For an Ising Model with a (2L+ 1) by (2L+ 1) square grid of magnetic particles, show that $$\pi(\xi)=\frac{1}{Z_\beta}e^{\beta\sum_{x,y=x}{\xi_x\xi_y}}$$ Is indeed a stationary distribution for the Metropolis-Hasting process. (Here $\beta > 0$ is a constant, and $Z_\beta > 0$ is a too-hard-to-compute constant that makes $π$ an actual distribution.) Recall the transition matrix for the process from Metropolis-Hasting is $$p(\xi,\xi')=q(\xi,\xi')r(\xi,\xi')=q(\xi,\xi')min\Bigg(\frac{\pi(\xi')q(\xi',\xi)}{\pi(\xi)q(\xi,\xi')},1\Bigg)$$where our particular choice of distribution $q$ is $q(\xi,\xi')=(2L+1)^{-2}$ if $\xi$ and $\xi'$ have only one magnetic particle with a different sign, and $q(\xi,\xi')=0$ otherwise


closed as off-topic by mdewey, usεr11852, mkt, jpmuc, Siong Thye Goh Apr 13 at 5:43

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Self-study questions (including textbook exercises, old exam papers, and homework) that seek to understand the concepts are welcome, but those that demand a solution need to indicate clearly at what step help or advice are needed. For help writing a good self-study question, please visit the meta pages." – mdewey, mkt, jpmuc, Siong Thye Goh
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 7
    $\begingroup$ What have you tried? Where are you stuck? Looks like a self-study tag is missing. $\endgroup$ – corey979 Apr 3 at 21:08

Typo: $$\pi(\xi)=\frac{1}{Z_\beta}e^{\beta\sum_{x,y=x}{\xi_x\xi_y}}$$ should be $$\pi(\xi)=\frac{1}{Z_\beta}e^{\beta\sum_{x,y\sim x}{\xi_x\xi_y}}$$ where $x\sim y$ denotes the neighbourhood relation.

Hint #1: Is there anything special about the Ising when applying a Metropolis-Hastings (not Hasting) step?

Hint #2: What are the generic conditions for the Markov chain generated by a Metropolis-Hastings algorithm to converge?


Not the answer you're looking for? Browse other questions tagged or ask your own question.