My actual question appears at the bottom of this posting.

Suppose $$ \Pr(\Lambda\in d\ell) = \frac 1{\Gamma(\alpha)} \left( \frac\ell m\right)^{\alpha-1} e^{-\ell/m}\, \left( \frac{d\ell} m \right) \text{ for } \ell>0, $$ so $\Lambda$ is a random variable with a gamma distribution with expected value $\alpha m$ and variance $\alpha m^2$.

Let the conditional distribution of the random variable $N$, given $\Lambda$, be $$ N\mid\Lambda \sim \operatorname{Poisson}(\Lambda). \tag 1 $$ Then the marginal (``unconditional'') distribution of $N$ is a negative binomial distribution: \begin{align*} & \Pr(N=n) \\[10pt] = {} & \operatorname E(\Pr(N=n\mid\Lambda)) = \operatorname E\left( \frac{\Lambda^n e^{-\Lambda}}{n!} \right) \\[10pt] = {} & \int_0^\infty \frac{\ell^n e^{-\ell}}{n!} \cdot \frac1{\Gamma(\alpha)} \left( \frac\ell m\right)^{\alpha-1} e^{-\ell/m} \, \left( \frac{d\ell} m \right) \\[8pt] = {} & \frac{(n+\alpha-1)(n+\alpha-2)(n+\alpha-3)\cdots\alpha}{n!} \left( \frac m{m+1} \right)^n \left( \frac1{m+1} \right)^\alpha \\[8pt] = {} & \binom{-\alpha}{\phantom{-}n} p^\alpha (-q)^n \text{ for } n\in\{0,1,2,3,\ldots\} \tag 2 \end{align*} (so that $p+q=1$), where $$ \binom{-\alpha}{\phantom{-}n} = \frac{\overbrace{-\alpha(-\alpha-1)(-\alpha-2)\cdots(-\alpha-n+1)}^\text{$n$ factors}}{n!}. $$ This has expected value $\alpha q/p$ and variance $\alpha q/p^2$.

Perhaps it is less widely known that the same negative binomial distribution arises as a compound Poisson distribution:

Suppose $\Pr(X=x) = \dfrac{-q^x}{x\log(1-q)}$ for $x=1,2,3,\ldots$, and let $X_1,X_2,X_3,\ldots$ be independent copies of this random variable. (This is called the logarithmic series distribution since $\sum_{x=1}^\infty q^x/x = -\log(1-q).$) Suppose $M\sim\operatorname{Poisson}(-\alpha\log(1-q))$.

Then $$ N = \sum_{i=1}^M X_i \tag 3 $$ also has the same negative binomial distribution that appears on line $(2)$ above.

My question is: How can we construct a single probability space that is the domain of all of the random variables mentioned here, in such a way that the $N$ defined on line $(1)$ above and the $N$ defined on line $(3)$ above are not just two random variables sharing the same distribution, but are just one and the same random variable?

My actual question appears at the bottom of this posting.

Suppose $$ \Pr(\Lambda\in d\ell) = \frac 1{\Gamma(\alpha)} \left( \frac\ell m\right)^{\alpha-1} e^{-\ell/m}\, \left( \frac{d\ell} m \right) \text{ for } \ell>0, $$ so $\Lambda$ is a random variable with a gamma distribution with expected value $\alpha m$ and variance $\alpha m^2$.

