3
$\begingroup$

I am trying to understand the detailed balance condition precisely.

To get it I went on this page : https://cs.adelaide.edu.au/~paulc/teaching/montecarlo/node22.html

It is written the following :

Let $P(A,t)$ be the probability distribution to be in state $A$ at time $t$.

We have :

$$P(A,t+1)=P(A,t)+\sum_B W(B \rightarrow A)P(B,t) - W(A \rightarrow B)P(A,t)$$

Where $W(B \rightarrow A)$ is the probability that the system goes from state $B$ to state $A$.

I don't understand this formula.

I would write this :

$$P(A,t+1)=\sum_B P(B,t) W(B \rightarrow A,t)$$

Indeed, consider I am at time $t+1$. The only things I need to know the probability of where I can be at time $t$ are given by the previous states + the probability of transition.

I don't get why we would substract things like the formula above...?

$\endgroup$
2
$\begingroup$

Let's say that in time $t+1$ you are in state $A$. You don't know where you have been before, but you have two options:

  • You were in state $B$ at $t$ and jumped into $A$. Probability you were in $B$ is $P(B,t)$. The jump happens with probability $W(B\rightarrow A)$, so the total one is $P(B,t)W(B\rightarrow A)$.

  • You were already in $A$ and nothing happened. The probability of being in $A$ at $t$ is $P(A,t)$. And the probability that nothing happened during the time interval (i.e. you did not jumped out of the state $A$ during the time interval) is $(1-W(A\rightarrow B))$. Then, the probability for this option is $P(A,t)\cdot(1-W(A\rightarrow B))$.

So, you have option one or option two:

$$P(A,t+1)=P(B,t)W(B\rightarrow A) + P(A,t)(1-W(A\rightarrow B))$$.

Add a sum over all possible $B$ states, make the product of the last term, and you recover your formula.

$\endgroup$
  • $\begingroup$ Thank you for your answer. In fact from my perspective the $W(B \rightarrow A,t)$ already contain the case when nothing change or when the state "when out" from $A$. In a sense, I have my $P(A,t)*W(A \rightarrow A)=P(A,t)*(1-\sum_B W(A \rightarrow B))$ where lhs is my point of view and rhs yours. So I guess that the summation from the link suppose that $B \neq A$ right (whereas in my case $B=A$ is included) ? $\endgroup$ – StarBucK Oct 12 '17 at 14:31
  • $\begingroup$ Ok my comment is probably not really clear. I would like to check if indeed the summation on the website assume $B \neq A$ and then they in fact say the exact same thing than me. That was probably what you meant by your last sentence but I am not sure if you talked about my own formula or you said that what you wrote was the same as the website formula. $\endgroup$ – StarBucK Oct 12 '17 at 14:45
  • 1
    $\begingroup$ The $B=A$ case is considered but drops out since there is a $+P(A\rightarrow A) P(A)$ (from first term in sum) and a $-P(A\rightarrow A) P(A)$ (from the second term in sum). $\endgroup$ – Dave Oct 12 '17 at 19:47
  • 1
    $\begingroup$ When I have faced these proofs, it is usually assumed that $B\neq A$. Basically each one is a different state, and you can jump between them (as an analogy, think in a quantum particle randomly jumping between energy levels). In addition to that, I would say $W(A\rightarrow A)=0$, since when you go out of $A$, you go somewhere else. The case probability you stayed is 1 minus the probability of going out: $1-W(A\rightarrow B)$. $\endgroup$ – VictorSeven Oct 13 '17 at 8:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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