Has anyone looked at the exercise at 16:00 ? Could it be that the right-hand side should read sum P* / sum Q* instead of sum w* ? Otherwise I'd have to think it through more thoroughly.
@@muhong9636 thanks for your reply. As you say, w* = P*/Q*. Therefore sum w* = sum P*/Q*. I thought it maybe should read sum P* / sum Q* instead, but I just found out that the version in the slides makes it even easier. Here's my proof: 1/S sum w* = 1/S sum P*/Q* -> E_Q[P*/Q*] (for S->infinity) = E_P[ Q/P P*/Q*] = E_P[ ZP/ZQ Q*/P* P*/Q*] = ZP/ZQ
This has to be the clearest explanation of MCMC I could find online. Thank you!
50:00 Auxiliary Variables
52:00 Swendsen Wang
55:00 Hamiltonian Monte Carlo
Has anyone looked at the exercise at 16:00 ? Could it be that the right-hand side should read
sum P* / sum Q* instead of sum w* ? Otherwise I'd have to think it through more thoroughly.
From my understanding, the w* = P*/Q*. And he normalized the w* divided by sum w* making up to 1.
@@muhong9636 thanks for your reply. As you say, w* = P*/Q*. Therefore sum w* = sum P*/Q*. I thought it maybe should read sum P* / sum Q* instead, but I just found out that the version in the slides makes it even easier. Here's my proof:
1/S sum w*
= 1/S sum P*/Q*
-> E_Q[P*/Q*] (for S->infinity)
= E_P[ Q/P P*/Q*]
= E_P[ ZP/ZQ Q*/P* P*/Q*]
= ZP/ZQ
@@dermitdembrot3091 Thanks for sharing👍
38:28 What does he mean by valid in "MCMC T is valid"? like T is a stationary distribution?