Changing Measure Inside a Conditional Expectation
Changing probability measures changes not only unconditional expectations but also the probabilities assigned to events after information has been observed. The density process records how the two measures weight the information available at each time. Dividing by its current value gives the conditional Bayes formula: a conditional expectation under the new measure is a ratio of conditional expectations under the reference measure.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\), and fix a finite horizon \(T\). The Brownian motion and stochastic integrals are not needed for the general formula, but will be useful in examples. When an Ito integral \(\int_0^t H_s\,dB_s\) is used, its integrand is progressively measurable and belongs to \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). No stochastic integral is required below. Let \(\mathbb{Q}\) be a probability measure equivalent to \(\mathbb{P}\) on \(\mathcal{F}_T\): the two measures have the same null events in \(\mathcal{F}_T\). This notation alone does not say that \(\mathbb{Q}\) is a risk-neutral measure.
The density satisfies \(Z_T>0\) \(\mathbb{P}\)-almost surely and \(\mathbb{E}[Z_T]=1\). Also, \(Z_t>0\) \(\mathbb{P}\)-almost surely for every fixed \(t\le T\). To see the positivity at time \(t\), if \(A=\{Z_t=0\}\in\mathcal{F}_t\), then \(\mathbb{Q}(A)=\mathbb{E}[Z_T\mathbf{1}_A]=\mathbb{E}[Z_t\mathbf{1}_A]=0\). Equivalence implies \(\mathbb{P}(A)=0\). These facts ensure that division by \(Z_t\) is legitimate almost surely. The process \(Z_t\) is a true \(\mathbb{P}\)-martingale, not merely a local martingale: it is integrable, and for \(s\le t\), the tower property gives \(\mathbb{E}[Z_t\mid\mathcal{F}_s]=\mathbb{E}[Z_T\mid\mathcal{F}_s]=Z_s\).
The Conditional Bayes Formula
Proof. Set \(N_t=\mathbb{E}[Z_TX\mid\mathcal{F}_t]\) and \(R_t=N_t/Z_t\) on the full-\(\mathbb{P}\)-measure event where \(Z_t>0\), defining \(R_t\) arbitrarily on its null complement. Since \(Z_TX\) is integrable, \(N_t\) is integrable and \(\mathcal{F}_t\)-measurable. Therefore \(R_t\) is \(\mathcal{F}_t\)-measurable. It is also integrable under \(\mathbb{Q}\): using the density on \(\mathcal{F}_t\), the tower property, and the conditional-expectation inequality, $$ \mathbb{E}^{\mathbb{Q}}[|R_t|] =\mathbb{E}[Z_t|R_t|] =\mathbb{E}[|N_t|] \le \mathbb{E}[Z_T|X|] =\mathbb{E}^{\mathbb{Q}}[|X|]<\infty. $$ For any \(A\in\mathcal{F}_t\), the same density identity and conditional expectation give $$ \mathbb{E}^{\mathbb{Q}}[\mathbf{1}_A R_t] =\mathbb{E}[Z_t\mathbf{1}_A R_t] =\mathbb{E}[\mathbf{1}_A N_t] =\mathbb{E}[\mathbf{1}_A Z_TX] =\mathbb{E}^{\mathbb{Q}}[\mathbf{1}_A X]. $$ Thus \(R_t\) is \(\mathcal{F}_t\)-measurable, \(\mathbb{Q}\)-integrable, and has the defining integrals of \(\mathbb{E}^{\mathbb{Q}}[X\mid\mathcal{F}_t]\) on every \(A\in\mathcal{F}_t\). Uniqueness of conditional expectation up to almost-sure equality proves the formula. \(\square\)
The theorem needs only integrability of \(X\) under \(\mathbb{Q}\); it does not require \(X\) to be integrable under \(\mathbb{P}\). Indeed, the condition it uses under \(\mathbb{P}\) is \(Z_TX\in L^1(\mathbb{P})\), which is exactly equivalent to \(\mathbb{E}^{\mathbb{Q}}|X|<\infty\). For a nonnegative \(X\), the same identity extends to conditional expectations with values in \([0,\infty]\), by applying the integrable version to \(X\wedge n\) and taking increasing limits. The ratio requires \(Z_t>0\); equivalence supplies this. If the measures were only absolutely continuous and \(Z_t\) could vanish, that ratio could not be used on \(\{Z_t=0\}\) without additional qualifications.
