1/3=1\), and each is strictly positive, so both measures assign positive probability to exactly the same outcomes.
Check: the expectation of the density under \(\mathbb{Q}^B\) is \(\frac13(2)+\frac23(\frac12)=1\), matching the sum of the new probabilities. This example also shows how the new measure gives more weight to the state in which the numeraire has the larger relative value.
Common Pitfalls and the Martingale Condition
Wrong calculation: if \(X=N/B^{\mathrm{bank}}\) is only known to be a positive local martingale, it is tempting to assert that \(\mathbb{E}^{\mathbb{Q}^B}[X_T]=X_0\) and then use \(X_T/X_0\) as a probability density. The first assertion does not follow from local-martingale status. A positive local martingale is a supermartingale, for which only \(\mathbb{E}^{\mathbb{Q}^B}[X_T]\le X_0\) is guaranteed. If the inequality is strict, then \(\mathbb{E}^{\mathbb{Q}^B}[X_T/X_0]<1\), so the proposed density fails to define a probability measure. The correction is to assume or establish that \(X\) is a true martingale on the horizon, or separately verify that the terminal candidate density is positive and has expectation one.
A second pitfall is to use an ex-dividend stock price as though its discounted price were always a martingale under the bank-account risk-neutral measure. For a stock with continuous dividend yield \(q>0\) satisfying \(dS_t/S_t=(r-q)dt+\sigma dB_t\), discounting by \(e^{rt}\) leaves drift \(-q\). The relevant martingale object, when the required integrability holds, is the discounted cum-dividend gains process. For the density theorem, the chosen numeraire’s discounted value must itself meet the true-martingale condition stated above.
| Condition on \(X=N/B^{\mathrm{bank}}\) | Consequence | What it does not establish |
|---|---|---|
| Strict positivity | A normalized density, if it has expectation one, is positive | Normalization by itself |
| True martingale with \(X_0>0\) | \(\mathbb{E}^{\mathbb{Q}^B}[X_T/X_0]=1\) | It is not implied by local-martingale status |
| Strictly positive terminal density | \(\mathbb{Q}^N\) and \(\mathbb{Q}^B\) are equivalent | Equivalence if the density can vanish |
The density process is useful because it records the change of measure at each time, not just at maturity. Its derivation is a conditional-expectation calculation once the relative numeraire is known to be a true martingale. The density formula itself does not require an explicit Brownian drift transformation; the essential ingredients here are normalization, conditional expectation, and strict positivity.
Check Your Understanding
For each question, distinguish the terminal density from the density process and state any martingale or positivity assumption being used.
- If \(N_t/B_t^{\mathrm{bank}}\) is a true martingale with initial value \(3\), write the terminal density of \(\mathbb{Q}^N\) relative to \(\mathbb{Q}^B\).
- Why does strict positivity of \(N_T/B_T^{\mathrm{bank}}\) imply equivalence once the normalized variable has expectation one?
- For \(X_t=e^{2B_t-2t}\), use the normal moment-generating function to check whether \(\mathbb{E}^{\mathbb{Q}^B}[X_t]=1\).
- What goes wrong with the density argument if \(X\) is a strict local martingale and \(\mathbb{E}^{\mathbb{Q}^B}[X_T]<X_0\)?
- For a dividend-paying stock, which process should be considered instead of the discounted ex-dividend price when checking the risk-neutral martingale property?