From a Change of Measure to a Process
Suppose two probability measures describe the same measurable events but assign them different probabilities. Their Radon–Nikodym derivative records the change of probability on the full sigma-field. On a filtration, conditioning that derivative on the information available at time \(t\) gives a time-indexed process. This process is the density process. Its martingale property follows from the tower property of conditional expectation; its positivity follows from equivalence of the measures.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{0\le t\le T},\mathbb{P})\), where \(T<\infty\). We do not require a Brownian motion for this construction. Let \(\mathbb{Q}\) be a probability measure on \((\Omega,\mathcal{F})\) equivalent to \(\mathbb{P}\), meaning that \(\mathbb{P}(A)=0\) if and only if \(\mathbb{Q}(A)=0\) for every \(A\in\mathcal{F}\). Assume \(\mathcal{F}_t\subseteq\mathcal{F}\). Define \(Y=\dfrac{d\mathbb{Q}}{d\mathbb{P}}\), the Radon–Nikodym derivative on \(\mathcal{F}\). Then \(Y\) is nonnegative, \(\mathcal{F}\)-measurable, and \(\mathbb{E}[Y]=1\). Equivalence implies \(Y>0\) \(\mathbb{P}\)-almost surely.
The conditioning is important: \(Z_t\) uses only information in \(\mathcal{F}_t\), whereas \(Y\) describes the change of measure on all of \(\mathcal{F}\). In particular, one should not assume \(Z_0=1\) without checking what \(\mathcal{F}_0\) contains.
The Density Process Is a Positive Martingale
Proof. Since \(Y\ge0\) and \(\mathbb{E}[Y]=1\), the defining property of conditional expectation gives \(Z_t\ge0\), \(\mathcal{F}_t\)-measurability, integrability, and \(\mathbb{E}[Z_t]=\mathbb{E}[Y]=1\). To prove strict positivity, let \(A=\{Z_t=0\}\), an event in \(\mathcal{F}_t\). Then \(\mathbb{E}[Y\mathbf{1}_A]=\mathbb{E}[Z_t\mathbf{1}_A]=0\). Since \(Y>0\) almost surely, this forces \(\mathbb{P}(A)=0\). For \(s\le t\), the tower property gives \(\mathbb{E}[Z_t\mid\mathcal{F}_s]=\mathbb{E}[\mathbb{E}[Y\mid\mathcal{F}_t]\mid\mathcal{F}_s]=\mathbb{E}[Y\mid\mathcal{F}_s]=Z_s\) almost surely. This is the martingale condition, and the preceding expectation calculation verifies integrability. Finally, if \(Y\) is \(\mathcal{F}_T\)-measurable, conditioning it on \(\mathcal{F}_T\) leaves it unchanged, so \(Z_T=Y\) almost surely. \(\square\)
Here “positive” means strictly positive almost surely at each fixed time. The theorem does not assert that every choice of versions has strictly positive paths simultaneously at every time. The martingale property is under \(\mathbb{P}\); it does not say that \(Z\) is a martingale under \(\mathbb{Q}\).
Worked Example: A Two-Outcome Filtration
Let \(\Omega=\{a,b\}\), with \(\mathcal{F}\) the set of all subsets, \(\mathbb{P}(\{a\})=\tfrac13\), and \(\mathbb{P}(\{b\})=\tfrac23\). Define \(\mathbb{Q}(\{a\})=\mathbb{Q}(\{b\})=\tfrac12\). Take \(\mathcal{F}_0=\{\varnothing,\Omega\}\) and \(\mathcal{F}_1=\mathcal{F}\).
Both measures give positive probability to each outcome, so they are equivalent. The density values are \(Y(a)=\mathbb{Q}(\{a\})/\mathbb{P}(\{a\})=(\tfrac12)/(\tfrac13)=\tfrac32\) and \(Y(b)=(\tfrac12)/(\tfrac23)=\tfrac34\). At time \(0\), conditioning on the trivial sigma-field gives the constant \(Z_0=\mathbb{E}[Y]=\tfrac13\cdot\tfrac32+\tfrac23\cdot\tfrac34=\tfrac12+\tfrac12=1\). At time \(1\), \(Z_1=Y\).
