# Are all change of measure operations between equivalent probability measures Doléans-Dade exponentials?

Let $$(\Omega, \mathcal{F}, \mathbb{F}, \mathbb{P})$$ be a filtered probability space, where $$\mathbb{F}=\left(\mathcal{F}\right)_{t\in[0;T]}$$ and $$\mathcal{F}=\mathcal{F}_T$$. Let $$(W_t)_{t\in[0;T]}$$ be a Brownian motion with respect to $$\mathbb{F}$$, in the given probability space.

We have the following theorem (Stochastic Calculus for Finance II, Continuous Time Models, p 212):

Theorem 5.2.3 Let $$\left(\Theta_t\right)_{t\in[0;T]}$$ be an $$\mathbb{F}$$-adapted process. Define: $$Z_t=e^{-\int_0^t\Theta_udu-\frac{1}{2}\int_0^t\Theta^2_udW_u} >0, Z:=Z_T$$ $$\widetilde{W}_t=W_t+\int_0^t\Theta_udu$$ and assume that (this is somehow weaker than Novikov condition): $$\mathbb{E}_{\mathbb{P}}\left[\int_0^T\Theta^2_uZ^2_udu\right]<+\infty.$$

THEN

1. $$\mathbb{E}_{\mathbb{P}}[Z]=1$$. (This, along with the fact that $$Z:=Z_T\geq 0$$ ensure that $$Z$$ can be a Radon-Nikodym derivative)
2. Under the probability measure defined by $$\widetilde{\mathbb{P}}(A)=\int_{A}Z(\omega)d\mathbb{P}(\omega), (\forall)A\in\mathcal{F}$$, $$\left(\widetilde{W}_t\right)_{t\in[0;T]}$$ is a standard Brownian motion with respect to filtration $$\mathbb{F}$$.

QUESTION: With the notation above, knowing only the fact that that $$\left(W_t\right)_{t\in[0;T]}$$ is a Brownian motion in $$(\Omega, \mathcal{F}, \mathbb{P})$$ generating filtration $$\mathbb{F}=(\mathcal{F}_t)_{t\in[0;T]}$$, that $$(\Omega, \mathcal{F}, \mathbb{F}, \widetilde{\mathbb{P}})$$ is another probability space and that $$\mathbb{P}\approx \widetilde{\mathbb{P}}$$, does this necessarily imply that the Radon-Nikodym derivative process $$\frac{d\widetilde{\mathbb{P}}}{d\mathbb{P}}|_{t}$$ must of the form: $$Z_t=e^{-\int_0^t\Theta_udu-\frac{1}{2}\int_0^t\Theta^2_udW_u} >0, Z:=Z_T$$ where $$\left(\Theta_t\right)_{t\in[0;T]}$$ is some $$\mathbb{F}$$-adapted process? If this is true, and $$\left(\widetilde{W}_t\right)_{t\in[0;T]}$$ is a Brownian motion in $$(\Omega, \mathcal{F}, \mathbb{F}, \widetilde{\mathbb{P}})$$, does the above necessarily imply that $$\widetilde{W}_t=W_t+\int_0^t\Theta_udu$$?

• In order to apply the Martingale Representation theorem, the filtration has to be the one generated by the Brownian motion (cf. Shreve). – SN76 Apr 13 '20 at 22:26
• I have now rephrased the question and changed the answer slightly to clarify I am only interested in those cases where $\mathbb{F}$ is generated by Brownian motion $W$ – fwd_T Apr 14 '20 at 8:49

Proof:

Theorem (Radon-Nikodym) Let $$(\Omega, \mathcal{F})$$ be a measurable space. Let $$\mathbb{P}$$ and $$\widetilde{\mathbb{P}}$$ be two $$\sigma$$-finite measures. Let $$\widetilde{\mathbb{P}}$$ be absolutely continuous w.r.t. $$\mathbb{P}$$ (i.e. $$\widetilde{\mathbb{P}}\ll\mathbb{P}$$). THEN: $$(\exists)$$ measurable function $$f:\Omega\to[0;+\infty)$$ such that: $$\widetilde{\mathbb{P}}(A)=\int_A f(\omega)d\mathbb{P}(\omega), (\forall)A\in\mathcal{F}.$$ $$f$$ is unique up to indistinguishability, i.e. if there is another $$g$$ with the same properties as above, then $$f=g, \mathbb{P}-a.s.$$ (or $$\mathbb{P}$$-a.e.).

