Wednesday 2 September 2026

The numeraire change (part 1)

Since $S_t^2=S_0^2\exp ((r-\frac{1}{2}{\sigma }_2^2)t+{\sigma }_2{W}_t^2)$, we have

$$\frac{S_t^2}{B_t}=S_0^2\exp (-\frac{1}{2}{\sigma }_2^{2}t+{\sigma }_2{W}_t^2).$$

This is $S_0^2$ times a Doléans–Dade exponential $ℰ({\sigma }_2{W}^2{)}_t$, hence a strictly positive $Q$-martingale (the drift $r$ cancels precisely because $S^2$ is a discounted tradeable). Its normalized version

$$Z_t:=\frac{S_t^2}{S_0^2{e}^{rt}}=\exp (-\frac{1}{2}{\sigma }_2^{2}t+{\sigma }_2{W}_t^2),Z_0=1,$$

is a legitimate Radon–Nikodym density process, so $Q_2$ is well defined and equivalent to $Q$.

The general change-of-numeraire pricing formula says: if $N$ is the price of a strictly positive self-financing tradeable and $Q_N$ is the associated measure, then for any ${ℱ}_T$-payoff $H_T$,

$$\frac{V_0}{N_0}={𝔼}^{Q_N}\left[\frac{H_T}{N_T}\right].$$

Take $N=S^2$, $H_T=(S_T^1-S_T^2{)}^{+}$. Then $H_T/S_T^2=(S_T^1/S_T^2-1{)}^{+}=(Y_T-1{)}^{+}$, so

$$V_0=S_0^2{𝔼}^{Q_2}[(Y_T-1{)}^{+}].$$

The key insight here — the one candidates miss — is to deflate by $S^2$ rather than by $B$. It collapses a two-asset problem to a one-dimensional option on the single ratio $Y$, and it will make $r$ vanish.

Dynamics of the ratio under $Q_2$ (part 2)

Under $Q$, applying Itô to the logs,

$$d\ln S_t^1=(r-\frac{1}{2}{\sigma }_1^2)dt+{\sigma }_{1}d{W}_t^1,$$

$$d\ln S_t^2=(r-\frac{1}{2}{\sigma }_2^2)dt+{\sigma }_{2}d{W}_t^2,$$

so that

$$d\ln Y_t=\frac{1}{2}({\sigma }_2^2-{\sigma }_1^2)dt+{\sigma }_{1}d{W}_t^1-{\sigma }_{2}d{W}_t^2.$$

Girsanov: the density $Z$ has log-martingale part ${\sigma }_2{W}^2$, so under $Q_2$ each Brownian motion is shifted by its covariation rate with ${\sigma }_2{W}^2$. Since $d⟨W^2,{\sigma }_2{W}^2⟩={\sigma }_{2}dt$ and $d⟨W^1,{\sigma }_2{W}^2⟩=\rho {\sigma }_{2}dt$,

$${\tilde{W}}_t^2=W_t^2-{\sigma }_{2}t,{\tilde{W}}_t^1=W_t^1-\rho {\sigma }_{2}t$$

are $Q_2$-Brownian motions (with the same correlation $\rho$). The step a hurried candidate botches is the shift of $W^1$: it must move by $\rho {\sigma }_{2}dt$, not zero. Substituting $d{W}^1=d{\tilde{W}}^1+\rho {\sigma }_{2}dt$ and $d{W}^2=d{\tilde{W}}^2+{\sigma }_{2}dt$,

$$d\ln Y_t=[\frac{1}{2}({\sigma }_2^2-{\sigma }_1^2)+\rho {\sigma }_1{\sigma }_2-{\sigma }_2^2]dt+{\sigma }_{1}d{\tilde{W}}_t^1-{\sigma }_{2}d{\tilde{W}}_t^2.$$

The drift is

$$-\frac{1}{2}({\sigma }_1^2-2\rho {\sigma }_1{\sigma }_2+{\sigma }_2^2)=-\frac{1}{2}{\widehat{\sigma }}^2,$$

while the martingale part ${\sigma }_{1}d{\tilde{W}}^1-{\sigma }_{2}d{\tilde{W}}^2$ has variance rate ${\sigma }_1^2-2\rho {\sigma }_1{\sigma }_2+{\sigma }_2^2={\widehat{\sigma }}^2$. Writing it as $\widehat{\sigma }d\tilde{W}$ for a single $Q_2$-Brownian motion $\tilde{W}$,

$$d\ln Y_t=-\frac{1}{2}{\widehat{\sigma }}^{2}dt+\widehat{\sigma }d{\tilde{W}}_t⟺d{Y}_t=\widehat{\sigma }{Y}_{t}d{\tilde{W}}_t.$$

So $Y$ is a driftless GBM — a $Q_2$-martingale, exactly as required, since $S^1/S^2$ is the price of a tradeable ($S^1$) deflated by the numeraire ($S^2$). The short rate $r$ has cancelled: it never appears in $\widehat{\sigma }$, in the drift, or anywhere in the reduced problem.

