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I have some trouble understanding changes of measure for jump processes. I guess I'm missing some important bit of the theory.

Consider a simple example. Let $N_t$ be a standard Poisson process with constant intensity $\lambda_t=1$ adopted to filtration $(\mathcal F_t)_{t\geq 0}$. We wish to change the intensity to $\mu_t=X_{t-}$ by performing a change of measure with Radon-Nikodym derivative $$ \left.\frac{d\tilde P}{dP}\right|_{\mathcal F_t} = L_t = \prod_{n\geq 1} \mu_{T_n}1_{T_n \leq t} \exp(- \int_0^t (1-\mu_s)\lambda_s ds ) $$

This yields $$ L_t= \prod_{n\geq1}^{N_t} X_{T_n-} \exp( \int_0^t X_{s-}ds - t). $$ This is an equivalent change of measure if $\mathbb E L_1 = 1$. My question is: is this an equivalent change of measure? And if yes, what is wrong with the following argumentation:

By changing measure we alter the intensity of the process $N_t$. We pass from constant $\lambda_t=1$ to non-constant $\mu_t=X_{t-}$, and clearly $P( \lambda_. >1)=0$ but $\tilde P( \mu_.>1)>0$ if we assume $X_0>0$.

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what is the process $X$ here? Moreover, for any finite $t$ it holds that $\{\lambda >1\}\notin \mathcal F_t$ and $\{\mu >1\}\notin \mathcal F_t$ i.e. these events are not measurable and hence the argumentation you've provided does not apply. –  Ilya Nov 13 '11 at 19:06
    
some more comments: what are $T_n$? I failed to understand when $1_{T_n\geq 1}(t) = 1$ and when is it $0$. You can also write $\int\limits X_{s-}\mathrm ds = \int\limits X_{s}\mathrm ds$. Btw, where did you get the formula for $L_t$ from? –  Ilya Nov 13 '11 at 19:16
    
The reference for this is P. Bremaud: Point Processes and Queues. Oh .. $X$ is actually $N$ under $P$ and the process with intensity $\mu$ under $\tilde P$. $T_n$ are jump times of $N_t$. And there is a typo in the first formula. It should be $1_{T_n \leq t}$. –  Julian Wergieluk Nov 13 '11 at 19:31
    
@Gortaur: In the last line I mean $P(\lambda_t>1)=0$ for some $t$. Or, alternatively, we could simply consider the probability that the intensity is not constant at some point. –  Julian Wergieluk Nov 13 '11 at 19:41
    
This event is still unobservable: in other words you cannot make any conclusion about intensity based on the observation of a single finite-time trajectory –  Ilya Nov 13 '11 at 19:50
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