Worked Applications
Worked Example: An Event Already Known at Time \(t\)
Let \(A\in\mathcal{F}_t\) and take \(X=\mathbf{1}_A\). Since \(A\) is already determined by the information at time \(t\), its conditional probability under \(\mathbb{Q}\) should be \(\mathbf{1}_A\). The formula confirms this. Because \(\mathbf{1}_A\) is \(\mathcal{F}_t\)-measurable,
Dividing by \(Z_t>0\) gives \(\mathbb{E}^{\mathbb{Q}}[\mathbf{1}_A\mid\mathcal{F}_t]=\mathbf{1}_A\). This is a check on both the density restriction and the denominator: information already observed must remain known after changing to an equivalent measure.
Worked Example: Conditional Mean Under a Brownian Exponential Tilt
Let the filtration be the natural Brownian filtration, and define \(Z_T=\exp(\theta B_T-\tfrac12\theta^2T)\) for a fixed real \(\theta\). For \(B_T\sim N(0,T)\), its moment generating function gives \(\mathbb{E}[Z_T]=e^{\theta^2T/2}e^{-\theta^2T/2}=1\). Since \(Z_T>0\), it defines a measure \(\mathbb{Q}\) equivalent to \(\mathbb{P}\) on \(\mathcal{F}_T\). This example uses the density directly; it does not invoke a change-of-measure theorem for Brownian motion.
For \(t\le T\), put \(h=T-t\). The increment \(Y=B_T-B_t\) is independent of \(\mathcal{F}_t\) and has distribution \(N(0,h)\). Factoring the density into its observed and future parts gives $$ Z_T=Z_t\exp(\theta Y-\tfrac12\theta^2h), \qquad Z_t=\exp(\theta B_t-\tfrac12\theta^2t). $$ The second identity follows by taking the conditional expectation of the first and using \(\mathbb{E}[e^{\theta Y}]=e^{\theta^2h/2}\). In particular, this calculation verifies directly that the displayed \(Z_t\) is the density process.
Apply conditional Bayes to \(X=B_T\). Completing the square in the normal density, or differentiating its moment generating function, gives \(\mathbb{E}[e^{\theta Y-\theta^2h/2}Y]=\theta h\). Hence $$ \mathbb{E}[Z_TB_T\mid\mathcal{F}_t] =Z_t\big(B_t+\theta h\big), \qquad \mathbb{E}^{\mathbb{Q}}[B_T\mid\mathcal{F}_t] =B_t+\theta(T-t). $$ For the exact parameters \(T=2\), \(t=1\), and \(\theta=\tfrac12\), the conditional mean is \(B_1+\tfrac12\). As an independent check, the same density gives \(\mathbb{E}^{\mathbb{Q}}[B_1]=\tfrac12\), so taking expectations of the conditional answer gives \(\tfrac12+\tfrac12=1\). Directly, \(\mathbb{E}^{\mathbb{Q}}[B_2]=\mathbb{E}[Z_2B_2]=\theta T=1\), in agreement.
Worked Example: Conditional Tail Probability
Continue with the exponential tilt above, using \(T=2\), \(t=1\), and \(\theta=\tfrac12\). Find \(\mathbb{Q}(B_2>0\mid\mathcal{F}_1)\). Here \(X=\mathbf{1}_{\{B_2>0\}}\) is bounded, so the integrability condition holds automatically. Given \(B_1=b\), the future increment \(Y=B_2-B_1\) has \(N(0,1)\) distribution under the reference conditional calculation. The density weight for this increment is \(e^{Y/2-1/8}\). Therefore Bayes' formula gives $$ \mathbb{Q}(B_2>0\mid\mathcal{F}_1) = \frac{\mathbb{E}[e^{Y/2-1/8}\mathbf{1}_{\{Y>-B_1\}}\mid\mathcal{F}_1]} {\mathbb{E}[e^{Y/2-1/8}\mid\mathcal{F}_1]}. $$ The denominator is \(e^{-1/8}e^{(1/2)^2/2}=1\). Multiplying the standard normal density at \(y\) by \(e^{y/2-1/8}\) gives the standard normal density at \(y-\tfrac12\), since $$ -\frac{y^2}{2}+\frac{y}{2}-\frac18=-\frac{(y-\frac12)^2}{2}. $$ Thus the ratio is \(\mathbb{P}(N(\tfrac12,1)>-B_1)=\Phi(B_1+\tfrac12)\), where \(\Phi\) is the standard normal CDF. At the value \(B_1=0\), the conditional-probability version evaluates to \(\Phi(\tfrac12)\), approximately \(0.6915\). This is evaluation of a conditional-probability function at \(b=0\), not conditioning on the probability-zero event \(\{B_1=0\}\).
