Radon-Nikodym derivative of pushforward measures and Girsanov theorem

Let $$\mu$$ and $$\nu$$ be two measures on a measure space $$(\Omega, \Sigma)$$, and $$\mu$$ is absolute continuous w.r.t. $$\nu$$. Also let $$X\colon \Omega \to H$$ be a measurable functions mapping to another measure space $$(H, Z)$$.

Suppose that we known the Radon-Nikodym derivative

$$\omega \mapsto \frac{d \mu}{d \nu}(\omega).$$

How to then find the Radon-Nikodym derivative of their pushforward measures? That is,

$$\frac{d \mu_X}{d \nu_X}\colon H \to [0, \infty),$$

where $$\mu_X$$ and $$\nu_Y$$ are the pushforward measures of $$\mu$$ and $$\nu$$ of the functions $$X$$, respectively.

A concrete example is given as follows. Let $$B(t, \omega)$$ be a Brownian motion under measure $$\nu$$. If

$$Z_T(\omega) = \exp\bigg(\int^T_0 a(B(s, \omega)) dB(s, \omega) - \frac{1}{2}\int^T_0 a^2(B(s, \omega)) ds\bigg)$$

satisfies certain conditions, then we can define the Radon-Nikodym derivative

$$\frac{d \mu}{d \nu}(\omega) := Z_T(\omega),$$

so that the process $$B$$ under the measure $$\mu$$ created from this derivative is now a weak solution to the SDE $$d X(t) = a(X(t)) dt + d\overline{B}(t)$$, where $$\overline{B}$$ is another Brownian motion under measure $$\mu$$.

This is the famous Girsanov theorem applied on SDEs.

The purpose that I want to obtain $$\frac{d \mu_B}{d \nu_B}$$ is to get the (finite-dimensional) distribution $$\mu_B$$ of $$B$$, since the distribution of Brownian motion $$\nu_B$$ is easy to compute.

I am purely guessing (by change of variable formula), is it true that

$$\frac{d \mu_B}{d \nu_B}(f) = \exp\bigg(\int^T_0 a(f(s)) df(s) - \frac{1}{2}\int^T_0 a^2(f(s)) ds\bigg)?$$

• Why would $\mu_X$ be absolutely continuous w.r.t. $\nu_Y$? Consider constant $X$, $Y$, for example.
– m7e
Apr 29, 2022 at 18:58
• @m7e thanks for pointing this out. That's a mistake, now corrected.
– null
May 1, 2022 at 14:13
• I don't know about the general case, but in your example looks correct, since we can work on the Wiener space. May 2, 2022 at 8:40
• @Chaos would you like to elaborate?
– null
May 2, 2022 at 9:12