Reverse Changes of Measure and Likelihood Ratios

Reversing a Change of Measure

A change of measure is often introduced by specifying a density under a reference probability measure $\mathbb{P}$. Under the new measure $\mathbb{Q}$, the Brownian motion acquires a drift adjustment. To compare path probabilities in the opposite direction, one needs the reciprocal density, expressed in terms of the Brownian motion that is actually Brownian under $\mathbb{Q}$. This reversal is also the basic calculation behind likelihood ratios for drifted Brownian paths.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. Fix $T>0$. Let $\theta_t$ be progressively measurable and satisfy $\mathbb{E}\int_0^T\theta_t^2\,dt<\infty$. Stochastic integrals under $\mathbb{P}$ below are Ito integrals with integrands in $\mathcal{L}^2(\mathbb{P})$, the progressively measurable processes $H$ for which $\mathbb{E}\int_0^T H_t^2\,dt<\infty$.

Definition: The forward density process is $Z_t=\exp\!\left(-\int_0^t\theta_s\,dB_s-\tfrac12\int_0^t\theta_s^2\,ds\right)$. Assume it is a true $\mathbb{P}$-martingale on $[0,T]$, with $Z_0=1$, and $Z_T>0$ almost surely. Define $\mathbb{Q}$ on $\mathcal{F}_T$ by $d\mathbb{Q}/d\mathbb{P}=Z_T$. Then $\mathbb{P}$ and $\mathbb{Q}$ are equivalent on $\mathcal{F}_T$, and $B_t^{\mathbb{Q}}=B_t+\int_0^t\theta_s\,ds$ is Brownian motion under $\mathbb{Q}$, by Girsanov’s theorem.

A sufficient condition for the assumed true-martingale property is Novikov’s condition, $\mathbb{E}\exp(\tfrac12\int_0^T\theta_s^2\,ds)<\infty$. It is a sufficient condition, not a necessary one. The reverse-density result below only needs the forward density to be a strictly positive true martingale; it does not require a second application of Novikov’s condition.

The Reverse Density Formula

Theorem: Under the assumptions above, the density process of $\mathbb{P}$ relative to $\mathbb{Q}$ is $Y_t=d\mathbb{P}/d\mathbb{Q}|_{\mathcal{F}_t}=1/Z_t$. If, in addition, $\theta$ is progressively measurable with $\mathbb{E}^{\mathbb{Q}}\int_0^T\theta_t^2\,dt<\infty$, then $$ Y_t=\exp\!\left(\int_0^t\theta_s\,dB_s^{\mathbb{Q}}-\frac12\int_0^t\theta_s^2\,ds\right). $$ Here the integral is an Ito integral under $\mathbb{Q}$ with an integrand in $\mathcal{L}^2(\mathbb{Q})$. The process $Y$ is a true $\mathbb{Q}$-martingale, not merely a local martingale.

Proof. For any $A\in\mathcal{F}_t$, the definition of $\mathbb{Q}$ and the martingale property of $Z$ give $\mathbb{Q}(A)=\mathbb{E}[Z_T\mathbf{1}_A]=\mathbb{E}[Z_t\mathbf{1}_A]$. Since $Z_t>0$, the reciprocal $1/Z_t$ is the Radon–Nikodym density of $\mathbb{P}$ relative to $\mathbb{Q}$ on $\mathcal{F}_t$. In particular, $\mathbb{E}^{\mathbb{Q}}[Y_t]=1$. For $0\le s\le t\le T$ and $A\in\mathcal{F}_s$, $\mathbb{E}^{\mathbb{Q}}[\mathbf{1}_A Y_t]=\mathbb{P}(A)=\mathbb{E}^{\mathbb{Q}}[\mathbf{1}_A Y_s]$. The defining property of conditional expectation therefore gives $\mathbb{E}^{\mathbb{Q}}[Y_t\mid\mathcal{F}_s]=Y_s$; this proves the true-martingale claim.

For the exponential representation, substitute $dB_s=dB_s^{\mathbb{Q}}-\theta_s\,ds$ into the logarithm of $1/Z_t$. The finite-variation integral $\int_0^t\theta_s\,ds$ is well defined almost surely by Cauchy–Schwarz and the assumed square integrability. Thus

\log(1/Z_t)=\int_0^t\theta_s\,dB_s+\frac12\int_0^t\theta_s^2\,ds =\int_0^t\theta_s\,dB_s^{\mathbb{Q}}-\frac12\int_0^t\theta_s^2\,ds.

