Identifying the distribution which represents a negative binomial distribution as a compound poisson distribution 
Suppose that the random variable $X$, which has a negative binomial distribution with probability $p$ and parameter $r$, can be represented as the summation of $N$ iid random variables $Y_1, Y_2, \ldots$:
$$X = Y_1 + Y_2 + \cdots + Y_N.$$
If $N$ has a Poisson distribution and is independent of $Y_1, Y_2, \ldots$, what is the probability distribution of each $Y_1, Y_2, \ldots$ and what is the value of the Poisson parameter?

I tried but have no idea. If I know the distribution of $Y_i$, I can get the distribution of $X$, which is their summation, a compound poisson distribution. But in this case, it is the reverse.
 A: I think you are looking for this wikipedia page. 
A: 
Suppose that the random variable $X$, which has a negative binomial distribution with probability $p$ and parameter $r$,

Unfortunately there are differing conventions about what this means, so one should be explicit:
(In the case where $r$ is a positive integer) $X$ is distributed as the number of failures before the $r$th success in independent Bernoulli trials with probability $p$ of success on each trial. Thus the distribution of $X$ is supported on the set $\{0,1,2,3,\ldots\}.$ Therefore
\begin{align}
\Pr(X=x) = {} & \binom {x+r-1} x p^r (1-p)^x = (-1)^x \binom{-r} x p^r q^x \\[12pt]
& \text{where } \binom{-r} x = \frac{(-r)(-r-1)(-r-2) \cdots(-r-x+1)}{x!}.
\end{align}
With $\binom{-r} x$ defined that way, the distribution is defined even when $r$ is not an integer, and is so defined that the sum of independent negative-binomially distributed random variables with parameters $r_1,r_2$ and the same $p$ is negative-binomially distributed with parameters $r_1+r_2$ and $p.$)
I don't know a way to find the distribution of each of the $Y$s given this information, but I know how to show that it's a "logarithmic" distribution. If that had never been pointed out to me, I might never have succeeded in solving this problem.
Definition: With $q=1-p$ as above, we have
$$
\sum_{y=1}^\infty \frac {q^y} y = -\log(1-q) = -\log p >0. \tag 1
$$
Thus $\Pr(Y=y) = \dfrac{q^y}{-y\log p}$ for $y=1,2,3,\ldots$ defines a probability distribution, conventionally called the logarithmic distribution (which I think is probably a name given to it by Ronald Fisher some time around 1930 or so).
Lemma: The probability-generating function of $Y_1+Y_2$ is the pointwise product of their separate probability-generating functions, and since they're identically distributed, is just the square of the probability-generating function of $Y_1.$ Corollary: By a trivial induction, this extends to sums of more than two i.i.d. of these random variables.  Proof:
$$
\sum_{y=1}^\infty \Pr(Y_1+Y_2 = y) s^y = \sum_{y=1}^\infty \sum_{k=1}^{y-1} \Pr(Y_1=k) s^k \cdot \Pr(Y_2=y-k) s^{y-k} 
$$
and this is a Cauchy product of two power series. $\quad\blacksquare$
Suppose $Y_1,Y_2,Y_3,\ldots$ are i.i.d., each with the distribution given in $(1)$ above,
and $N\sim\operatorname{Poisson}(\lambda),$ independent of the $Y\text{s},$ and let
$$
X = \sum_{n=1}^N Y_n.
$$
Let us find the probability-generating function $g$ of $X.$
\begin{align}
g(s) = {} & \sum_{x=0}^\infty \Pr(X=x) s^x
= \sum_{x=0}^\infty \operatorname E(\Pr(X=x\mid N)) s^x \\[10pt]
= {} & \operatorname E\left( \sum_{x=0}^\infty \Pr(X=x\mid N) s^x \right) \\[10pt]
= {} & \operatorname E\left( \sum_{x=0}^\infty \Pr(Y_1+\cdots + Y_N=x\mid N) s^x \right) \\[10pt]
= {} & \operatorname E\left( h(s)^N \right) \tag 2 \\[10pt]
& \text{where the last equality follows from the } \textbf{Lemma} \text{ above,} \\
& \text{the function $h$ being the probability-generating function of } Y_1\text{:} \\[10pt]
h(s) = {} & \sum_{y=1}^\infty \Pr(Y_1=y)s^y.
\end{align}
So evaluate line $(2){:}$
$$
\operatorname E\left(h(s)^N\right) = \sum_{n=0}^\infty \frac{e^{-\lambda} \lambda^n}{n!} h(s)^n = e^{-\lambda} e^{\lambda h(s)} = e^{\lambda(h(s)-1)}. \tag 3
$$
So we need to know what $h(s)$ is.
\begin{align}
h(s) = {} & \sum_{y=1}^\infty \Pr(Y_1=y)s^y \\[8pt]
= {} & \sum_{y=1}^\infty \frac{q^y}{-y\log p} s^y
= \frac{\log(1-qs)}{\log p}. \tag 4
\end{align}
Plug $(4)$ into $(3)$:
\begin{align}
& h(s)-1 = \frac{\log\left( \frac{1-qs}{1-q} \right)}{\log p} \\[10pt]
g(s) = {} & e^{\lambda(h(s)-1)} = \left( \frac {1-qs}{1-q} \right)^{\lambda/\log p}. \tag 5
\end{align}
It remains only to show that this is the probability-generating function of a negative binomial distribution.
\begin{align}
\sum_{x=0}^\infty \Pr(X=x) s^x = {} & \sum_{x=0}^\infty \binom{-r} x p^r (-qs)^x \\[10pt]
= {} & p^r (1-qs)^{-r} = \left( \frac{1-q}{1-qs} \right)^r. \tag 6
\end{align}
Finally, observe that $(5)$ agrees with $(6)$ if
$$
r = \frac{-\lambda}{\log p}, \qquad \lambda = -r\log p, \qquad p = e^{-\lambda/r}.
$$
A: Each $Y_1, Y_2, \ldots$ follows a logarithmic distribution:
$$\Pr(Y=k) = -\frac{(1-p)^k}{k\log(p)}, \quad \text{where } k\geqslant1, 0<p<1.$$
If $N \sim \mathsf{Poi}(\mu)$ independently of $Y_1, Y_2, \ldots$, then we find that $X$ has a negative binomial distribution with parameter $r = -\mu/\log(p)$ and success probability $p$.
Hence, the parameter of the Poisson distribution is $\mu = -r\log(p)$.
