I try to read paper, MCMC using Hamiltonian dynamics). The author, Neal states(P28):

To begin, Cruetz nodes that the following relationship holds when any Metropolis-style algorithm is used to sample a density $P(x) = (1/Z)\exp(-E(x))$:

$$ 1 = \mathbb{E}(P(x^*)/P(x)) = \mathbb{E}(\exp(-(E(x^*)-E(x)))) = \mathbb{E}(\exp(-\Delta)) \quad (4.17) $$

I can't see why $\mathbb{E}(P(x^*)/P(x))=1$ so I follow the reference, Global Monte Carlo algorithms for many-fermion systems, and found this strange substitution(P3):

Some rather useful identities follow from considering expectation values over this distributions. Consider the paritition function

$$ Z = \int dA' dp' e^{-H'} = \int dAdp e^{-H}e^{H-H'}, \quad (16) $$

where $H$ and $H'$ denote $H(p,A)$ and $H(p',A')$, respectively.

I can't see why it is valid. $A = A', p = p'$ may works but it will imply $\mathbb{E}(P(x^*)/P(x))=\mathbb{E}(1)=1$? It sounds not the point Neal want to present.

  • 1
    $\begingroup$ Are you using E both to denote expectation and the energy? $\endgroup$ Sep 11, 2019 at 7:43
  • $\begingroup$ Yes, I have fixed the latex code. $\endgroup$
    – user137795
    Sep 11, 2019 at 7:52
  • $\begingroup$ If not exactly a duplicate, quite correlated with this question of a few weeks ago. $\endgroup$
    – Xi'an
    Sep 12, 2019 at 4:48

1 Answer 1


If $K$ is a Markov kernel(with density $k$) with stationary distribution $P$ (with density $p$), then, if $(X_t)_t$ is a stationary Markov chain associated with $K$, \begin{align*}\mathbb E\left[\frac{p(X_{t+1})}{p(X_t)}\right] &=\int_{\mathfrak{X^2}} \frac{p(x_{t+1})}{p(x_t)} p(x_t)k(x_t,x_{t+1})\text{d}\lambda(x_t)\text{d}\lambda(x_{t+1}) \\ &= \int_{\mathfrak{X^2}} p(x_{t+1}) k(x_t,x_{t+1})\text{d}\lambda(x_t)\text{d}\lambda(x_{t+1})\\ \end{align*} Unless $k$ is symmetric and with a support at least equal to $\mathfrak X$, the support of $p$, there is no reason for the ratio to be of expectation one.


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