Closed form and number (part 3)

${𝔼}^{Q_2}[(Y_T-1{)}^{+}]$ is a Black–Scholes call on an underlying starting at $Y_0=S_0^1/S_0^2$, with strike $1$, volatility $\widehat{\sigma }$, and zero interest rate (there is no discounting left, because pricing is already expressed in units of the numeraire). Hence

$${𝔼}^{Q_2}[(Y_T-1{)}^{+}]=Y_{0}N(d_1)-N(d_2),$$

$$d_1=\frac{\ln Y_0+\frac{1}{2}{\widehat{\sigma }}^{2}T}{\widehat{\sigma }\sqrt{T}},d_2=d_1-\widehat{\sigma }\sqrt{T}.$$

Multiplying by $S_0^2$ (and using $S_0^2{Y}_0=S_0^1$),

$$\boxed{V_0=S_0^{1}N(d_1)-S_0^{2}N(d_2)},d_1=\frac{\ln (S_0^1/S_0^2)+\frac{1}{2}{\widehat{\sigma }}^{2}T}{\widehat{\sigma }\sqrt{T}},d_2=d_1-\widehat{\sigma }\sqrt{T},$$

with $\widehat{\sigma }=\sqrt{{\sigma }_1^2-2\rho {\sigma }_1{\sigma }_2+{\sigma }_2^2}$. This is Margrabe's formula, and it is manifestly independent of $r$.

Numerics. With the given data,

$${\widehat{\sigma }}^2=0.04-2(0.4)(0.2)(0.3)+0.09=0.082,\widehat{\sigma }=0.2864.$$

$$\ln (100/95)=0.05129,\frac{1}{2}{\widehat{\sigma }}^{2}T=0.0410,\widehat{\sigma }\sqrt{T}=0.2864.$$

$$d_1=\frac{0.05129+0.0410}{0.2864}=0.3223,d_2=0.3223-0.2864=0.0359.$$

$$N(d_1)=0.6264,N(d_2)=0.5143.$$

$$V_0=100(0.6264)-95(0.5143)=62.64-48.86\approx \$13.78.$$

Closing note

The two things that separate candidates: (i) choosing $S^2$ as numeraire so the payoff becomes a plain vanilla call on $Y=S^1/S^2$, and (ii) getting both Girsanov shifts right, which is what produces the cross term in ${\widehat{\sigma }}^2={\sigma }_1^2-2\rho {\sigma }_1{\sigma }_2+{\sigma }_2^2$. The tempting wrong answers are $\widehat{\sigma }=\sqrt{{\sigma }_1^2+{\sigma }_2^2}$ (forgetting the correlation cross term) and any formula containing $r$ (from discounting under $B$ instead of pricing in units of $S^2$). A useful sanity check: at $\rho \to 1$ with ${\sigma }_1={\sigma }_2$ the two stocks are perfectly locked together, $\widehat{\sigma }\to 0$, and $V_0\to (S_0^1-S_0^2{)}^{+}$, the deterministic payoff — as it must.

Part 1: the absolute score

Since $p\in [0,1]$ we have $|p-1|=1-p$ and $|p-0|=p$, so

$$\overline{\ell }(p;q)=q|p-1|+(1-q)|p-0|=q(1-p)+(1-q)p.$$

Expanding,

$$\overline{\ell }(p;q)=q+(1-2q)p.$$

This is affine in $p$ — the point a candidate can miss is that the modulus collapses to linear once $p$ is restricted to $[0,1]$. A linear objective is minimized at an endpoint:

$$\mathrm{\backslash arg}{min}_{p\in [0,1]}\overline{\ell }(p;q)=\{\begin{array}{cc}\{1\},\hfill & q>1/2,\hfill \\ [0,1],\hfill & q=1/2,\hfill \\ \{0\},\hfill & q<1/2.\hfill \end{array}$$

So the optimal report is a hard decision (predict the more likely label with certainty), not the true probability $q$. The honest report $p=q$ is optimal only when $q\in \{0,1\}$ (endpoints, where the decision and the probability coincide) and, degenerately, when $q=1/2$ (every $p$ ties). For all other $q$, reporting $q$ is strictly worse than reporting $0$ or $1$.

Hence $\ell (p,y)=|p-y|$ is not a proper scoring rule: it rewards confident classification, not calibrated probabilities.

Part 2: the Brier score

$$\overline{\ell }(p;q)=q(1-p{)}^2+(1-q)p^2.$$

Expand:

$$\overline{\ell }(p;q)=q-2qp+q{p}^2+(1-q)p^2=q-2qp+p^2.$$

Differentiate in $p$:

$$\frac{d}{dp}\overline{\ell }(p;q)=-2q+2p=2(p-q),$$

which vanishes at $p=q$, and $\frac{d^2}{d{p}^2}\overline{\ell }=2>0$, so this is the unique global minimum. The Brier score is therefore strictly proper.

The optimal value is $\overline{\ell }(q;q)=q-2q^2+q^2=q(1-q)$, and

$$\overline{\ell }(p;q)-\overline{\ell }(q;q)=(q-2qp+p^2)-(q-q^2)=p^2-2qp+q^2=(p-q{)}^2.$$

So the excess expected score is

$$\boxed{\overline{\ell }(p;q)-\overline{\ell }(q;q)=(p-q{)}^2}$$

the squared Euclidean distance between the report and the truth.

Closing note

The quadratic in $p$ being minimized at $p=q$ with excess $(p-q{)}^2$ is the binary instance of the general fact that every strictly proper score induces a Bregman divergence as its excess loss: for the Brier score the generating convex function is $-x(1-x)$ (giving squared distance), and for the log score $\ell (p,y)=-y\log p-(1-y)\log (1-p)$ the excess is the KL divergence $q\log \frac{q}{p}+(1-q)\log \frac{1-q}{1-p}$. The absolute score fails precisely because $|p-y|$ has no strictly convex generator — its expectation is affine in $p$, so no interior point can be a strict minimizer.