Timeline for is the variance of a test function of a markov chain always increasing?
Current License: CC BY-SA 2.5
9 events
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| Mar 12, 2011 at 16:46 | history | edited | John Jiang | CC BY-SA 2.5 |
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| Mar 12, 2011 at 16:45 | comment | added | John Jiang | @James: I agree that even if starting at deterministic initial state, counterexample exists if $f$ is not required to be an eigenfunction. But I don't quite understand your last 2-state example. The starting variance is always 0 right? | |
| Mar 12, 2011 at 16:39 | history | edited | John Jiang | CC BY-SA 2.5 |
added 6 characters in body; added 84 characters in body
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| Mar 12, 2011 at 16:34 | history | edited | John Jiang | CC BY-SA 2.5 |
Change the function $f$ from arbitrary to eigenfunction, specified that the chain starts at a single state, restricted to finite state ergodic markov chains
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| Mar 12, 2011 at 15:33 | answer | added | André Schlichting | timeline score: 1 | |
| Mar 12, 2011 at 14:40 | history | edited | Nate Eldredge | CC BY-SA 2.5 |
Fill in mathbb's
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| Mar 12, 2011 at 13:07 | comment | added | camomille | Take any finite chain with an absorbing state and where all other states are transient. Then the variance converges to $0$. | |
| Mar 12, 2011 at 8:39 | comment | added | James Martin | Not sure I understand the question. What do you take for the initial distribution? If the chain is stationary, the quantity doesn't depend on $t$. On the other hand, if you allow a general initial distribution, just start in some distribution where the variance of the test function is higher than it is under the stationary distribution. Even if you insist on a deterministic initial state, pretty much any 2-state chain will give a counterexample - take a function whose variance is not maximised by the stationary distribution, and consider approaching stationarity from one extreme or the other. | |
| Mar 12, 2011 at 7:43 | history | asked | John Jiang | CC BY-SA 2.5 |