I have trouble understanding Radon-Nikodym theorem. Here's an example.
Consider the measurable space $(\Omega,F)$, where $\Omega = R$, $F~$ is the $\sigma$-algebra of Borel sets. Let $P[dw] = \frac{1}{\sqrt(2\pi)}e^{-w^2/2}dw$ and $\tilde{P}[dw] = e^{-w}1_{[0,\infty)(w)}dw$ be two probability measures on $\omega$. How to test whether $\tilde{P} << P$ and $P << \tilde{P}$. How do we compute Radon-Nikodym derivative $\frac{d\tilde{P}}{dP}$?
Best Answer
$\tilde P << P$ is true but $P << \tilde P$ is false. We have $$\frac{d\tilde P}{dP}(w)=\sqrt{2\pi}\frac{e^{-w} 1_{[0,\infty)}}{e^{-w^2/2}}$$ Proving these facts is an exercise that you should really do on your own, as it is not that hard and you would not learn anything if you were just told the proof.