Timeline for Comparing mixing time of lazy and non-lazy Markov chains
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 28, 2019 at 18:08 | vote | accept | Josh R | ||
| Apr 28, 2019 at 18:04 | comment | added | Josh R | Thanks for the comments, you're quite right. I should have specified the question more carefully. | |
| Apr 28, 2019 at 7:33 | comment | added | Algernon | @RW: You are right, I was too hasty. And of course it makes sense: if we remove $p$ from $Q(x,x)$ and remove $q<p$ from $Q(y,y)$, we are favoring $y$ over $x$, which means in the long run the chain will spend more time in $y$, hence a bias towards $y$ in the stationary distribution. | |
| Apr 28, 2019 at 0:39 | comment | added | R W | @Algernon - This is precisely what is false. The new chain has the transition probabilities $\tilde Q(x,y)=Q(x,y)/(1-Q(x,x))$, and it is reversible with respect to $\pi$ if and only if $Q(x,x)$ is the same for all $x\in X$. | |
| Apr 27, 2019 at 22:33 | answer | added | Algernon | timeline score: 1 | |
| Apr 27, 2019 at 20:25 | comment | added | Algernon | @JoshR: with your definition, for $t>t_x(Q,\varepsilon)$, the distribution $Q^t(x,\cdot)$ may again be far from $\pi$, which I guess is not what you want. Note that $\|Q^t(x,\cdot)-\pi\|$ is not monotonic. | |
| Apr 27, 2019 at 19:29 | comment | added | Algernon | @RW: As the OP pointed out, if $\pi$ is reversible for the original kernel, it is also reversible for the "non-lazy" version. | |
| Apr 26, 2019 at 22:05 | comment | added | R W | Why do you claim that the stationary distributions are the same? They are not, generally speaking! | |
| Apr 26, 2019 at 17:30 | review | First posts | |||
| Apr 26, 2019 at 18:22 | |||||
| Apr 26, 2019 at 17:28 | history | asked | Josh R | CC BY-SA 4.0 |