Let the conditional distribution of the random variable $N$, given $\Lambda$, be $$ N\mid\Lambda \sim \operatorname{Poisson}(\Lambda). \tag 1 $$ Then the marginal (``unconditional'') distribution of $N$ is a negative binomial distribution: \begin{align*} & \Pr(N=n) \\[10pt] = {} & \operatorname E(\Pr(N=n\mid\Lambda)) = \operatorname E\left( \frac{\Lambda^n e^{-\Lambda}}{n!} \right) \\[10pt] = {} & \int_0^\infty \frac{\ell^n e^{-\ell}}{n!} \cdot \frac1{\Gamma(\alpha)} \left( \frac\ell m\right)^{\alpha-1} e^{-\ell/m} \, \left( \frac{d\ell} m \right) \\[8pt] = {} & \frac{(n+\alpha-1)(n+\alpha-2)(n+\alpha-3)\cdots\alpha}{n!} \left( \frac m{m+1} \right)^n \left( \frac1{m+1} \right)^\alpha \\[8pt] = {} & \binom{-\alpha}{\phantom{-}n} p^\alpha (-q)^n \text{ for } n\in\{0,1,2,3,\ldots\} \tag 2 \end{align*} (so that $p+q=1$), where $$ \binom{-\alpha}{\phantom{-}n} = \frac{\overbrace{-\alpha(-\alpha-1)(-\alpha-2)\cdots(-\alpha-n+1)}^\text{$n$ factors}}{n!}. $$ This has expected value $\alpha q/p$ and variance $\alpha q/p^2$.

Perhaps it is less widely known that the same negative binomial distribution arises as a compound Poisson distribution:

Suppose $\Pr(X=x) = \dfrac{-q^x}{x\log(1-q)}$ for $x=1,2,3,\ldots$, and let $X_1,X_2,X_3,\ldots$ be independent copies of this random variable. (This is called the logarithmic series distribution since $\sum_{x=1}^\infty q^x/x = -\log(1-q).$) Suppose $M\sim\operatorname{Poisson}(-\alpha\log(1-q))$ and $M$ is independent of $X_1,X_2,X_3,\ldots.$

Then $$ N = \sum_{i=1}^M X_i \tag 3 $$ also has the same negative binomial distribution that appears on line $(2)$ above.

My question is: How can we construct a single probability space that is the domain of all of the random variables mentioned here, in such a way that the $N$ defined on line $(1)$ above and the $N$ defined on line $(3)$ above are not just two random variables sharing the same distribution, but are just one and the same random variable?

My actual question will appear at the bottom of this posting.

Suppose $$ \Pr(\Lambda\in d\ell) = \frac 1{\Gamma(\alpha)} \left( \frac\ell m\right)^{\alpha-1} e^{-\ell/m}\, \left( \frac{d\ell} m \right) \text{ for } \ell>0, $$ so $\Lambda$ is a random variable with a gamma distribution with expected value $\alpha m$ and variance $\alpha m^2$.

Let the conditional distribution of the random variable $N$, given $\Lambda$, be $$ N\mid\Lambda \sim \operatorname{Poisson}(\Lambda). \tag 1 $$ Then the marginal (``unconditional'') distribution of $N$ is a negative binomial distribution: \begin{align*} & \Pr(N=n) \\[10pt] = {} & \operatorname E(\Pr(N=n\mid\Lambda)) = \operatorname E\left( \frac{\Lambda^n e^{-\Lambda}}{n!} \right) \\[10pt] = {} & \int_0^\infty \frac{\ell^n e^{-\ell}}{n!} \cdot \frac1{\Gamma(\alpha)} \left( \frac\ell m\right)^{\alpha-1} e^{-\ell/m} \, \left( \frac{d\ell} m \right) \\[8pt] = {} & \frac{(n+\alpha-1)(n+\alpha-2)(n+\alpha-3)\cdots\alpha}{n!} \left( \frac m{m+1} \right)^n \left( \frac1{m+1} \right)^\alpha \\[8pt] = {} & \binom{-\alpha}{\phantom{-}n} p^\alpha (-q)^n \text{ for } n\in\{0,1,2,3,\ldots\} \tag 2 \end{align*} (so that $p+q=1$), where $$ \binom{-\alpha}{\phantom{-}n} = \frac{\overbrace{-\alpha(-\alpha-1)(-\alpha-2)\cdots(-\alpha-n+1)}^\text{$n$ factors}}{n!}. $$ This has expected value $\alpha q/p$ and variance $\alpha q/p^2$.

