Numeraires and Relative Asset Prices

Prices Depend on the Unit of Account

A price is always expressed in some unit. A house price quoted in dollars and the same price quoted in euros are not different houses; they are different numerical representations of the same value. In a financial market, a positive traded asset can play the role of that unit. Prices divided by the value of this asset are relative prices: they report how many units of the chosen asset are needed to represent another asset’s value.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ under $\mathbb{P}$. An Ito integral $\int_0^t H_s\,dB_s$ below is understood for a progressively measurable integrand $H$ in $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$ on each finite horizon $T$ where the integral is used. The measures $\mathbb{P}$ and $\mathbb{Q}$ are probability measures; a traded asset is a process of prices. These are different kinds of objects.

Definition: On a time interval $[0,T]$, a numeraire is a traded asset with price process $N_t$ that is strictly positive almost surely at every time on the interval. The price of another traded asset $S_t$, expressed in units of $N$, is the relative price $S_t/N_t$. A numeraire-associated measure is a probability measure under which the relevant traded-asset prices, when divided by $N$, are martingales, whenever such a measure is available under the model’s assumptions.

Strict positivity matters: division by $N_t$ must be defined. A money-market account $A_t=e^{rt}$, for constant interest rate $r$, is a common numeraire. But the numeraire need not be risk-free; a strictly positive stock or a bond that remains positive over the interval can also serve. Choosing a numeraire changes the unit of account. It does not, by itself, change the probabilities assigned to events.

Why Relative Prices Are the Natural Quantities

Suppose one unit of an asset costs $S_t$ in currency and one unit of the numeraire costs $N_t$ in the same currency. Then $S_t/N_t$ is the number of numeraire units equivalent in value to one unit of the asset. If an investor holds $S_t/N_t$ units of the numeraire, the value is $(S_t/N_t)N_t=S_t$. This is an accounting identity, but it identifies the quantities that matter when the numeraire is used as the unit of value.

Proposition: Let $S_t$ and $N_t$ be asset prices and suppose $N_t>0$. The value $S_t$, expressed in units of $N$, is $S_t/N_t$. Converting that relative price back into currency recovers $S_t$, and converting a currency value $V_t$ into numeraire units and back recovers $V_t$.

Proof. By definition, one unit of the numeraire is worth $N_t$ in currency. Therefore $S_t/N_t$ units of the numeraire have currency value $(S_t/N_t)N_t=S_t$. For any currency value $V_t$, the same calculation gives $(V_t/N_t)N_t=V_t$. The divisions are valid because $N_t>0$. $\square$

In a model where the money-market account $A_t$ is the numeraire, a risk-neutral measure $\mathbb{Q}$ is an equivalent probability measure under which traded prices divided by $A_t$ are martingales, subject to the market’s modeling and integrability assumptions. “Risk-neutral” describes a property of the measure relative to the chosen numeraire; it does not name the numeraire itself. A different positive asset can be selected as numeraire, but the associated probability measure is then a separate object. The construction of that new measure is a distinct step.

Worked Example: Currency Prices in Money-Market Units

Let the money-market account be $A_t=e^{t/20}$. At time $t=2$, another asset is worth $120$ currency units. Express its price in units of $A$, then convert back.

At $t=2$, the numeraire value is $A_2=e^{2/20}=e^{1/10}$. Thus the asset’s relative price is $120/e^{1/10}$ units of the account. Converting back gives

\left(\frac{120}{e^{1/10}}\right)e^{1/10}=120.

Check: the account’s value appears once in the denominator when converting into account units and once in the numerator when converting back. The currency value is therefore exactly $120$, not $120e^{1/10}$ or $120e^{-1/10}$.

How Ratios of Stochastic Asset Prices Move

Relative prices need not be martingales under every probability measure. Their drift depends on both asset processes and on their common fluctuations. The following result gives the exact condition in a useful one-Brownian-motion setting.

