贝叶斯公式

codex --dangerously-bypass-approvals-and-sandbox

x_0 ——> x_t-1 ——> x_t

0 ——> A ——> B

逆转时空

p(A|B)=p(B|A)*P(A)  /  P(B)

p( x_t-1 | x_t , x_0 ) = p( x_t | x_t-1 ) * P( x_t-1 | x_0 )  /  P( x_t | x_0 )

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