Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

We have a random walk process $X_{k+1} = X_{k} + Y_{k}$, where $Y_k$ are independent random values with the same distribution, $X_{0}$ have some fixed distribution $\pi$. Let $P^{k,s}(x,B)$ be a transition kernel from $X_{k}$ to $X_{s}$. A measure $\mu$ is called invariant with respect to the transition kernel, if $\mu P^{k,s} = \mu$, where $$ \mu P^{k,s} (B) = \int P^{k,s}(x,B) \mu(dx) $$ Here $P^{k,s} = P^{0,s-k} \equiv P^{s-k}$ and I received that $P^1(x,B) = Q(B-x)$. Furthermore, $$ \pi P^{1} (B) = \int Q(B-x) \pi(dx) = \int \pi(B-x) Q(dx) $$ So we can see that if $\pi$ is rotation invariant, then $\pi$ is invariant measure ($\pi P^1 = \pi$). But Lebesgue measure is not probability measure. So how can I show that untrivial probability invariant measures dont exist?

share|improve this question
    
Well, what about a case when $Y_k = 0$ a.s.? If there are any nontrivial invariant measures? –  Ilya Jan 12 '12 at 13:00
    
It is trivial case. So, if our random walk takes place in compact space (such as $S^2$), then uniform distribution will be invariant. But what in the case of noncompact spaces (but localy compact)? I think we can use that there is only one shift-invariant measure (Haar measure) but it is unbounded, so it can't be probability measure. –  Nimza Jan 13 '12 at 15:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.