RM算法的地下宫殿
证： X n + 1 = X n + β n ( ξ n − X n ) = ( 1 − β n ) X n + β n ξ n X_{n+1}=X_n+\beta_n(\xi_n-X_n)=(1-\beta_n)X_n+\beta_n\xi_n Xn+1=Xn+βn(ξn−Xn)=(1−βn)Xn+βnξn。由数学归纳法可得 X n + 1 = ∑ j = 1 n ξ j β j ∏ i = j n − 1 ( 1 − β i + 1 ) + X 1 ∏ i = 1 n ( 1 − β i ) X_{n+1}=\sum_{j=1}^{n}\xi_j\beta_j\prod_{i=j}^{n-1}{(1-\beta_{i+1})}+X_1\prod_{i=1}^{n}{(1-\beta_i)} Xn+1=∑j=1nξjβj∏i=jn−1(1−βi+1)+X1∏i=1n(1−βi)。由前言的基本定理的推论， ∏ i = 1 ∞ ( 1 − β i ) = 0 \prod_{i=1}^{\infty}{(1-\beta_i)}=0 ∏i=1∞(1−βi)=0， lim n → ∞ X n = lim n → ∞ ∑ j = 1 n ξ j β j ∏ i = j + 1 n ( 1 − β i ) = lim n → ∞ ∑ j = 1 n ξ j β j ∏ i = j + 1 n ( 1 − β i ) 1 − 0 = lim n → ∞ ∑ j = 1 n ξ j β j ∏ i = j + 1 n ( 1 − β i ) 1 − ∏ i = 1 ∞ ( 1 − β i ) = lim n → ∞ ∑ j = 1 n ξ j β j ∏ i = j + 1 n ( 1 − β i ) lim n → ∞ [ 1 − ∏ i = 1 n ( 1 − β i ) ] = lim n → ∞ ∑ j = 1 n ξ j β j ∏ i = j + 1 n ( 1 − β i ) 1 − ∏ i = 1 n ( 1 − β i ) = lim n → ∞ ∑ j = 1 n ξ j β j ∏ i = j + 1 n ( 1 − β i ) 1 − ∏ i = 1 n ( 1 − β i ) \underset{n\to\infty}{\lim}{X_n}=\underset{n\to\infty}{\lim}\sum_{j=1}^{n}\xi_j\beta_j\prod_{i=j+1}^{n}{(1-\beta_{i})}=\frac{\underset{n\to\infty}{\lim}\sum_{j=1}^{n}\xi_j\beta_j\prod_{i=j+1}^{n}{(1-\beta_{i})}}{1-0}=\frac{\underset{n\to\infty}{\lim}\sum_{j=1}^{n}\xi_j\beta_j\prod_{i=j+1}^{n}{(1-\beta_{i})}}{1-\prod_{i=1}^{\infty}{(1-\beta_i)}}=\frac{\underset{n\to\infty}{\lim}\sum_{j=1}^{n}\xi_j\beta_j\prod_{i=j+1}^{n}{(1-\beta_{i})}}{\underset{n\to\infty}{\lim}[1-\prod_{i=1}^{n}{(1-\beta_i)}]}=\underset{n\to\infty}{\lim}\frac{\sum_{j=1}^{n}\xi_j\beta_j\prod_{i=j+1}^{n}{(1-\beta_{i})}}{1-\prod_{i=1}^{n}{(1-\beta_i)}}=\underset{n\to\infty}{\lim}\sum_{j=1}^{n}\xi_j\frac{\beta_j\prod_{i=j+1}^{n}{(1-\beta_{i})}}{1-\prod_{i=1}^{n}{(1-\beta_i)}} n→∞limXn=n→∞lim∑j=1nξjβj∏i=j+1n(1−βi)=1−0n→∞lim∑j=1nξjβj∏i=j+1n(1−βi)=1−∏i=1∞(1−βi)n→∞lim∑j=1nξjβj∏i=j+1n(1−βi)=n→∞lim[1−∏i=1n(1−βi)]n→∞lim∑j=1nξjβj∏i=j+1n(1−βi)=n→∞lim1−∏i=1n(1−βi)∑j=1nξjβj∏i=j+1n(1−βi)=n→∞lim∑j=1nξj1−∏i=1n(1−βi)βj∏i=j+1n(1−βi)。又 { ξ n } \{\xi_n\} {ξn}有界，所以