Density Processes and Equivalent Probability Measures

From a Martingale to a New Probability Measure

Changing probability measure begins with a nonnegative random variable that reweights outcomes. Outcomes assigned a larger value receive greater probability under the new measure; outcomes assigned value zero receive none. A martingale packages the conditional versions of these weights at earlier times. This makes it possible to calculate conditional expectations under the new measure using expectations under the original one.

Fix a finite horizon $T>0$ and a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{0\le t\le T},\mathbb{P})$ carrying a Brownian motion $B_t$. No stochastic integral is needed in this tutorial. When stochastic integrals are used, their integrands $H_t$ are understood to be progressively measurable and to satisfy $\mathbb{E}\int_0^T H_t^2\,dt<\infty$, the class $\mathcal{L}^2$. All equalities between random variables or conditional expectations below are understood to hold almost surely under the measure in question.

Definition: Let $Z_T$ be an $\mathcal{F}_T$-measurable random variable such that $Z_T\ge0$ and $\mathbb{E}[Z_T]=1$. Define $\mathbb{Q}$ on $\mathcal{F}_T$ by $\mathbb{Q}(A)=\mathbb{E}[\mathbf{1}_A Z_T]$ for $A\in\mathcal{F}_T$. The associated density process is $Z_t=\mathbb{E}[Z_T\mid\mathcal{F}_t]$, $0\le t\le T$.

The normalization makes $\mathbb{Q}$ a probability measure, and nonnegativity makes it countably additive. Conditional expectation gives $\mathbb{E}[Z_t]=1$ for each $t$, and the tower property gives $\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s$ for $s\le t$. Thus $Z_t$ is a nonnegative, true martingale. In particular, for every $A\in\mathcal{F}_t$,

\mathbb{Q}(A)=\mathbb{E}[\mathbf{1}_A Z_T] =\mathbb{E}[\mathbf{1}_A\,\mathbb{E}[Z_T\mid\mathcal{F}_t]] =\mathbb{E}[\mathbf{1}_A Z_t].

So $Z_t$ is the density of the restriction of $\mathbb{Q}$ to $\mathcal{F}_t$ relative to the restriction of $\mathbb{P}$. This statement does not require $Z_0=1$. In general, $Z_0=\mathbb{E}[Z_T\mid\mathcal{F}_0]$. It equals one if $\mathcal{F}_0$ is trivial up to null sets, or if the condition $\mathbb{E}[Z_T\mid\mathcal{F}_0]=1$ is separately imposed.

Worked Example: A Nontrivial Initial Sigma-Field

Let $\Omega=\{\omega_1,\omega_2,\omega_3,\omega_4\}$, with each outcome having $\mathbb{P}$-probability $1/4$. Let $\mathcal{F}_0$ distinguish the two sets $G_1=\{\omega_1,\omega_2\}$ and $G_2=\{\omega_3,\omega_4\}$, and let $\mathcal{F}_T$ distinguish all four outcomes. Set $Z_T(\omega_1,\omega_2,\omega_3,\omega_4)=(1/4,3/4,1,2)$.

The terminal mean is $(1/4+3/4+1+2)/4=1$, so these values define a probability measure $\mathbb{Q}$. Conditional expectation given $\mathcal{F}_0$ averages within each two-outcome set. Therefore $Z_0=1/2$ on $G_1$, since $(1/4+3/4)/2=1/2$, and $Z_0=3/2$ on $G_2$, since $(1+2)/2=3/2$. In particular, $Z_0$ is not almost surely one.

The $\mathbb{Q}$-probabilities of the four outcomes are their original probabilities multiplied by their density values: $1/16,3/16,1/4,1/2$. They sum to one. Thus $\mathbb{Q}(G_1)=1/4$ and $\mathbb{Q}(G_2)=3/4$. The density formula at time zero checks directly: $\mathbb{E}[\mathbf{1}_{G_1}Z_0]=(1/4)(1/2)=1/8$ would be incorrect if the factor $1/4$ were treated as the probability of $G_1$; here $\mathbb{P}(G_1)=1/2$, so the correct value is $(1/2)(1/2)=1/4=\mathbb{Q}(G_1)$. Likewise, $\mathbb{E}[\mathbf{1}_{G_2}Z_0]=(1/2)(3/2)=3/4=\mathbb{Q}(G_2)$.

Check: the two $\mathbb{Q}$-probabilities sum to $1/4+3/4=1$, while $Z_0$ takes two distinct values. Normalization of the terminal density gives $\mathbb{E}[Z_0]=1$, not $Z_0=1$ pointwise.

