Equilibration vs statistics accumulation steps in DMC
Posted: Sun May 31, 2015 9:12 am
Hi, all.
CASINO suggests that the DMC calculation is as follows:
We do n1 - equilibration steps, and then n2 - statistics accumulation steps.
So OK, let's assume that DMC energy - E (n) depends on step number - n as follows:
E(n) = E(infinity) * (1 + exp(-alpha*n)) + Noise(n)
where:
(1 + exp(-alpha*n)) - describes equilibration process
and
Noise(n) - Gaussian noise
Mean(Noise(n)) = 0
Var(Noise(n)) is independent of n.
After n1 equilibration steps and n2 statistics accumulation steps we've got:
E = SUM[E(n)]<from n1 to n2>/(n2-n1)
average over the ensemble of realizations:
Mean(E) = E(infinity) + SUM[exp(-alpha*n)]<from n1 to n2>/(n2-n1) + Mean(Noise(n))
recall that Mean(Noise(n)) = 0
we've got:
Mean(E) = E(infinity) + SUM[exp(-alpha*n)]<from n1 to n2>/(n2-n1)
that is, we always have a biased estimate of E(infinity) with:
bias(n1, n2) = SUM[exp(-alpha*n)]<from n1 to n2>/(n2-n1)
The question is how should relate n1 and n2, to satisfy:
bias(n1, n2) < Variance(Noise(n))/sqrt(n2-n1)
I'm not sure that the noise is Gaussian and equilibration is exponential, but who knows maybe n1/n2 have to be constant?
Vladimir
CASINO suggests that the DMC calculation is as follows:
We do n1 - equilibration steps, and then n2 - statistics accumulation steps.
So OK, let's assume that DMC energy - E (n) depends on step number - n as follows:
E(n) = E(infinity) * (1 + exp(-alpha*n)) + Noise(n)
where:
(1 + exp(-alpha*n)) - describes equilibration process
and
Noise(n) - Gaussian noise
Mean(Noise(n)) = 0
Var(Noise(n)) is independent of n.
After n1 equilibration steps and n2 statistics accumulation steps we've got:
E = SUM[E(n)]<from n1 to n2>/(n2-n1)
average over the ensemble of realizations:
Mean(E) = E(infinity) + SUM[exp(-alpha*n)]<from n1 to n2>/(n2-n1) + Mean(Noise(n))
recall that Mean(Noise(n)) = 0
we've got:
Mean(E) = E(infinity) + SUM[exp(-alpha*n)]<from n1 to n2>/(n2-n1)
that is, we always have a biased estimate of E(infinity) with:
bias(n1, n2) = SUM[exp(-alpha*n)]<from n1 to n2>/(n2-n1)
The question is how should relate n1 and n2, to satisfy:
bias(n1, n2) < Variance(Noise(n))/sqrt(n2-n1)
I'm not sure that the noise is Gaussian and equilibration is exponential, but who knows maybe n1/n2 have to be constant?
Vladimir