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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 $\frac{d\tilde{P}}{dP}$?

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Like another question you have asked, this one has no context or motivation. Frankly, it appears to be simply a homework question. What have you tried? What background are you assuming? –  Carl Mummert Oct 24 '10 at 23:34
    
I got this question from the internet by searching contents on Radon-Nikodym theorem... –  user957 Oct 25 '10 at 0:18
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1 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.

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