Check: The martingale condition from time \(0\) to time \(1\) is \(\mathbb{E}[Z_1\mid\mathcal{F}_0]=\mathbb{E}[Y]=1=Z_0\). Also, the probabilities recovered from the density are \(\mathbb{P}(\{a\})Y(a)=\tfrac13\cdot\tfrac32=\tfrac12=\mathbb{Q}(\{a\})\) and \(\mathbb{P}(\{b\})Y(b)=\tfrac23\cdot\tfrac34=\tfrac12=\mathbb{Q}(\{b\})\).
Worked Example: Information Revealed in Two Stages
Let \(\Omega=\{a,b,c\}\), with \(\mathcal{F}\) the set of all subsets. Set \(\mathbb{P}(a)=\tfrac14\), \(\mathbb{P}(b)=\tfrac14\), \(\mathbb{P}(c)=\tfrac12\), and \(\mathbb{Q}(a)=\tfrac12\), \(\mathbb{Q}(b)=\tfrac18\), \(\mathbb{Q}(c)=\tfrac38\). Let \(\mathcal{F}_0\) be trivial, let \(\mathcal{F}_1\) distinguish the group \(\{a,b\}\) from \(\{c\}\), and let \(\mathcal{F}_2=\mathcal{F}\).
The atomwise densities are \(Y(a)=\tfrac{1/2}{1/4}=2\), \(Y(b)=\tfrac{1/8}{1/4}=\tfrac12\), and \(Y(c)=\tfrac{3/8}{1/2}=\tfrac34\). At time \(1\), conditional expectation averages within each group using \(\mathbb{P}\). Given \(\{a,b\}\), each outcome has conditional probability \((\tfrac14)/(\tfrac12)=\tfrac12\), so \(Z_1=\tfrac12(2)+\tfrac12(\tfrac12)=\tfrac54\) on that group. On \(\{c\}\), \(Z_1=\tfrac34\). At time \(0\), \(Z_0=\mathbb{E}[Y]=\tfrac14(2)+\tfrac14(\tfrac12)+\tfrac12(\tfrac34)=\tfrac12+\tfrac18+\tfrac38=1\).
Check: On \(\{a,b\}\), \(\mathbb{E}[Z_2\mid\mathcal{F}_1]=\tfrac12(2)+\tfrac12(\tfrac12)=\tfrac54=Z_1\); on \(\{c\}\), that conditional expectation is \(\tfrac34=Z_1\). From time \(0\) to time \(1\), \(\mathbb{E}[Z_1]=\tfrac12(\tfrac54)+\tfrac12(\tfrac34)=\tfrac58+\tfrac38=1=Z_0\). The terminal expectation also checks directly: \(\mathbb{E}[Z_2]=\tfrac12+\tfrac18+\tfrac38=1\).
When Is the Initial Value One?
The common normalization \(Z_0=1\) is not a consequence of equivalence alone. It holds when the two measures agree on the information already present at time zero. The following criterion makes this precise.
