2
$\begingroup$

I am looking for an MCMC algorithm that leaves a target $\pi$ invariant but that overestimates the mode.

Basically I am looking for an algorithm that whose transition kernel leaves $\pi$ stationary (can be either reversible or non-reversible) but that mixes poorly in the sense that it tends to spend too much time on the mode rather than the tails.

Do you know of any such method? Intuitively I was thinking that some gradient-based algorithm could do the job. However HMC mixes very well of course. I know this is an odd question!

Improve this question
$\endgroup$
8
  • 2
    $\begingroup$ Why you need it? Context would be helpful. $\endgroup$
    – Tim ♦
    May 21, 2021 at 12:57
  • 1
    $\begingroup$ Can you clarify exactly what you want? By definition, if the algorithm spends "too much time in the mode" it won't have the posterior as its stationary distribution. It's easy to come up with settings where you end up in the mode for an extremely long time, then move to the tail and get stuck there for an extremely long time, but the fact that the posterior is stationary means that you get stuck in each part for exactly the correct amount of time in the long run. $\endgroup$
    – guy
    May 21, 2021 at 15:35
  • 1
    $\begingroup$ For instance could it be that a MALA algorithm with a quite large step size will tend to overestimate modes in a finite (and not too large) number of iterations? My intuition tells me that gradient-based methods might consistently overestimate the mode in the short-run $\endgroup$
    – Physics_Student
    May 21, 2021 at 19:11
  • 1
    $\begingroup$ My point is that most natural algorithms which "overestimates the mode" in the short-run will also "underestimate the mode" it in the short-run as well, depending on where you start the chain. That must be the case, because otherwise the stationary distribution would have to be wrong. So you can't design an algorithm which overshoots on the mode by design without introducing the exact same behavior in the tails. Composing two chains with the same flaw probably won't give any progress on your problem. $\endgroup$
    – guy
    May 21, 2021 at 20:45
  • 1
    $\begingroup$ A MALA is attracted by the mode due to its gradient component but unless there are several modes that are far away from each other and the energy of the MALA is low, the chain will eventually explore all modes. $\endgroup$
    – Xi'an
    May 22, 2021 at 5:37

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.