Check: under the tilted increment density, \(Y\) has mean \(\tfrac12\) and variance \(1\). Consequently \(B_2=B_1+Y\), conditional on \(B_1\), has mean \(B_1+\tfrac12\) and variance \(1\); its probability of being positive is exactly \(\Phi(B_1+\tfrac12)\), as obtained from the ratio.
Second Moments and a Common Mistake
Worked Example: Conditional Second Moment
With the same parameters \(T=2\), \(t=1\), and \(\theta=\tfrac12\), calculate \(\mathbb{E}^{\mathbb{Q}}[B_2^2\mid\mathcal{F}_1]\). For a standard normal \(Y\), completing the square as above shows that the normalized tilted density is that of \(N(\tfrac12,1)\). Thus, writing \(b=B_1\), the weighted conditional second moment is $$ \mathbb{E}\big[e^{Y/2-1/8}(b+Y)^2\big] =\mathbb{E}\big[(b+N(\tfrac12,1))^2\big] =(b+\tfrac12)^2+1. $$ The denominator in the Bayes ratio is \(1\), so \(\mathbb{E}^{\mathbb{Q}}[B_2^2\mid\mathcal{F}_1]=(B_1+\tfrac12)^2+1\).
Check by taking \(\mathbb{Q}\)-expectations. Under this tilt, \(B_1\) has mean \(\tfrac12\) and variance \(1\), so \(B_1+\tfrac12\) has mean \(1\) and variance \(1\). Therefore \(\mathbb{E}^{\mathbb{Q}}[(B_1+\tfrac12)^2+1]=1^2+1+1=3\). Directly, \(B_2\) under the tilted terminal density has mean \(1\) and variance \(2\), so its second moment is \(1^2+2=3\) as well.
Common mistake: putting \(Z_T\) in the denominator. In the conditional mean example, the correct answer at \(t=1\) is \(B_1+\tfrac12\). Replacing \(Z_t\) by \(Z_T\) would instead produce $$ \frac{\mathbb{E}[Z_2B_2\mid\mathcal{F}_1]}{Z_2} =\frac{Z_1(B_1+\frac12)}{Z_2}. $$ This expression depends on the future increment \(B_2-B_1\), so it is generally not \(\mathcal{F}_1\)-measurable. A conditional expectation given \(\mathcal{F}_1\) must be \(\mathcal{F}_1\)-measurable. The denominator must be the density of the information available at the conditioning time, namely \(Z_1\). A related error is to use \(\mathbb{E}[X\mid\mathcal{F}_t]\) under \(\mathbb{P}\) without the density weighting: in this example at \(t=0\), that gives \(\mathbb{E}[B_2]=0\), whereas the tilted mean is \(\mathbb{E}^{\mathbb{Q}}[B_2]=1\).
The formula is useful whenever a payoff or random variable is easier to evaluate under one measure but the desired conditional expectation is under another. In applications to asset pricing, the measure may be risk-neutral \(\mathbb{Q}\), but that financial interpretation is an additional modeling assumption, not part of the conditional Bayes identity itself. In every application, check the same three items: the terminal density defines the measure, the payoff is integrable under that measure, and the density process at the conditioning time is positive.
Check Your Understanding
Use the theorem and its hypotheses to answer each question.
- Why does \(\mathbb{E}^{\mathbb{Q}}|X|<\infty\) imply that \(Z_TX\) is integrable under \(\mathbb{P}\)?
- Show that if \(A\in\mathcal{F}_t\), then the formula gives \(\mathbb{Q}(A\mid\mathcal{F}_t)=\mathbf{1}_A\).
- For the exponential tilt with parameter \(\theta\), what are \(Z_t\) and \(\mathbb{E}^{\mathbb{Q}}[B_T\mid\mathcal{F}_t]\)?
- Under the parameters \(T=2\), \(t=1\), \(\theta=\tfrac12\), express \(\mathbb{Q}(B_2>a\mid\mathcal{F}_1)\) for a fixed real \(a\).
- Explain why replacing \(Z_t\) by \(Z_T\) in the conditional Bayes formula generally fails.