Proof. For every \(A\in\mathcal{F}_0\), the defining property of conditional expectation and the Radon–Nikodym derivative gives \(\mathbb{E}[\mathbf{1}_A Z_0]=\mathbb{E}[\mathbf{1}_A Y]=\mathbb{Q}(A)\). If \(Z_0=1\) almost surely, the left side is \(\mathbb{P}(A)\), proving agreement on \(\mathcal{F}_0\). Conversely, suppose \(\mathbb{Q}(A)=\mathbb{P}(A)\) for every \(A\in\mathcal{F}_0\). Then \(\mathbb{E}[\mathbf{1}_A(Z_0-1)]=0\) for every such \(A\). If \(\mathbb{P}(Z_0>1)>0\), some event \(A_n=\{Z_0\ge1+\tfrac1n\}\in\mathcal{F}_0\) has positive probability, and \(\mathbb{E}[\mathbf{1}_{A_n}(Z_0-1)]\ge\tfrac1n\mathbb{P}(A_n)>0\), a contradiction. If \(\mathbb{P}(Z_0<1)>0\), some event \(C_n=\{Z_0\le1-\tfrac1n\}\in\mathcal{F}_0\) has positive probability, giving \(\mathbb{E}[\mathbf{1}_{C_n}(Z_0-1)]\le-\tfrac1n\mathbb{P}(C_n)<0\), also a contradiction. Therefore \(Z_0=1\) almost surely. If \(\mathcal{F}_0\) is \(\mathbb{P}\)-trivial, both probability measures assign probability zero to its null events and one to its probability-one events; hence they agree there. \(\square\)
Worked Example: Equivalent Measures with a Nonunit Initial Density
Let \(\Omega=\{a,b\}\), with \(\mathbb{P}(a)=\mathbb{P}(b)=\tfrac12\), \(\mathbb{Q}(a)=\tfrac34\), and \(\mathbb{Q}(b)=\tfrac14\). Take \(\mathcal{F}_0=\mathcal{F}_1=\mathcal{F}\). Both outcomes have positive probability under both measures, so the measures are equivalent. The density is \(Y(a)=(\tfrac34)/(\tfrac12)=\tfrac32\) and \(Y(b)=(\tfrac14)/(\tfrac12)=\tfrac12\). Because all information is already in \(\mathcal{F}_0\), \(Z_0=Y\): it equals \(\tfrac32\) on \(a\) and \(\tfrac12\) on \(b\), not one almost surely.
Check: Its expectation is \(\tfrac12\cdot\tfrac32+\tfrac12\cdot\tfrac12=\tfrac34+\tfrac14=1\), as required. But the measures differ on the initial event \(\{a\}\): \(\mathbb{P}(\{a\})=\tfrac12\), whereas \(\mathbb{Q}(\{a\})=\tfrac34\). This is exactly why the proposition does not give \(Z_0=1\).
Why the Density Process Matters
The process \(Z_t\) gives the density associated with the information available at each time. Its martingale property expresses consistency: conditioning a later density on an earlier sigma-field recovers the earlier density. Its strict positivity at each fixed time is useful when comparing measures, while its expected value remains one at every time. These facts make the density process a basic tool in change-of-measure arguments. They do not, by themselves, give a formula for conditional expectations under \(\mathbb{Q}\); that requires a separate conditional change-of-measure identity.
Common pitfall: confusing unconditional and conditional expectation. A frequent incorrect calculation is \(Z_0=\mathbb{E}[Y]=1\) for every filtration. In the preceding example, \(\mathbb{E}[Y]=1\), but \(\mathbb{E}[Y\mid\mathcal{F}_0]=Y\), which takes the values \(\tfrac32\) and \(\tfrac12\). The unconditional expectation determines \(Z_0\) only when \(\mathcal{F}_0\) is trivial up to null sets (or when the measures agree on \(\mathcal{F}_0\)). Always retain the conditioning sigma-field until its information has been accounted for.
| Claim | What ensures it |
|---|---|
| \(\mathbb{E}[Z_t]=1\) | \(\mathbb{E}[Y]=1\) and the conditional-expectation identity |
| \(Z_t>0\) almost surely at each fixed \(t\) | \(Y>0\) almost surely, which follows from equivalence |
| \(\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s\) for \(s\le t\) | The tower property of conditional expectation |
| \(Z_0=1\) almost surely | Agreement of \(\mathbb{P}\) and \(\mathbb{Q}\) on \(\mathcal{F}_0\) |
Check Your Understanding
For each question, distinguish carefully between the full sigma-field and the information available at the indicated time.
- Why does equivalence of \(\mathbb{P}\) and \(\mathbb{Q}\) imply that \(Y>0\) almost surely under \(\mathbb{P}\)?
- Use the tower property to show that \(\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s\) whenever \(s\le t\).
- In the two-stage example, what are the conditional probabilities of \(a\) and \(b\) under \(\mathbb{P}\), given the event \(\{a,b\}\), and how do they give \(Z_1=\tfrac54\) there?
- State a necessary and sufficient condition for \(Z_0=1\) almost surely.
- In the nonunit-initial-density example, calculate \(Z_0\) and \(\mathbb{E}[Z_0]\), and explain why those two quantities do not contradict each other.