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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~~


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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Do you have any specific sequence $(b_i)$ in mind? Otherwise it seems difficult to go much further than the closed form formula for $G$ you computed. – Did Nov 11 '12 at 7:55
Hi @did, the expression of $b_i$ is $b_i=\gamma^{\delta}\int_{\gamma^{-\delta}}^{\infty}\frac{\delta x^{\delta}}{\left(1+x\right)^{i+1}}dx$ for $i\geq 1$, where $\gamma>0$ and $0<\delta<1$. But I am not sure whether it is helpful. Thank you~~ – Chang Nov 11 '12 at 9:10
Indeed, one can get more or less explicit formulas for $b_0$ and $\sum\limits_{i\geqslant1}b_iz^i$ (by the way, are you sure the lower bound in the integral defining $b_i$ is not $\gamma^{-1}$ instead of $\gamma^{-\delta}$?) but to use these in $G(z)$ does not seem straightforward. – Did Nov 11 '12 at 12:52
Wow, it is a mistake, sorry. It should be $\gamma^{-1}$. Thank you so much. Let me rewrite it: $b_i=\gamma^\delta \int_{\gamma^{-1}}^{\infty}\frac{\delta x^\delta}{\left(1+x\right)^{i+1}}dx$ for $i\geq1$, where $\gamma>0$ and $0<\delta<1$. – Chang Nov 11 '12 at 13:06
And I have no idea how to use this $b_i$ in $G(z)$ or the recurrence form....Sigh... – Chang Nov 11 '12 at 13:12

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