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I'll formulate the question via an example.

On $( C[0,1], \mathcal{C} )$, where $C[0,1]$ is the set of continuous functions on $[0,1]$ and $\mathcal{C}$ the Borel $\sigma$-algebra given by uniform topology, consider Wiener measure $\mathbb{P}$. Denote by $t \mapsto W_t$ the Brownian paths given by $\mathbb{P}$.

Let $X$ be specified via $$ dX = \mu(X) dt + dW, $$ where $\mu: \mathbb{R} \rightarrow \mathbb{R}$ is fixed.

Girsanov's theorem says, under the measure $\mathbb{Q}$ given by (assume, e.g., Novikov's condition holds) $$ \frac{d \mathbb{Q} }{ d\mathbb{P} } = e^{ \int_0^1 \mu(X) dW - \frac{1}{2} \int_0^1 \mu(X)^2 dt }, $$ the $\mathbb{Q}$-law of $W$ is the $\mathbb{P}$-law of $X$.

Question

The Radon-Nikodym derivative $\frac{d \mathbb{Q} }{ d\mathbb{P} }$ is specified via stochastic integral. But in principle, a Radon-Nikodym derivative is an object one must be able to define $\omega$-by-$\omega$, $\mathbb{P}$-almost surely in this case. So what is $\frac{d \mathbb{Q} }{ d\mathbb{P} }$ as a functional on $C[0,1]$ (strictly speaking on the support of $\mathbb{P}$)?

Conjecture

Ignore that Brownian paths do not have finite variation, etc. Formally, it is the functional $\phi : C[0,1] \rightarrow \mathbb{R}$ given by $$ f(\cdot) \stackrel{\phi}{\mapsto} e^{\int_0^1 \mu(x(t)) df(t) - \frac{1}{2} \int_0^1 \mu^2(x(t)) dt } $$ where $dx = \mu(x) dt + df$, and $\int_0^1 \mu(x(t)) df(t)$ is a Riemann-Stieltjes integral with respect to $df$. In other words, given $f \in C[0,1]$, one acts as if $f$ is a realization of Brownian path and substitute formally into the expression for $\frac{d \mathbb{Q} }{ d\mathbb{P} }$. Is this correct in some sense---e.g. discretize into step functions and taking weak limit in the Skorohod space $D[0,1]$...?

Suggestive Example (or not)

Suppose $dX = a \, dt + dW$ where $a$ is a real number. Then, $\omega$-by-$\omega$, the Radon-Nikodym derivative $$ \frac{d \mathbb{Q} }{ d\mathbb{P} } = e^{ \int_0^1 a dW - \frac{1}{2} \int_0^1 a^2 dt }, $$ is given by the functional $$ f(\cdot) \stackrel{\phi}{\mapsto} e^{ a \int_0^1 df(t) - \frac{1}{2} \int_0^1 a^2 dt } $$ where $\int_0^1 df(t)$ is interpreted as $f(1) - f(0)$.

(First asked on Math SE.)

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  • $\begingroup$ First asked on Math SE. $\endgroup$ – Michael Aug 9 at 7:53
  • $\begingroup$ When you crosspost from Math.SE, please include a link to the original version, and also update your Math.SE post with a link to the MO post. $\endgroup$ – Nate Eldredge Aug 9 at 15:11
  • $\begingroup$ How can you "ignore that Brownian paths do not have finite variation"? If, say, $\mu(x) = x$, then how is $\phi$ defined? $\endgroup$ – Mateusz Kwaśnicki Aug 9 at 19:43

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