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TaPaKaH
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Suppose we have a Markov chain with stationary distributions ##p_n=\frac{a}{nb+c}p_{n-1}## for ##n\in\mathbb{N}## where ##a,b## and ##c## are some positive constants.
It follows that ##p_n=p_0\prod_{i=1}^n\frac{a}{ib+c}##. Normalisation yields ##1=p_0\sum_{n=0}^\infty\prod_{i=1}^n\frac{a}{ib+c}## so ##p_0=\left(\sum_{n=0}^\infty\prod_{i=1}^n\frac{a}{ib+c}\right)^{-1}##.
Question: how can one compute the sum in the brackets?
It follows that ##p_n=p_0\prod_{i=1}^n\frac{a}{ib+c}##. Normalisation yields ##1=p_0\sum_{n=0}^\infty\prod_{i=1}^n\frac{a}{ib+c}## so ##p_0=\left(\sum_{n=0}^\infty\prod_{i=1}^n\frac{a}{ib+c}\right)^{-1}##.
Question: how can one compute the sum in the brackets?
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