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We begin with example. For the Poisson process with an intensity $\lambda_1$ there is an equivalent change of measure which makes it intensity to $\lambda_2$. I would like to find the conditions when is it possible to do with a homogeneous Markov process in a continuous time? It seems to be true for the Markov chains, but it's not true for piecewise deterministic Markov processes (because of pure deterministic part between jumps).

Maybe we can consider at first the diffusion case. Let $$ dX_t = \mu(X_t)dt+\sigma(X_t)dw_t. $$

Is there a change of measure $P\to Q$ such that under the new measure $$ dX_t = a\mu(X_t)dt+\sigma(X_t)\sqrt{a}\cdot dw_t $$ where $a>0$?

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Why are the probability distributions of Poisson processes with different (constant) intensities equivalent? –  Did Jan 5 '11 at 16:59
    
See e.g. Shreve, Stochastic Calculus for Finance II, chapter 11 - change of measure for the Poisson process. There he provide an equivalent change of measure which changes an intensity of the Poisson process. –  Ilya Jan 5 '11 at 18:25
    
On a bounded time interval, sure (and the Radon-Nykodym density is straightforward). But for Poisson processes in the usual sense, i.e. indexed by the whole time interval $[0,+\infty[$, the probability distributions are mutually singular. This observation is similar to a remark Alekk made about the "diffusion" part of the former version of your question. –  Did Jan 5 '11 at 22:54
    
(cont'd) Consider $X_t=w_t$ and $Y_t=\sqrt{a}w_t$ for $t$ in $[0,T]$. To see that, even in this most simple case, the probability distributions are mutually singular, see Alekk's answer to your previous question mathoverflow.net/questions/51090. –  Did Jan 5 '11 at 23:14
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@Ilya, most of these measure changes are defined on a finite time interval, I think that was the point of Did. The same is true with the Girsanov theorem though. –  Grzenio Oct 18 '13 at 14:58
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2 Answers

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To address a question asked by the OP in a comment: yes, the LLN yields almost sure results (no expectations here) which show that there exist disjoint sets of complete trajectories such that each Poisson distribution "sees" only one of them (namely $Q_\lambda$ "sees" only the set $D_\lambda$ and $Q_\lambda(D_\mu)=0$ if $\lambda\ne\mu$).

About diffusions with different drifts: as you know, if $(W_t)_{t\ge0}$ is a Brownian motion under $P$ and if $X_t=W_t+\mu t$, then for every $T$, $(X_t)_{0\le t\le T}$ is a Brownian motion (without drift) under a measure $Q_T$ on the space of paths indexed by $[0,T]$. Furthermore, $Q_T$ is absolutely continuous with respect to the restriction $P_T$ of $P$ to the space of paths indexed by $[0,T]$ and the density $dQ_T/dP_T$ is $Z_T=\exp(-\mu W_T-\frac12\mu^2T)$ (this is Cameron-Martin-Girsanov theorem). Thus there exists what you call an equivalent change of measure between $P_T$ and $Q_T$.

But this does not imply that $P$ and $Q$ are mutually absolutely continuous and in fact, they are mutually singular. Note that when $T\to+\infty$, the $P$-martingale $(Z_T)$ converges almost surely when $T\to+\infty$ to $Z_\infty=0$... which is not a Radon-Nykodym density. A more direct proof that $P$ and $Q$ are mutually singular is as follows: for every $\mu$, let $L_\mu$ denote the set of continuous functions $f$ defined on $[0,+\infty[$ such that $f(t)/t\to\mu$ when $t\to+\infty$. If $\mu\ne0$, then $L_0$ and $L_\mu$ are disjoint and $P(X\in L_\mu)=Q(X\in L_0)=1$.

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You mean that we can make an equivalent change of measure only on a finite horizon? Ok, it will be enough for me. Unfortunately as I understand the change of the diffusion term is impossible even on the finite horizon because it will change a quadratic variation which is a local characteristic of a process. –  Ilya Jan 8 '11 at 14:57
    
Took me quite some time to get it. Thanks (a bit too late, though) anyway! –  Ilya Jan 17 '13 at 9:35
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(This is to answer a question about Poisson processes asked by the OP in the comments.)

Here is an analog of Alekk's argument about the mutual singularity of the probability distributions of diffusions on bounded intervals, but for Poisson processes on $[0,+\infty[$. For every positive $\lambda$, call $Q_\lambda$ the probability distribution of the homogeneous Poisson process with intensity $\lambda$ on $[0,+\infty[$, and $D_\lambda$ the set of locally finite subsets $S$ of $[0,+\infty[$ with asymptotic density $\lambda$: these are the sets $S$ such that $x^{-1}\#(S\cap[0,x])\to\lambda$ when $x\to+\infty$.

Then, if $\lambda\ne\mu$, sets $D_\lambda$ and $D_\mu$ are disjoint by definition and $Q_\lambda(D_\lambda)=Q_\mu(D_\mu)=1$ by the law of large numbers. Hence $Q_\lambda$ and $Q_\mu$ are mutually singular.

This is in contrast to Poisson processes of finite total intensity. Let $I$ denote any measurable subset of $[0,+\infty[$ of finite Lebesgue measure $|I|$, and $Q^I_\lambda$ the probability distribution of the homogenous Poisson process with intensity $\lambda$ on $I$. Then all the measures $Q^I_\lambda$ are mutually absolutely continuous and the density of $Q^I_\lambda$ with respect to $Q^I_\mu$ at a finite subset $S$ of $I$ is $\mathrm{e}^{(\mu-\lambda)|I|}(\lambda/\mu)^{\#S}$.

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Do I understand correct, roughly speaking calculating expectations with the Law of Large Numbers will give us different results P-a.s.? But what about diffusion with drift? Can't we use the same argument to prove that on the interval $[0,+\infty)$ there is no equivalent cahnge of measure? –  Ilya Jan 6 '11 at 15:31
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