The function \(B\) is a kernel for \((\N, \le)\) since \(B(k, n) = 0\) for \(k \gt n\). It has an inverse since \(B(k, k) = 1 \ne 0\) for \(k \in \N\). The functions \(B\) and \(B^{-1}\) as given above are inverses with respect to the usual product operation: \[\sum_{j=k}^n (-1)^{n-j} \binom{n}{j} \binom{j}{k} = \bs 1(k = n), \quad (k, n) \in \N^2\] As always, the kernel \(I\) given by \(I(k, n) = \bs 1(k = n)\) for \((k, n) \in \N^2\) is the identity.

The condition in is \[ \alpha f(k) = \E\left[(-1)^{N - k} \binom{N}{k}\right] = \sum_{n = k}^\infty (-1)^{n - k} \binom{n}{k} f(n), \quad k \in \N \] with the underlying assumption that the series on the right converges absolutely. Equivalently, \(f\) is an eigenfunction of \(B^{-1}\) corresponding to the eigenvalue \(\alpha \in (0, \infty)\), on the space \(\ms L_1\). so the result follows from the Möbius inversion formula discussed in Section 1.2.

The computation is a familiar one, in a variety of contexts. It suffices to show that the function \(g\) is given by \(g(n) = \beta^n / n!\) for \(n \in \N\) is an eigenfunction of \(B\) corresponding to the eigenvalue \(e^\beta\). \[\sum_{n=k}^\infty \binom{n}{k} g(n) = \sum_{n = k}^\infty \binom{n}{k} \frac{\beta^n}{n!} = \frac{\beta^k}{k!} \sum_{n = k}^\infty \frac{\beta^{n - k}}{(n - k)!} = e^\beta \frac{\beta^k}{k!} = e^\beta g(k), \quad k \in \N\]

Let \(f\) denote the probability density function of \(N\). Recall that the probability generating function \(P\) is defined by \[P(t) = \E(t^N) = \sum_{n = 0}^\infty f(n) t^n\] for \(t\) in the interval of absolute convergence, which must be at least \((-1, 1]\), since \(P(1) = 1\). Suppose that \(N\) satisfies the equivalent conditions. Then using the condition in , a sum interchange, and the binomial theorem we have \begin{align*} P(t) &= \sum_{k = 0}^\infty t^k f(k) = \sum_{k = 0}^\infty t^k \alpha \sum_{n = k}^\infty \binom{n}{k} f(n) \\ &= \alpha \sum_{n = 0}^\infty f(n) \sum_{k = 0}^n \binom{n}{k} t^k = \alpha \sum_{n = 0}^\infty f(n) (1 + t)^n = \alpha P(t + 1) \end{align*} In addition, \(t + 1\) must be in the interval of convergence, so this interval must be \(\R\). Conversely, suppose that \(P(t) = P(t + 1)\) for \(t \in \R\). Then using the same tricks as before, \[\sum_{k = 0}^\infty t^k f(k) = \sum_{k = 0}^\infty t^k \alpha \sum_{n = k}^\infty \binom{n}{k} f(n), \quad t \in \R\] Equating coefficients gives \[f(k) = \alpha \sum_{n = k}^\infty \binom{n}{k} f(n), \quad k \in \N\]

The generating function \(P_r\) of \(f_r\) is given by \[P_r(t) = \sum_{n = 0}^\infty f_r(n) t^n = \alpha^{r - 1} \sum_{n = 1}^\infty f(n) r^n t^n = \alpha^{r -1} P(r t), \quad t \in R \] Using the condition in , \[P_r(1) = \alpha^{r - 1} P(r) = \alpha^{r - 1} \frac{1}{\alpha^r} P(0) = \frac{\alpha^r}{\alpha^r} = 1\] Hence \(P_r\) is a probability generating function and so \(f_r\) is a probability density function. Finally, note that \(P_r(0) = f_r(0) = \alpha^{r - 1} P(0) = \alpha^{r}\). Thus, \[P_r(t) = \alpha^{r - 1} P(r t) = \alpha^{r - 1} \alpha^{r} P(r t + r) = \alpha^r \alpha^{r - 1} P[r (t + 1)] = \alpha^r P_r(t + 1), \quad t \in \R\]

Suppose that \(f_r\) is Poisson for some \(r \in \N_+\). Note that \(f_r(0) = \alpha^{r - 1} f(0) = \alpha^r\). Hence the Poisson parameter is \(\beta r\) where \(\beta = - \ln \alpha\). So \[f_r(n) = e^{-r \beta} \frac{(r \beta)^n}{n!} = e^{-(r - 1) \beta} r^n f(n), \quad n \in \N\] It follows that \(f(n) = e^{-\beta} \beta^n / n!\) for \(n \in \N\) so that \(f\) is the Poisson density function with parameter \(\beta\). Then clearly \(f_r\) is the Poisson density with parameter \(\beta r\) for every \(r \in \N_+\).

1. From the definition, \[\sum_{n = k}^\infty g_r(k, n) = \alpha^r r^k \sum_{n = k}^\infty \binom{n}{k} f(n) = \alpha^r r^k \frac{1}{\alpha} f(k) = \alpha^{r - 1} r^k f(k) = f_r(k), \quad k \in \N\] It follows that \(g_r\) is a valid probability density function on \(\N^2\) and that \(X\) has probability density function \(f_r\).
2. Next, \[\sum_{k = 0}^n g_r(k, n) = \alpha^r f(n) \sum_{k = 0}^n \binom{n}{k} r^k = \alpha^r (r + 1)^n f(n) = f_{r + 1}(n), \quad n \in \N\] so \(N\) has probability density function \(f_{r+1}\).
3. For \(n \in \N\), the conditional density of \(X\) given \(N = n\) is \[\frac{g_r(k, n)}{f_{r + 1}(n)} = \frac{\alpha^r r^k \binom{n}{k} f(n)}{\alpha^r (r + 1)^n f(n)} = \binom{n}{k}\left(\frac{r}{r + 1}\right)^k \left(\frac{1}{r + 1}\right)^{n - k}, \quad k \in \{0, 1, \ldots, n\}\]

We already showed in that the Poisson distribution satisfies the equivalent conditions, so we just need to show the converse. We will use the Rao-Rubin characterization of the Poisson distribution: Suppose that \((X, N)\) takes values in \(\N^2\). If the conditional distribution of \(X\) given \(N\) is binomial with parameters \(N\) and \(p\) and if the distribution of \(X\) is the same as the conditional distribution of \(X\) given \(X = N\), then \(N\) has a Poisson distribution.

Towards that end, suppose that \(f\) satisfies the equivalent conditions and let \(P\) denote the corresponding probability generating function. Let \(r \in \N_+\) and let \((X, N)\) have the probability density function \(g_r\) given in . Then \[\P(X = N) = \sum_{n = 0}^\infty \alpha^r r^n \binom{n}{n} f(n) = \alpha^r P(r) = \alpha^r \frac{1}{\alpha^r} P(0) = \alpha\] Hence \begin{align*} \P(X = k \mid X = N) &= \frac{\P(X = k, X = N)}{\P(X = N)} = \frac{\P(X = k, N = k)}{\P(X = N)}\\ &= \frac{\alpha^r r^k \binom{k}{k} f(k)}{\alpha} = \alpha^{r - 1} r^k f(k) = f_r(k) = \P(X = k), \quad k \in \N \end{align*} It follows that \(N\) has the Poisson distribution. From the function \(f\) is the Poisson density function with parameter \(\beta = - \ln \alpha\).