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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 $\widetilde{W}_t=W_t+\int_0^t\Theta_udu$?

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  • $\begingroup$ In order to apply the Martingale Representation theorem, the filtration has to be the one generated by the Brownian motion (cf. Shreve). $\endgroup$ – SN76 Apr 13 at 22:26
  • $\begingroup$ 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 $W$ $\endgroup$ – Gabe Apr 14 at 8:49
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The answer is yes.

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.

| improve this answer | |
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  • $\begingroup$ Welcome to Quant Stackexchange! $\endgroup$ – noob2 Apr 13 at 19:33
  • $\begingroup$ noob2, Thank you! $\endgroup$ – Gabe Apr 13 at 21:01
  • $\begingroup$ +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. $\endgroup$ – ir7 Apr 16 at 1:33
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    $\begingroup$ @ir7 I edited the answer to show the intermediary step. $\endgroup$ – Gabe Apr 16 at 17:20

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