Perhaps it is less widely known that the same negative binomial distribution arises as a compound Poisson distribution:

Suppose $\Pr(X=x) = \dfrac{-q^x}{x\log(1-q)}$ for $x=1,2,3,\ldots$, and let $X_1,X_2,X_3,\ldots$ be independent copies of this random variable. (This is called the logarithmic series distribution since $\sum_{x=1}^\infty q^x/x = -\log(1-q).$) Suppose $M\sim\operatorname{Poisson}(-\alpha\log(1-q))$.

Then $$ N = \sum_{i=1}^M X_i \tag 3 $$ also has the same negative binomial distribution that appears on line $(2)$ above.

My question is: How can we construct a single probability space that is the domain of all of the random variables mentioned here, in such a way that the $N$ defined on line $(1)$ above and the $N$ defined on line $(3)$ above are not just two random variables sharing the same distribution, but are just one and the same random variable?

My actual question appears at the bottom of this posting.

Suppose $$ \Pr(\Lambda\in d\ell) = \frac 1{\Gamma(\alpha)} \left( \frac\ell m\right)^{\alpha-1} e^{-\ell/m}\, \left( \frac{d\ell} m \right) \text{ for } \ell>0, $$ so $\Lambda$ is a random variable with a gamma distribution with expected value $\alpha m$ and variance $\alpha m^2$.

Let the conditional distribution of the random variable $N$, given $\Lambda$, be $$ N\mid\Lambda \sim \operatorname{Poisson}(\Lambda). \tag 1 $$ Then the marginal (``unconditional'') distribution of $N$ is a negative binomial distribution: \begin{align*} & \Pr(N=n) \\[10pt] = {} & \operatorname E(\Pr(N=n\mid\Lambda)) = \operatorname E\left( \frac{\Lambda^n e^{-\Lambda}}{n!} \right) \\[10pt] = {} & \int_0^\infty \frac{\ell^n e^{-\ell}}{n!} \cdot \frac1{\Gamma(\alpha)} \left( \frac\ell m\right)^{\alpha-1} e^{-\ell/m} \, \left( \frac{d\ell} m \right) \\[8pt] = {} & \frac{(n+\alpha-1)(n+\alpha-2)(n+\alpha-3)\cdots\alpha}{n!} \left( \frac m{m+1} \right)^n \left( \frac1{m+1} \right)^\alpha \\[8pt] = {} & \binom{-\alpha}{\phantom{-}n} p^\alpha (-q)^n \text{ for } n\in\{0,1,2,3,\ldots\} \tag 2 \end{align*} (so that $p+q=1$), where $$ \binom{-\alpha}{\phantom{-}n} = \frac{\overbrace{-\alpha(-\alpha-1)(-\alpha-2)\cdots(-\alpha-n+1)}^\text{$n$ factors}}{n!}. $$ This has expected value $\alpha q/p$ and variance $\alpha q/p^2$.

Perhaps it is less widely known that the same negative binomial distribution arises as a compound Poisson distribution:

Suppose $\Pr(X=x) = \dfrac{-q^x}{x\log(1-q)}$ for $x=1,2,3,\ldots$, and let $X_1,X_2,X_3,\ldots$ be independent copies of this random variable. (This is called the logarithmic series distribution since $\sum_{x=1}^\infty q^x/x = -\log(1-q).$) Suppose $M\sim\operatorname{Poisson}(-\alpha\log(1-q))$.

Then $$ N = \sum_{i=1}^M X_i \tag 3 $$ also has the same negative binomial distribution that appears on line $(2)$ above.

My question is: How can we construct a single probability space that is the domain of all of the random variables mentioned here, in such a way that the $N$ defined on line $(1)$ above and the $N$ defined on line $(3)$ above are not just two random variables sharing the same distribution, but are just one and the same random variable?