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Sep 28, 2015 at 16:37 comment added Did "Can you outline the steps to show that the chaotic map generates r.v which forms markov chain?" I already did. Perhaps you are missing the point but every deterministic sequence $(x_n)$ such that $x_{n+1}=M(x_n)$ for some function $M$ is a (very degenerate) Markov chain. Proof: Look up the definitions.
Sep 28, 2015 at 0:11 comment added Ria George How to show that K(x) is a dirac mass? Is it needed to show this in order to prove that the sequence output from the chaotic map is a Markov chain? Can you outline the steps to show that the chaotic map generates r.v which forms markov chain?
Sep 27, 2015 at 21:21 comment added Did "can you provide links where I can find more information?" More information about what? "how to show what is a Dirac mass" One does not "show what is a Dirac mass", either one reads the definition to know what is a Dirac mass or one shows that something is a Dirac mass. For your information, the Dirac mass at $y$ is the unique measure $\mu$ such that $\mu(A)=1$ for every $A$ such that $y\in A$ and $\mu(A)=0$ for every $A$ such that $y\notin A$.
Sep 27, 2015 at 21:13 comment added Ria George Thank you for your answer, can you provide links where I can find more information? I am not aware of this proof and how to show what is a Dirac mass and hence the proof.
Sep 26, 2015 at 15:01 comment added Did Indeed the sequence is a Markov chain, but a very special one since its transition kernel $$K(x,A):=P(x_{n+1}\in A\mid x_n=x,x_{n-1},\ldots,x_0)$$ is such that, for every $x$, $K(x,\ )$ is a Dirac mass (at $M(x)$). Not your typical Markov chain, but a Markov chain yes.
Sep 19, 2015 at 16:49 history edited Ria George CC BY-SA 3.0
edited title
Sep 19, 2015 at 16:48 vote accept Ria George
S Sep 18, 2015 at 9:41 history suggested Dawny33 CC BY-SA 3.0
Fixed hyperlinks
Sep 18, 2015 at 9:21 answer added RUser4512 timeline score: 3
Sep 18, 2015 at 8:08 review Suggested edits
S Sep 18, 2015 at 9:41
Sep 18, 2015 at 7:29 history asked Ria George CC BY-SA 3.0