Conditional Expectations Under the New Measure

For an $\mathcal{F}_T$-measurable random variable $X$ with $\mathbb{E}^{\mathbb{Q}}[|X|]<\infty$, the conditional expectation under $\mathbb{Q}$ is obtained by weighting with $Z_T$ and dividing by the current density. The division is meaningful $\mathbb{Q}$-almost surely: the event $\{Z_t=0\}$ has $\mathbb{Q}$-probability zero.

Theorem: Under the definition above, for each $t\le T$ and each $\mathcal{F}_T$-measurable $X$ with $\mathbb{E}[Z_T|X|]<\infty$, $$ \mathbb{E}^{\mathbb{Q}}[X\mid\mathcal{F}_t] =\frac{\mathbb{E}[Z_T X\mid\mathcal{F}_t]}{Z_t} \quad\mathbb{Q}\text{-almost surely}. $$ The ratio can be assigned any value on $\{Z_t=0\}$, a $\mathbb{Q}$-null set.

Proof. Write $N_t=\mathbb{E}[Z_T X\mid\mathcal{F}_t]$. Since $\mathbb{E}[Z_T|X|]<\infty$, this conditional expectation is defined and finite almost surely. Also, $\mathbb{Q}(Z_t=0)=\mathbb{E}[\mathbf{1}_{\{Z_t=0\}}Z_t]=0$. On $\{Z_t=0\}$, the nonnegative conditional expectation $\mathbb{E}[Z_T\mid\mathcal{F}_t]$ is zero; hence $Z_T=0$ almost surely on that event, conditionally on $\mathcal{F}_t$, and $N_t=0$ there as well. Define $Y=N_t/Z_t$ on $\{Z_t>0\}$, with any value elsewhere.

To verify the defining property of conditional expectation, take $A\in\mathcal{F}_t$. First consider nonnegative $X$; truncating $Y$ if necessary and using the density formula on $\mathcal{F}_t$ gives $\mathbb{E}^{\mathbb{Q}}[\mathbf{1}_A Y]=\mathbb{E}[\mathbf{1}_A Z_tY]=\mathbb{E}[\mathbf{1}_A N_t]=\mathbb{E}[\mathbf{1}_A Z_TX]=\mathbb{E}^{\mathbb{Q}}[\mathbf{1}_A X]$. For signed $X$, apply the same argument to its positive and negative parts; both are integrable under $\mathbb{Q}$ by the hypothesis. The defining test-set identity holds for every $A\in\mathcal{F}_t$, and $Y$ is $\mathcal{F}_t$-measurable, so $Y$ is the claimed conditional expectation. $\square$

If $X$ is already $\mathcal{F}_t$-measurable, the formula reduces to $X$: the numerator is $X\mathbb{E}[Z_T\mid\mathcal{F}_t]=XZ_t$. At $t=T$, it also reduces to $X$ on $\{Z_T>0\}$, as expected. The condition $\mathbb{E}[Z_T|X|]<\infty$ is precisely integrability of $X$ under $\mathbb{Q}$.

Worked Example: Computing a Conditional Expectation

Use the four-outcome model above and let $X(\omega_1,\omega_2,\omega_3,\omega_4)=(2,-1,4,0)$. Compute its expectation under $\mathbb{Q}$, and its conditional expectation given $\mathcal{F}_0$.

The unconditional expectation is $\mathbb{E}^{\mathbb{Q}}[X]=(1/16)2+(3/16)(-1)+(1/4)4+(1/2)0=1/8-3/16+1=15/16$. For $G_1$, the conditional numerator is the $\mathbb{P}$-conditional average of $Z_TX$: $((1/4)2+(3/4)(-1))/2=(1/2-3/4)/2=-1/8$. Dividing by $Z_0=1/2$ gives $-1/4$. For $G_2$, the numerator is $(1\cdot4+2\cdot0)/2=2$; dividing by $Z_0=3/2$ gives $4/3$.

Thus $\mathbb{E}^{\mathbb{Q}}[X\mid\mathcal{F}_0]$ is $-1/4$ on $G_1$ and $4/3$ on $G_2$. Check: averaging these conditional values with the $\mathbb{Q}$-probabilities of the groups gives $(1/4)(-1/4)+(3/4)(4/3)=-1/16+1=15/16$, matching the direct calculation.

Absolute Continuity and Equivalence

Theorem: For $\mathbb{Q}(A)=\mathbb{E}[\mathbf{1}_A Z_T]$ with $Z_T\ge0$ and $\mathbb{E}[Z_T]=1$, $\mathbb{Q}$ is absolutely continuous with respect to $\mathbb{P}$. The measures are equivalent, meaning they have the same null events, if and only if $Z_T>0$ $\mathbb{P}$-almost surely.

