- #1
kulimer
- 9
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I digged out this old tread, but it is closed. I'll repost, but with my question.
https://www.physicsforums.com/showthread.php?t=74004&highlight=metropolis
[itex]\pi(x)[/itex]
and
[itex]\pi(y)[/itex]
and
[itex]q(y,x)[/itex] is the jump distribution
in the relation:
[itex]\alpha(x,y)= \min \left( 1,\frac{\pi (y)q(y,x)}{\pi (x)q(x,y)} \right)[/itex]
Say, my jump distribution(aka transition prob) is normal(0,1). How do you write out [itex]q(y,x)[/itex]? Is it [itex]\frac{1}{\sqrt{2\pi }\sigma }{{e}^{\frac{{{(x-0)}^{2}}}{2{{\sigma }^{2}}}}}[/itex]?
But this doesn't make sense, because it doesn't involve y, since [itex]q(y,x)[/itex] means given y, the transition probability of getting x. We are suppose to relate y to x in the equation.
https://www.physicsforums.com/showthread.php?t=74004&highlight=metropolis
[itex]\pi(x)[/itex]
and
[itex]\pi(y)[/itex]
and
[itex]q(y,x)[/itex] is the jump distribution
in the relation:
[itex]\alpha(x,y)= \min \left( 1,\frac{\pi (y)q(y,x)}{\pi (x)q(x,y)} \right)[/itex]
Say, my jump distribution(aka transition prob) is normal(0,1). How do you write out [itex]q(y,x)[/itex]? Is it [itex]\frac{1}{\sqrt{2\pi }\sigma }{{e}^{\frac{{{(x-0)}^{2}}}{2{{\sigma }^{2}}}}}[/itex]?
But this doesn't make sense, because it doesn't involve y, since [itex]q(y,x)[/itex] means given y, the transition probability of getting x. We are suppose to relate y to x in the equation.
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