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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

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

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

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