The stochastic-integral substitution is valid for the specified Ito integrals: under $\mathbb{Q}$, $\theta\in\mathcal{L}^2(\mathbb{Q})$, and the finite-variation part contributes $-\int\theta_s^2\,ds$. This proves the formula. $\square$

The sign can be remembered by first fixing the forward convention: the density $Z$ has a minus sign before $\int\theta\,dB$, while $B^{\mathbb{Q}}=B+\int\theta\,dt$. Reversing the density reciprocates $Z$; expressing that reciprocal with $B^{\mathbb{Q}}$ gives a plus sign before the stochastic integral and retains the negative half-variance correction.

Takeaway: Under $\mathbb{Q}$, $dB_t^{\mathbb{Q}}\,dB_t^{\mathbb{Q}}=dt$, $dB_t^{\mathbb{Q}}\,dt=0$, and $dt\,dt=0$. The reverse exponential’s negative half-integrated-square term is essential: it compensates for the quadratic variation of its stochastic integral.

Worked Reverse-Density Calculations

Worked Example: A Constant Drift Shift

Let $T=2$, $\theta_t=\tfrac12$, and define $\mathbb{Q}$ by $Z_2=\exp(-\tfrac12 B_2-\tfrac18\cdot2)$. Find the reverse density in terms of $B_2^{\mathbb{Q}}$, and check that its $\mathbb{Q}$-expectation is one.

Because the integrand is deterministic, it belongs to $\mathcal{L}^2(\mathbb{P})$ and satisfies Novikov’s condition: $\mathbb{E}\exp(\tfrac12\int_0^2\theta_s^2\,ds)=\exp(\tfrac12\cdot\tfrac14\cdot2)=e^{1/4}<\infty$. Girsanov’s theorem gives $B_t^{\mathbb{Q}}=B_t+\tfrac12t$, and the reverse formula gives

\frac{d\mathbb{P}}{d\mathbb{Q}}\bigg|_{\mathcal{F}_2} =\exp\!\left(\frac12B_2^{\mathbb{Q}}-\frac12\cdot\frac14\cdot2\right) =\exp\!\left(\frac12B_2^{\mathbb{Q}}-\frac14\right).

Under $\mathbb{Q}$, $B_2^{\mathbb{Q}}\sim N(0,2)$. Its normal moment-generating function gives $\mathbb{E}^{\mathbb{Q}}[\exp(\tfrac12B_2^{\mathbb{Q}}-\tfrac14)]=\exp(\tfrac12(\tfrac12)^2\cdot2-\tfrac14)=e^0=1$, as required for a density. For a path with $B_2^{\mathbb{Q}}=1$, the likelihood ratio is $e^{1/2-1/4}=e^{1/4}$. This is a pathwise value, not the expectation.

Worked Example: A Time-Dependent Deterministic Kernel

On $[0,1]$, let $\theta_t=t$. Compute the reverse density and verify its expectation under $\mathbb{Q}$. The deterministic kernel is in both $\mathcal{L}^2(\mathbb{P})$ and $\mathcal{L}^2(\mathbb{Q})$, since $\int_0^1t^2\,dt=\tfrac13$. Novikov’s expectation is $\exp(\tfrac12\cdot\tfrac13)=e^{1/6}<\infty$.

The reverse density is

Y_1=\exp\!\left(\int_0^1s\,dB_s^{\mathbb{Q}}-\frac12\int_0^1s^2\,ds\right) =\exp\!\left(\int_0^1s\,dB_s^{\mathbb{Q}}-\frac16\right).

The Ito integral is centered normal with variance $\int_0^1s^2\,ds=\tfrac13$, so its exponential moment is $\mathbb{E}^{\mathbb{Q}}[\exp(\int_0^1s\,dB_s^{\mathbb{Q}}-\tfrac16)]=\exp(\tfrac12\cdot\tfrac13-\tfrac16)=1$. This independently checks normalization. The deterministic-integrand integration-by-parts identity also gives $\int_0^1s\,dB_s^{\mathbb{Q}}=B_1^{\mathbb{Q}}-\int_0^1B_s^{\mathbb{Q}}\,ds$, since the boundary term at zero vanishes. Hence the same likelihood can be computed from the transformed path as $\exp(B_1^{\mathbb{Q}}-\int_0^1B_s^{\mathbb{Q}}\,ds-\tfrac16)$.