Proof. If $\mathbb{P}(A)=0$, then $\mathbf{1}_A Z_T=0$ almost surely, so $\mathbb{Q}(A)=0$. This proves $\mathbb{Q}\ll\mathbb{P}$. Now suppose $Z_T>0$ $\mathbb{P}$-almost surely and $\mathbb{Q}(A)=0$. Then $\mathbb{E}[\mathbf{1}_A Z_T]=0$. For every positive integer $n$, on $A\cap\{Z_T\ge1/n\}$ the integrand is at least $1/n$, so $0\ge (1/n)\mathbb{P}(A\cap\{Z_T\ge1/n\})$. Each such event has probability zero. Since $Z_T>0$ almost surely, $A$ is contained, up to a null set, in the union of these events over $n$; hence $\mathbb{P}(A)=0$. Therefore $\mathbb{P}\ll\mathbb{Q}$ as well. Conversely, if $\mathbb{P}(Z_T=0)>0$, then $A=\{Z_T=0\}$ has positive $\mathbb{P}$-probability but $\mathbb{Q}(A)=0$, so the measures are not equivalent. $\square$

The same reasoning applies at any time $t$: the restriction of $\mathbb{Q}$ to $\mathcal{F}_t$ is equivalent to the restriction of $\mathbb{P}$ exactly when $Z_t>0$ $\mathbb{P}$-almost surely. Terminal equivalence requires strict positivity of $Z_T$, not merely nonnegativity. A nonnegative density may remove events; a strictly positive density only reweights them.

Worked Example: Absolute Continuity Without Equivalence

Take two outcomes of equal $\mathbb{P}$-probability, let $\mathcal{F}_t$ be trivial for $t<T$, let $\mathcal{F}_T$ distinguish the outcomes, and set $Z_T=(0,2)$. Its mean is $(0+2)/2=1$, so it defines a probability measure. The new probabilities are $\mathbb{Q}(\omega_1)=0$ and $\mathbb{Q}(\omega_2)=1$. The event $\{\omega_1\}$ has $\mathbb{P}$-probability $1/2$ and $\mathbb{Q}$-probability zero, so the measures are not equivalent, although $\mathbb{Q}\ll\mathbb{P}$.

For $t<T$, $Z_t=\mathbb{E}[Z_T]=1$; at $T$, the process takes the values $0,2$. This is a nonnegative martingale, but its terminal value is not strictly positive. Check: the density probabilities sum to $0+1=1$, while the event of zero density witnesses the failure of equivalence.

Practical Checks and a Common Pitfall

A density process must be a true martingale on the time interval in question. A local martingale is only a process that becomes a martingale after suitable stopping; by itself, it need not preserve expectations and cannot automatically be used as the conditional density $\mathbb{E}[Z_T\mid\mathcal{F}_t]$. Earlier results on stochastic exponentials and Novikov's condition provide useful sufficient conditions for a stochastic exponential to be a true martingale. Once that is established, the construction and conditional formulas here apply. The Brownian motion under $\mathbb{Q}$, and the drift change that may accompany it, are separate questions.

Common mistake: assuming that unit terminal mean forces $Z_0=1$ almost surely. In the four-outcome example, $\mathbb{E}[Z_T]=1$, but $Z_0=1/2$ on $G_1$ and $Z_0=3/2$ on $G_2$. The correct identity is $Z_0=\mathbb{E}[Z_T\mid\mathcal{F}_0]$, and its unconditional mean is one. A claim that $Z_0=1$ requires, for example, trivial $\mathcal{F}_0$ or the additional assumption $\mathbb{E}[Z_T\mid\mathcal{F}_0]=1$.

In applications, the measure $\mathbb{P}$ is the original probability law and $\mathbb{Q}$ may be a risk-neutral measure, but the measure change alone does not establish any financial pricing result. Its immediate role is precise: it changes the weights of events, provides a time-indexed density, and turns conditional expectations under $\mathbb{Q}$ into ratios of conditional expectations under $\mathbb{P}$.

Check Your Understanding

Use the definitions and conditions above to answer each question.

If $\mathbb{E}[Z_T]=1$ but $\mathcal{F}_0$ is not trivial, what formula determines $Z_0$?
For an $\mathcal{F}_t$-measurable integrable $X$, simplify the conditional-expectation formula under $\mathbb{Q}$.
What condition on $Z_T$ is necessary and sufficient for $\mathbb{P}$ and $\mathbb{Q}$ to be equivalent?
Why does $Z_T\ge0$ suffice for $\mathbb{Q}\ll\mathbb{P}$ but not for equivalence?
Why is a nonnegative local martingale not automatically a density process for the conditional-expectation formulas?

Density Processes and Equivalent Probability Measures

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