Note that if $$\mathbb{P}$$ and $$\widetilde{\mathbb{P}}$$ are equivalent measures (denoted by $$\mathbb{P}\approx\widetilde{\mathbb{P}}$$), then $$\widetilde{\mathbb{P}}\ll\mathbb{P}$$ and $$\mathbb{P}\ll\widetilde{\mathbb{P}}$$.

Let now $$(\Omega, \mathcal{F}, \mathbb{F}, \mathbb{P})$$ be a filtered probability space, where $$\mathbb{F}=(\mathcal{F}_t)_{t\geq0}$$ is the filtration. We use the Radon-Nikodym theorem to prove the next proposition:

Proposition. Let $$\mathbb{P}\approx\widetilde{\mathbb{P}}$$ be two equivalent probability measures on $$(\Omega, \mathcal{F}_T)$$, a measurable space from the notation above. THEN, $$(\exists)$$ a strictly positive $$(\mathbb{P}, \mathbb{F})$$-martingale $$(L_t)_{t\geq 0}$$ such that $$\widetilde{\mathbb{P}}(A)=\int_A L_t(\omega)d\mathbb{P}(\omega), (\forall) A\in\mathcal{F}_t, (\forall) t\leq T$$ with the properties that:

1. $$\mathbb{E}_{\widetilde{\mathbb{P}}}[X]=\mathbb{E}_{\mathbb{P}}[L_tX]$$, for all $$\mathcal{F}_t$$-measurable, non-negative, random variables $$X$$, when $$t\leq T$$.
2. $$L_0 = 1$$
3. $$\mathbb{E}_{\mathbb{P}}[L_t]=1, (\forall) t\leq T$$.

Proof: We know from the Radon-Nikodym theorem above that since $$\mathbb{P}\approx\widetilde{\mathbb{P}}$$ on $$(\Omega, \mathcal{F}_T)$$, then there must exist a non-negative, $$\mathcal{F}_T$$-measurable random variable $$Z$$ with the property that $$\widetilde{\mathbb{P}}(A)=\int_AZ(\omega)d\mathbb{P}(\omega), (\forall)A\in\mathcal{F}_T$$ Since we have already assumed that $$\widetilde{\mathbb{P}}$$ is a probability measure, we have that: $$\widetilde{\mathbb{P}}(\Omega)=1=\int_{\Omega}Z(\omega)d\mathbb{P}(\omega)=\mathbb{E}_{\mathbb{P}}[Z].$$ Since we now know that $$\mathbb{E}_{\mathbb{P}}[Z]=1$$, we can apply (Steve Shreve, Stochastic Calculus for Finance II - Continuous Models, p. 33, Theorem 1.6.1) to reach the conclusion that for any wandom variable $$X$$ that is a non-negative and $$\mathcal{F}_T$$-measurable we have: $$\mathbb{E}_{\widetilde{\mathbb{P}}}[X]=\mathbb{E}_{\mathbb{P}}[ZX].$$ In particular, for $$X=1$$ this leads to: $$\mathbb{E}_{\mathbb{P}}[Z]=1.$$ Let us define $$L_t=\mathbb{E}_{\mathbb{P}}[Z|\mathcal{F}_t]$$. Clearly, $$(L_t)_{t\geq 0}$$ is a $$(\mathbb{P}, \mathbb{F})$$-martingale because for all $$s\leq t$$: $$\mathbb{E}_{\mathbb{P}}[L_t|\mathcal{F}_s]=\mathbb{E}_{\mathbb{P}}[\mathbb{E}_{\mathbb{P}}[Z|\mathcal{F}_t]|\mathcal{F}_s]= \mathbb{E}_{\mathbb{P}}[L_t|\mathcal{F}_s]=L_s,$$ where the first equality is from the definition of $$L_t$$, the second inequality is due to the tower law, and the third equality is due to the definition of $$L_s$$. Taking expectation in the above we get the property that $$\mathbb{E}_{\mathbb{P}}[L_t]=1, (\forall)t\leq T$$. If we take $$\mathcal{F}_0=\{\emptyset, \Omega\}$$, as is usual, then $$L_0$$ is deterministic and $$L_0=1$$. This proves items (2.) and (3.) of the proposition.