Likelihood Ratios for Drifted Brownian Paths

Proposition: Let $\mathbb{P}_a$ and $\mathbb{P}_b$ be the laws on $\mathcal{F}_T$ of a coordinate process $X_t$ with constant drifts $a$ and $b$, respectively, and unit diffusion coefficient. They are equivalent for finite $T$, and $$ \frac{d\mathbb{P}_a}{d\mathbb{P}_b}\bigg|_{\mathcal{F}_T} =\exp\!\left((a-b)X_T-\frac12(a^2-b^2)T\right) =\exp\!\left((a-b)(X_T-bT)-\frac12(a-b)^2T\right). $$

Proof. Take $\mathbb{P}_0$, the law under which $X$ is standard Brownian motion. The constant-drift Girsanov density gives $d\mathbb{P}_a/d\mathbb{P}_0=\exp(aX_T-\tfrac12a^2T)$, and likewise $d\mathbb{P}_b/d\mathbb{P}_0=\exp(bX_T-\tfrac12b^2T)$. Both densities are strictly positive and have expectation one, so both drift laws are equivalent to $\mathbb{P}_0$ and hence to each other. Dividing the two densities yields the first expression. Under $\mathbb{P}_b$, $X_T-bT$ is centered normal with variance $T$. Expanding the second exponent gives $(a-b)X_T-(a-b)bT-\tfrac12(a-b)^2T=(a-b)X_T-\tfrac12(a^2-b^2)T$, proving the second expression and the identity. $\square$

Worked Example: Comparing Two Drifted Brownian Laws

On $[0,2]$, calculate the likelihood ratio of drift $a=1$ relative to drift $b=\tfrac12$. Under $\mathbb{P}_{1/2}$, define $B_t^{\mathbb{P}_{1/2}}=X_t-\tfrac12t$, which is Brownian motion.

Substitution in the proposition gives

\frac{d\mathbb{P}_1}{d\mathbb{P}_{1/2}}\bigg|_{\mathcal{F}_2} =\exp\!\left(\frac12X_2-\frac12(1-\tfrac14)\cdot2\right) =\exp\!\left(\frac12X_2-\frac34\right).

Since $X_2=B_2^{\mathbb{P}_{1/2}}+1$, the same ratio is $\exp(\tfrac12B_2^{\mathbb{P}_{1/2}}-\tfrac14)$. Under $\mathbb{P}_{1/2}$, $B_2^{\mathbb{P}_{1/2}}\sim N(0,2)$, so its expectation is $\exp(\tfrac12(\tfrac12)^2\cdot2-\tfrac14)=1$. For the observed path value $X_2=2$, the ratio is $e^{1-\tfrac34}=e^{1/4}$; using $B_2^{\mathbb{P}_{1/2}}=2-1=1$ gives $e^{1/2-1/4}=e^{1/4}$ again. The two forms agree exactly.

Why the Reverse View Matters

Likelihood ratios let one evaluate the relative support of a fixed observed path under competing drift models. In statistics, this is the pathwise basis for likelihood-based comparisons. In finance, the same logic distinguishes expectations under the physical measure $\mathbb{P}$ from those under the risk-neutral measure $\mathbb{Q}$: changing measures alters both the density used to weight paths and the Brownian motion used to describe their dynamics. The density is not a drift coefficient; it is a random variable that reweights probabilities.

The reciprocal identity also prevents a common sign error. The next calculation is wrong: from $B^{\mathbb{Q}}=B+\theta t$, one might assert that the reverse density is $\exp(-\int\theta\,dB^{\mathbb{Q}}-\tfrac12\int\theta^2dt)$. That keeps the forward stochastic-integral sign after changing both the measure and the Brownian motion. The correction is to start from $1/Z$ and substitute $dB=dB^{\mathbb{Q}}-\theta\,dt$. The result is $\exp(\int\theta\,dB^{\mathbb{Q}}-\tfrac12\int\theta^2dt)$, with the stochastic-integral sign reversed and the half-square term still negative. For $\theta=\tfrac12$ and $T=2$, the incorrect expression has expectation $\exp(\tfrac14-\tfrac14)=1$ too, so normalization alone cannot diagnose the sign error; the reciprocal-density derivation is essential.

Finally, equivalence is a finite-horizon statement under the assumptions made here. Strict positivity of $Z_T$ ensures that the two measures have the same null events on $\mathcal{F}_T$. If a proposed density can vanish or fail to be a true martingale, one cannot infer this equivalence or use the reciprocal formula without further hypotheses.

Check Your Understanding

For each question, keep track of which measure makes the stated Brownian motion standard.

If $Z_t=\exp(-\int_0^t\theta_s\,dB_s-\tfrac12\int_0^t\theta_s^2\,ds)$, write $d\mathbb{P}/d\mathbb{Q}|_{\mathcal{F}_t}$ using $B_t^{\mathbb{Q}}$.
For constant $\theta=c$ on $[0,T]$, verify directly from the normal moment-generating function that the reverse density has $\mathbb{Q}$-expectation one.
Write the likelihood ratio of drift $a$ relative to drift $b$ in terms of the centered process $X_T-bT$.
For $a=0$, $b=1$, and $T=1$, simplify the likelihood ratio $d\mathbb{P}_a/d\mathbb{P}_b$ as a function of $X_1$, then express it using $X_1-b$.
Why does the true-martingale property of the forward density imply that the reverse density is a true martingale under $\mathbb{Q}$, rather than merely a local martingale?