Timeline for is the variance of a test function of a markov chain always increasing?
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Mar 12, 2011 at 18:02 | comment | added | John Jiang | sorry about that. I will start a new post. | |
| Mar 12, 2011 at 16:47 | comment | added | camomille | Ahem. The new question is actually very different from the initial one. | |
| Mar 12, 2011 at 16:44 | comment | added | camomille | This is not related to the question. Moreover, the assumption of reversibility is useless. The exponential convergence will hold for any finite state irreductible aperiodic Markov Chain. In addition, the aperiodicity is useless to get the existence and unicity of a stationary distribution... | |
| Mar 12, 2011 at 16:39 | comment | added | John Jiang | @Andre: remember the variance is defined by $\text{var} f = \mathbb{E}_t(f^2) - \mathbb{E}_t(f)^2$ so you also had to subtract off the mean, and the mean is taken not with respect to the stationary measure, but rather the running measure at time $t$. If the mean is the stationary measure mean, then what you wrote is correct. But I really need the variance at time t without apriori knowledge of the stationary distribution. | |
| Mar 12, 2011 at 15:49 | history | edited | André Schlichting | CC BY-SA 2.5 |
typos
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| Mar 12, 2011 at 15:33 | history | answered | André Schlichting | CC BY-SA 2.5 |