Changing a Probability Measure at a Fixed Horizon
Changing probability measures begins with a comparison on the same measurable space. One measure may assign different probabilities to events and different expectations to payoffs, while still agreeing with the other measure about which events are impossible. The Radon–Nikodym derivative encodes this change as a measurable weight. At a fixed horizon, that weight is a terminal random variable; it is not yet a process indexed by time.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\), and fix \(T>0\). The measures in this tutorial are defined on \((\Omega,\mathcal{F}_T)\). No stochastic integrals are used here. In this course, an Ito integral \(\int_0^t H_s\,dB_s\), when used, has a progressively measurable integrand \(H\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\); that class is denoted by \(\mathcal{L}^2\). The measure \(\mathbb{P}\) is the reference measure, and \(\mathbb{Q}\) will be another probability measure on the same sigma-algebra. No risk-neutral assumption is imposed on \(\mathbb{Q}\).
Absolute continuity is one-way: \(\mathbb{P}\) may regard an event as possible even when \(\mathbb{Q}\) assigns it probability zero. Equivalence rules out that loss of possible events in either direction. The probabilities of non-null events need not be the same.
The Radon–Nikodym Density
The random variable \(Z\) is the terminal density: it weights outcomes according to how their probabilities change. The identity \(\mathbb{E}[Z]=1\) is essential. A nonnegative weight with expectation other than one does not define a probability measure through the displayed formula. Conversely, if \(Z\ge0\) is \(\mathcal{F}_T\)-measurable and \(\mathbb{E}[Z]=1\), then \(\mathbb{Q}(A)=\mathbb{E}[Z\mathbf{1}_A]\) defines a probability measure absolutely continuous with respect to \(\mathbb{P}\). Indeed, it gives zero to the empty set, gives one to \(\Omega\), and is countably additive by the monotone convergence theorem applied to indicators of disjoint unions.
Proof. First suppose \(\mathbb{Q}\sim\mathbb{P}\). Since \(Z\ge0\), the event \(A=\{Z=0\}\) satisfies \(\mathbb{Q}(A)=\mathbb{E}[Z\mathbf{1}_A]=0\). Equivalence implies \(\mathbb{P}(A)=0\), so \(Z>0\) \(\mathbb{P}\)-almost surely. Conversely, suppose \(Z>0\) \(\mathbb{P}\)-almost surely. We already have \(\mathbb{Q}\ll\mathbb{P}\). To prove the reverse absolute continuity, take any \(A\in\mathcal{F}_T\) with \(\mathbb{Q}(A)=0\). Then \(\mathbb{E}[Z\mathbf{1}_A]=0\). For each positive integer \(n\), on \(A\cap\{Z\ge1/n\}\) the integrand \(Z\mathbf{1}_A\) is at least \(1/n\), so $$ 0=\mathbb{E}[Z\mathbf{1}_A]\ge\frac{1}{n}\mathbb{P}\big(A\cap\{Z\ge1/n\}\big). $$ Thus each of these sets has \(\mathbb{P}\)-probability zero. Since \(Z>0\) almost surely, \(A\) is contained, up to a \(\mathbb{P}\)-null set, in the countable union \(\bigcup_{n=1}^{\infty}(A\cap\{Z\ge1/n\})\). Hence \(\mathbb{P}(A)=0\), proving \(\mathbb{P}\ll\mathbb{Q}\) and therefore equivalence. \(\square\)
This criterion distinguishes equivalence from absolute continuity. A density may equal zero on a set of positive \(\mathbb{P}\)-probability and still define a probability measure \(\mathbb{Q}\ll\mathbb{P}\); it simply cannot define a measure equivalent to \(\mathbb{P}\). All statements about \(Z\) are understood up to changes on \(\mathbb{P}\)-null sets.
