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Now I have a question, in which I need to find the probability mass function and the cumulative distribution function. But now I only have the recurrence relation. Here is the details:

Assume $p_n=\Pr\left(X=n\right)$ for $n\in\mathbb{N}$, $p_0=\frac{1}{1+ab_0}$, and for $n\geq1$, $ p_n=\sum_{i=0}^{n-1}ab_{n-i}p_i, $ where the constants $a>0$, $b_i>0$ for $i\in\mathbb{N}$ and $\sum_{i=1}^{\infty}b_i=b_0$. Also, we know that $\sum_{n=0}^{\infty}p_n=1$.

From these conditions, can we obtain the closed-form expression of $p_n$? And its CDF $F(M)=\Pr\left(X\leq M\right)$?

If the closed-form expression is complicated, could we find some approximation for it?

Thank you so much~~


FTI:

Actually, now I have derived the probability generating function: $ G\left(z\right)=\sum_{i=0}^{\infty}p_iz^i=\frac{1}{1+ab_0-a\sum_{i=1}^\infty b_iz^i}. $ But I have no idea whether it will help to derive the closed-form expression of $p_n$.

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    From the probability generating function you can get lots of information. But your $b_i$'s are too wild to go much further.2014-05-16

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