We can then use (Steve Shreve, Stochastic Calculus for Finance II - Continuous Models, p. 211, Lemma 5.2.1) to prove item (1.) of the proposition, namely that: $$\mathbb{E}_{\widetilde{\mathbb{P}}}[X]=\mathbb{E}_{\mathbb{P}}[L_tX],\text{ for all } \mathcal{F}_t\text{-measurable, non-negative, random variables }X,\text{ when }t\leq T.$$ In the above, let us substitute $$1_A$$ for $$X$$ and T for t. This proves immediately the rest of the proposition. Note that from this answer, $$L_t$$ is $$\mathbb{P}$$-a.s. non-negative.

Also note that we can take $$Z$$ to be strictly positive since the two measures are equivalent. Therefore, we can also take a version of $$L_t$$ that is strictly positive and this changes nothing. We will consider in what follows that we use such $$L_t$$.$$\Box$$

We have constructed above the Radom-Nikodym derivative process $$(L_t)_{t\geq 0}$$ , which is a $$(\mathbb{P}, \mathbb{F})$$-martingale. Because $$\mathbb{F}$$ is generated by $$(W_t)_{t\in[0;T]}$$ we can apply the martingale representation theorem $$\Rightarrow (\exists) (\psi_t)_{t\geq 0}$$ an $$\mathbb{F}$$-measurable process s.t.: $$L_t=1+\int_0^t \psi_udW_u.$$ or, alternatively, that: $$dL_t=\psi_tdW_t, L_0=1.$$ Since the Radon-Nikodym derivarive process is strictly positive, using Ito lemma we get: $$d\log(L_t)=\frac{1}{L_t}dL_t-\frac{1}{2}\frac{1}{L^2_t}d\langle L \rangle_t=\frac{\psi_t}{L_t}dW_t-\frac{1}{2}\frac{\psi^2_t}{L^2_t}dt$$ Since $$L_t$$ is strictly positive, we can simplify things a bit by introducing $$\Theta_t=-\frac{\psi_t}{L_t}.$$ This is also an $$\mathbb{F}$$-adapted process. With this notation, by integrating the result of the application of the Ito lemma and exponentiating we get: $$L_t=e^{-\int_0^t\Theta_udu-\frac{1}{2}\int_0^t\Theta^2_u dWu}.$$

The result also rests on the uniqueness (up to indistinguishability) of the Radon-Nikodym derivative (in the R-N theorem).

So yes, all changes of measure must be of this form.

• Welcome to Quant Stackexchange! – noob2 Apr 13 '20 at 19:33
• noob2, Thank you! – fwd_T Apr 13 '20 at 21:01
• +1 It might help to simply write dlog(L_t)=psi_t/L_tdW_t - 1/21/L_t^2psi_t^2dt right under your dlog(L_t) equation so its crystal clear where Theta_t is really coming from. – ir7 Apr 16 '20 at 1:33
• @ir7 I edited the answer to show the intermediary step. – fwd_T Apr 16 '20 at 17:20