Probability density under new measure Let $P$ and $Q$ be two equivalent probability measures, and let $dP=ZdQ$ for some $Z$.
If I know the probability density of a random variable $X$ under measure $P$, how can I calulate the density of $X$ under measure $Q$? In other words, if I know that
$$P(X\le x)=\int\limits_{-\infty}^x f(s)\, ds,$$
for all real $x$, is it possible to find a function $g$ such that
$$Q(X\le x)=\int\limits_{-\infty}^x g(s)\, ds.$$
 A: The definition of the Radon-Nykodym derivative $\dfrac{\mathrm dQ}{\mathrm dP}=U$  (your $Z$ being $Z=1/U$) is that, for every event $A$, one should have $Q(A)=\mathbb E_P(U;A)$. 
When $A=[X\leqslant x]$, this is $Q(X\leqslant x)=\mathbb E_P(U;X\leqslant x)$ hence one looks for some density function $g$ such that, for every $x$, 
$$
\mathbb E_P(U;X\leqslant x)=\displaystyle\int_{-\infty}^xg(t)\mathrm dt. 
$$
In the special case when $U=\varphi(X)$, calling $f_X$ the density of $X$, one gets
$$
Q(X\leqslant x)=\mathbb E_P(\varphi(X);X\leqslant x)=\displaystyle\int_{-\infty}^x\varphi(t)f_X(t)\mathrm dt,
$$ 
hence the function $g=\varphi f_X$ is a density (nonnegative, integrates to $1$) and solves your question.
In the general case, calling $f_{U,X}$ the density of the joint distribution of $(U,X)$, the solution is
$$
g(x)=\int_{0}^{+\infty} uf_{U,X}(u,x)\mathrm du.
$$
This specializes to $g=\varphi f_X$ when $U=\varphi(X)$ and, at the other extreme, to $g=f_X$ when $U$ and $X$ are independent. However, these cases are the exception rather than the rule since, in general, $g$ does not involve $f_X$ only but the joint distribution of $(U,X)$.