Converting Expectations
Proof. For an indicator \(X=\mathbf{1}_A\), the identity is the defining property of \(Z\). By linearity it follows for nonnegative simple random variables. For a general nonnegative measurable \(X\), choose an increasing sequence of nonnegative simple random variables converging pointwise to \(X\). The monotone convergence theorem, applied under \(\mathbb{Q}\) on the left and under \(\mathbb{P}\) to the products with \(Z\) on the right, gives the identity, including when the expectations are infinite. For real-valued \(X\) with \(\mathbb{E}[Z|X|]<\infty\), apply the nonnegative identity to \(X^+=\max(X,0)\) and \(X^-=\max(-X,0)\). Both resulting expectations are finite, so subtraction gives the stated formula. \(\square\)
The integrability condition is important for a signed payoff: it ensures that the positive and negative parts can be subtracted without an undefined difference of infinities. If \(X\ge0\), no finiteness condition is needed for the identity itself, but the expectation may be infinite. When \(\mathbb{Q}\sim\mathbb{P}\), \(X\) is finite \(\mathbb{Q}\)-almost surely exactly when it is finite \(\mathbb{P}\)-almost surely; integrability, however, can still differ between the measures.
Worked Example: A Density on a Finite Space
Let \(\Omega=\{a,b,c\}\) and let \(\mathcal{F}_T\) contain every subset of \(\Omega\). Set \(\mathbb{P}(a)=\tfrac12\), \(\mathbb{P}(b)=\tfrac13\), \(\mathbb{P}(c)=\tfrac16\), and \(\mathbb{Q}(a)=\tfrac14\), \(\mathbb{Q}(b)=\tfrac12\), \(\mathbb{Q}(c)=\tfrac14\). Find \(Z\), check equivalence, and transfer the expectation of \(X(a)=2\), \(X(b)=-1\), \(X(c)=4\).
For a point \(\omega\), the density is \(Z(\omega)=\mathbb{Q}(\{\omega\})/\mathbb{P}(\{\omega\})\). Thus \(Z(a)=(\tfrac14)/(\tfrac12)=\tfrac12\), \(Z(b)=(\tfrac12)/(\tfrac13)=\tfrac32\), and \(Z(c)=(\tfrac14)/(\tfrac16)=\tfrac32\). Each value is positive, so the proposition gives \(\mathbb{Q}\sim\mathbb{P}\). The normalization check is $$ \mathbb{E}[Z]=\tfrac12\cdot\tfrac12+\tfrac13\cdot\tfrac32+\tfrac16\cdot\tfrac32 =\tfrac14+\tfrac12+\tfrac14=1. $$ Directly under \(\mathbb{Q}\), \(\mathbb{E}^{\mathbb{Q}}[X]=\tfrac14\cdot2+\tfrac12\cdot(-1)+\tfrac14\cdot4=\tfrac12-\tfrac12+1=1\). Using the density instead gives \(\mathbb{E}[ZX]=\tfrac12\cdot\tfrac12\cdot2+\tfrac13\cdot\tfrac32\cdot(-1)+\tfrac16\cdot\tfrac32\cdot4=\tfrac12-\tfrac12+1=1\), as required.
Check: the three probabilities under \(\mathbb{Q}\) sum to \(\tfrac14+\tfrac12+\tfrac14=1\), and the transferred expectation agrees with the direct calculation.
Worked Example: An Exponential Terminal Density
Let \(T>0\) and define \(Z=\exp(B_T-T/2)\). Use it to define \(\mathbb{Q}\) on \(\mathcal{F}_T\), and calculate \(\mathbb{E}^{\mathbb{Q}}[e^{uB_T}]\) for real \(u\).
Under \(\mathbb{P}\), \(B_T\) has distribution \(N(0,T)\), whose moment-generating function is \(\mathbb{E}[e^{vB_T}]=e^{v^2T/2}\) for real \(v\). Therefore $$ \mathbb{E}[Z]=e^{-T/2}\mathbb{E}[e^{B_T}] =e^{-T/2}e^{T/2}=1. $$ Since \(Z>0\) for every finite value of \(B_T\), it defines a probability measure equivalent to \(\mathbb{P}\). The expectation conversion formula gives $$ \mathbb{E}^{\mathbb{Q}}[e^{uB_T}] =\mathbb{E}[e^{(u+1)B_T-T/2}] =e^{-T/2}e^{(u+1)^2T/2} =e^{uT+u^2T/2}. $$ At \(u=0\), this equals \(1\), as it must for the expectation of the constant one. Differentiating this moment-generating function at \(u=0\) gives \(\mathbb{E}^{\mathbb{Q}}[B_T]=T\); its second derivative at zero is \(T^2+T\), so \(\operatorname{Var}^{\mathbb{Q}}(B_T)=T^2+T-T^2=T\).