Theorem: Suppose, under $\mathbb{P}$, positive asset prices satisfy $dS_t=\mu_S S_t\,dt+\sigma_S S_t\,dB_t$ and $dN_t=\mu_N N_t\,dt+\sigma_N N_t\,dB_t$, with constant real coefficients and strictly positive initial values. Then $R_t=S_t/N_t$ satisfies $$ dR_t=R_t\left(\mu_S-\mu_N+\sigma_N^2-\sigma_S\sigma_N\right)dt +R_t(\sigma_S-\sigma_N)dB_t. $$ On every finite time interval, $R$ is a true $\mathbb{P}$-martingale if and only if $\mu_S-\mu_N+\sigma_N^2-\sigma_S\sigma_N=0$.

Proof. The explicit solutions of the two geometric Brownian motion equations are positive exponentials, so $R_t>0$. Apply Ito’s formula to $f(s,n)=s/n$. Its derivatives are $f_s=1/n$, $f_n=-s/n^2$, $f_{ss}=0$, $f_{sn}=-1/n^2$, and $f_{nn}=2s/n^3$. The second-order correction, including the cross variation because both assets are driven by the same $B$, is

\frac12 f_{ss}(dS_t)^2+f_{sn}\,d[S,N]_t+\frac12 f_{nn}(dN_t)^2 =0-\frac{\sigma_S\sigma_N S_t}{N_t}dt +\frac{\sigma_N^2 S_t}{N_t}dt.

Here $d[S,N]_t=\sigma_S\sigma_N S_tN_t\,dt$, $(dS_t)^2=\sigma_S^2S_t^2dt$, and $(dN_t)^2=\sigma_N^2N_t^2dt$. The first-order terms are $\mu_S(S_t/N_t)dt+\sigma_S(S_t/N_t)dB_t-\mu_N(S_t/N_t)dt-\sigma_N(S_t/N_t)dB_t$. Adding these terms and the displayed Ito correction gives the stated dynamics.

Write $\alpha=\mu_S-\mu_N+\sigma_N^2-\sigma_S\sigma_N$ and $\delta=\sigma_S-\sigma_N$. The explicit ratio is $R_t=R_0\exp((\alpha-\tfrac12\delta^2)t+\delta B_t)$. It is integrable at every finite time because a Gaussian random variable has finite exponential moments. For $0\le s\le t$, the increment $B_t-B_s$ is independent of $\mathcal{F}_s$ and is normal with mean $0$ and variance $t-s$. Hence

\mathbb{E}[R_t\mid\mathcal{F}_s] =R_s e^{\alpha(t-s)} \mathbb{E}\!\left[e^{\delta(B_t-B_s)-\frac12\delta^2(t-s)}\right] =R_s e^{\alpha(t-s)}.

The final expectation equals $1$, by the moment-generating function of a centered normal variable with variance $t-s$. If $\alpha=0$, the conditional expectation is $R_s$, proving the martingale property, including integrability. If $\alpha\ne0$ and $t>s$, then $e^{\alpha(t-s)}\ne1$; since $R_s>0$, the conditional expectation differs from $R_s$, so $R$ is not a martingale. $\square$

Worked Example: A Relative Price Under the Original Measure

Under $\mathbb{P}$, let $dS_t=0.08S_t\,dt+0.3S_t\,dB_t$ and $dN_t=0.03N_t\,dt+0.2N_t\,dB_t$, with $S_0=100$ and $N_0=50$. Find the dynamics and decide whether $S_t/N_t$ is a $\mathbb{P}$-martingale.

The drift coefficient for the ratio is $0.08-0.03+(0.2)^2-(0.3)(0.2)=0.05+0.04-0.06=0.03$. Its diffusion coefficient is $0.3-0.2=0.1$, and $R_0=100/50=2$. Therefore

dR_t=0.03R_t\,dt+0.1R_t\,dB_t.

The ratio is not a $\mathbb{P}$-martingale because its drift coefficient is nonzero. To check the calculation independently, the explicit ratio has log drift $(0.08-\tfrac12(0.3)^2)-(0.03-\tfrac12(0.2)^2)=0.035-0.010=0.025$. A process with diffusion coefficient $0.1$ and log drift $0.025$ has ordinary drift $0.025+\tfrac12(0.1)^2=0.025+0.005=0.03$, as above. The initial ratio is $2$, and the theorem gives $\mathbb{E}[R_t]=2e^{0.03t}$, which also rules out a martingale.

