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The probability distribution of a random variable X is $P(X=k)=p_k,k=0,1,2,...$ and $\frac{p_k}{p_{k-1}}=a+\frac{b}{k}$. If $p_0=\frac{1}{9},p_1=p_2=\frac{4}{27}$,what can be said about the sequence of probabilities ${p_0,p_1,p_2,...}$

My attempt:Putting the values of $p_0,p_1,p_2$ in $\frac{p_k}{p_{k-1}}=a+\frac{b}{k}$,we get $a=\frac{2}{3},b=\frac{2}{3}$

But is the sequence are terms of any A.P. or G.P. or any known sequence?

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    $\begingroup$ Just a note for readers who may not have seen this class before -- the class of distributions with a recurrence in the pmf of the form $\frac{p_k}{p_{k-1}}=a+\frac{b}{k}$ has been studied and encompasses the negative binomial ($a>0$), the binomial ($a<0$) and the Poisson ($a=0$). One important application is that it allows for a simple recursive calculation of pmfs for discrete compound distributions with number of terms, $N$, drawn from this class. $\endgroup$
    – Glen_b
    May 28 '16 at 23:05
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So the infinite sequence here will be:

$$ p_0,p_1,p_2,p_3,...,p_n,... $$

Given the values:

$$ \frac{1}{9},\frac{4}{27},\frac{4}{27},\frac{32}{243},\frac{80}{729},... $$

As far as i know, this sequence belongs to none of the following progressions: arithmetic, geometric or harmonic progression. In fact,

for $k \in{1,2,3,...}$ we have: $$ p_k =p_{k-1} (a+\frac{b}{k}) $$

The common sequence of this progrssion $(a+\frac{b}{k})$ varies with $k$ and when $k$ tends toward infinite, the ratio of $\frac{p_k}{p_{k-1}}$ will converge toward $a=\frac{2}{3}$.

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