Check: the normalization computed directly from the normal moment-generating function is exactly \(1\). This calculation concerns the terminal random variable \(B_T\); by itself it makes no claim about the process \(B_t\) under \(\mathbb{Q}\).
Worked Example: A Density That Is Not Equivalent
Let \(\Omega=\{a,b,c,d\}\), with all subsets measurable, and let \(\mathbb{P}\) assign probability \(\tfrac14\) to each point. Define \(Z(a)=Z(b)=2\) and \(Z(c)=Z(d)=0\), and set \(\mathbb{Q}(A)=\mathbb{E}[Z\mathbf{1}_A]\). Does \(\mathbb{Q}\) define a probability measure equivalent to \(\mathbb{P}\)?
The normalization is \(\mathbb{E}[Z]=\tfrac14(2+2+0+0)=1\), so the formula defines a probability measure. Its point probabilities are \(\mathbb{Q}(a)=\tfrac14\cdot2=\tfrac12\), \(\mathbb{Q}(b)=\tfrac12\), and \(\mathbb{Q}(c)=\mathbb{Q}(d)=0\). The event \(A=\{c,d\}\) has \(\mathbb{P}(A)=\tfrac12\) but \(\mathbb{Q}(A)=0\). Hence \(\mathbb{Q}\ll\mathbb{P}\), but \(\mathbb{Q}\not\sim\mathbb{P}\), in agreement with the fact that \(Z\) vanishes on a set of positive \(\mathbb{P}\)-probability.
Check: for \(X=\mathbf{1}_{\{c,d\}}\), direct calculation gives \(\mathbb{E}^{\mathbb{Q}}[X]=0\), and the density formula gives \(\mathbb{E}[ZX]=0\). The expectation identity remains valid even though the measures are not equivalent.
A Common Mistake and Why the Distinction Matters
A common error is to use the reciprocal density in the expectation formula. In the finite-space example, this would give \(\mathbb{E}[X/Z]\) instead of \(\mathbb{E}[ZX]\). For the stated payoff, the incorrect expression is $$ \mathbb{E}[X/Z] =\tfrac12\frac{2}{1/2}+\tfrac13\frac{-1}{3/2}+\tfrac16\frac{4}{3/2} =2-\tfrac29+\tfrac49=\tfrac{20}{9}, $$ which is not the directly computed \(\mathbb{E}^{\mathbb{Q}}[X]=1\). The orientation matters: \(Z=d\mathbb{Q}/d\mathbb{P}\) weights an expectation under \(\mathbb{P}\) to produce one under \(\mathbb{Q}\). When the measures are equivalent, \(1/Z\) is the reciprocal density \(d\mathbb{P}/d\mathbb{Q}\), interpreted \(\mathbb{Q}\)-almost surely, and it converts in the opposite direction.
The null-set criterion and expectation identity serve different purposes. Strict positivity of \(Z\) says the two measures agree on which events are null. The weighted expectation formula applies already under absolute continuity, even if \(Z\) vanishes on a set of positive \(\mathbb{P}\)-probability. In applications, first check the normalization \(\mathbb{E}[Z]=1\), then check whether strict positivity is needed, and finally verify the relevant integrability before transferring a signed payoff’s expectation.
Check Your Understanding
Use the definitions and results above. When testing equivalence, distinguish a zero-density event of positive probability from a null event.
- If \(Z=d\mathbb{Q}/d\mathbb{P}\), why must \(\mathbb{E}[Z]=1\)?
- Suppose \(Z\ge0\), \(\mathbb{E}[Z]=1\), and \(\mathbb{P}(Z=0)>0\). Can \(\mathbb{Q}\) still be absolutely continuous with respect to \(\mathbb{P}\)? Can it be equivalent?
- For the finite-space example, compute \(\mathbb{Q}(\{a,c\})\) both directly and using \(Z\).
- For the Brownian density example, evaluate \(\mathbb{E}^{\mathbb{Q}}[e^{2B_T}]\).
- State an integrability condition under which the expectation-conversion formula applies to a real-valued, possibly signed payoff \(X\).