A Numeraire Is Not a Probability Measure

A probability measure assigns probabilities to events. A numeraire is an asset whose price supplies the unit used to report values. The two concepts interact in pricing, but they cannot be substituted for one another. In particular, dividing prices by an asset does not automatically change $\mathbb{P}$ into a new measure. Nor does a martingale property under one measure guarantee that ratios are martingales under that same measure when the numeraire changes.

Worked Example: A Risk-Neutral Measure and a Stock Numeraire

Consider a one-period market with two states, up and down. The money-market account has value $1$ at both dates. A stock costs $1$ initially and pays $1/2$ in the down state and $3/2$ in the up state. Let the physical probabilities be $\mathbb{P}(\text{up})=1/4$ and $\mathbb{P}(\text{down})=3/4$.

A risk-neutral measure for the money-market numeraire must make the expected terminal stock price equal its initial price. If its up probability is $q$, then $q(3/2)+(1-q)(1/2)=1/2+q=1$, so $q=1/2$. Thus $\mathbb{Q}(\text{up})=\mathbb{Q}(\text{down})=1/2$. Both probabilities are positive, so $\mathbb{Q}$ and $\mathbb{P}$ assign zero probability to the same events in this two-state model; they are equivalent.

The stock is also strictly positive in both states and can be chosen as a numeraire. In stock units, the money-market account’s terminal relative price is $1/(3/2)=2/3$ in the up state and $1/(1/2)=2$ in the down state. Under the money-market risk-neutral measure $\mathbb{Q}$, its expected relative price is

\mathbb{E}^{\mathbb{Q}}\!\left[\frac{1}{S_1}\right] =\frac12\cdot\frac23+\frac12\cdot2 =\frac13+1=\frac43.

The initial relative price is $1/S_0=1$, so this ratio is not a $\mathbb{Q}$-martingale. This does not contradict the definition of $\mathbb{Q}$: it was chosen for the money-market account, not for the stock. The probabilities of the two states under $\mathbb{Q}$ remain $1/2$ and $1/2$; choosing the stock as numeraire does not itself alter them. Check: under $\mathbb{Q}$, the stock’s expected terminal value is $\tfrac12(\tfrac12)+\tfrac12(\tfrac32)=\tfrac14+\tfrac34=1$, exactly its initial value, confirming the risk-neutral condition for the money-market numeraire.

Common Pitfall: Dividing Martingales

A frequent error is to reason that if $S_t$ and $N_t$ are martingales under some measure, then $S_t/N_t$ must also be a martingale under that measure. Martingales are not generally closed under division. The ratio theorem shows the missing terms: its drift includes the drift difference and the volatility correction $\sigma_N^2-\sigma_S\sigma_N$. The finite-state example gives the same warning without stochastic calculus: the stock is a martingale under the money-market risk-neutral measure, yet the money-market account divided by the stock is not.

The useful distinction is therefore:

Numeraire: a strictly positive traded asset used as a unit of account.
Measure: a probability law on the underlying events.
Numeraire-associated martingale property: a relationship between a chosen asset unit and a chosen probability measure.

To use relative prices correctly, identify the asset serving as the unit and the measure under which the relevant martingale statement is intended. Neither choice should be inferred from the other.

Check Your Understanding

For each question, specify the numeraire and probability measure separately when both are involved.

If $N_t>0$ and $S_t$ is an asset price, what does $S_t/N_t$ measure, and how do you recover $S_t$?
For two assets driven by the same Brownian motion, write the drift coefficient of their price ratio.
Why does the nonzero ratio drift in the worked geometric Brownian motion example rule out a martingale under $\mathbb{P}$?
In the two-state example, verify that the risk-neutral up probability is $1/2$ and that both states have positive probability under $\mathbb{Q}$.
Explain why choosing the stock as numeraire does not, by itself, change the probabilities of the up and down states.