**Proof:**

1. From (2), taking the expected value of Λ, conditioning on *N t*ð Þ, we get

$$\mathbb{E}[\Lambda|N(t)=n] = \int\_0^\infty \frac{\lambda e^{-\lambda t} (\lambda t)^n f(\lambda)}{n! P[N(t)=n]} d\lambda = \frac{n+1}{t} \frac{P\_{n+1}(t)}{P\_n(t)}.\tag{32}$$

By substituting (24) into (32), we have

$$\mathbb{E}[\Lambda|\mathcal{N}(t) = \mathfrak{n}] = \lambda\_{\mathfrak{n}}(t; a).$$

Analogously, we can show that

$$\mathbb{E}\left[\Lambda^{2}|\mathcal{N}(t)=n\right]=\int\_{0}^{\infty}\frac{\lambda^{2}e^{-\lambda t}(\lambda t)^{n}f(\lambda)}{n!P[N(t)=n]}d\lambda=\frac{(n+2)(n+1)}{t^{2}}\frac{P\_{n+2}(t)}{P\_{n}(t)}.\tag{33}$$

By substituting (24) into (33), we have

$$\mathbb{E}\left[\Lambda^2|\mathcal{N}(t)=n\right] = \lambda\_{n+1}(t;a)\lambda\_n(t;a)\dots$$

Then the conditional variance of Λ, given that *N t*ðÞ¼ *n*, is

$$\operatorname{Var}[\Lambda|\mathbf{N}(t) = n] = \lambda\_{n+1}(t; a)\lambda\_n(t; a) - \lambda\_n^2(t; a),$$

and substituting Eq. (25) into the above yields the result.

2. We use the law of total expectation to find the expected value

$$\begin{aligned} \mathbb{E}[\Lambda] &= \mathbb{E}[|\mathbb{E}(\Lambda|N(t) = n)| = \sum\_{n=0}^{\infty} \mathbb{E}(|\Lambda|N(t) = n) P[N(t) = n] \\ &= \sum\_{n=0}^{\infty} \lambda\_n(t; a) P\_n(t) \end{aligned}$$

By substituting (24) into the above expression, we get

$$\mathbb{E}[\Lambda] = \sum\_{n=0}^{\infty} \frac{n+1}{t} P\_{n+1}(t) = \sum\_{j=0}^{\infty} \frac{r\_j(t; a)}{t} = \frac{1}{t} \mathbb{E}[N(t)].$$

And the proof is completed.

3. The pgf of *N t*ð Þ is defined as

$$\begin{split} \underbrace{\mathbf{G}\_{\mathbf{G}}(\mathbf{z};t)}\_{=\mathbf{0}} &= \sum\_{n=0}^{\infty} z^{n} P\_{n}(t) = \sum\_{n=0}^{\infty} z^{n} \int\_{0}^{\infty} \frac{\left(\lambda t\right)^{n}}{n!} e^{-\lambda t} f(\lambda) d\lambda \\ P\_{0}[(1-\mathbf{z})t] &= \int\_{0}^{\infty} \left[ \sum\_{n=0}^{\infty} \frac{\left(\mathbf{z}\lambda t\right)^{n}}{n!} \right] e^{-\lambda t} f(\lambda) d\lambda = \int\_{0}^{\infty} e^{\lambda(x-1)t} f(\lambda) d\lambda \\ &= M\_{\lambda}[(\mathbf{z}-\mathbf{1})t]. \end{split} \tag{34}$$

We make *z* ¼ 0 in the above expression and we have

$$P\_0(t) = \mathcal{M}\_\Lambda(-t)$$

Now, if we differentiate both sides with respect to *t*, we obtain

$$P\_0'(t) = -\mathcal{M}\_\Lambda'(-t)$$

We complete the proof by substituting *<sup>t</sup>* <sup>¼</sup> 0 in the above expression. □

According to Walhin and Paris in ref. [20], the intensity of the stochastic process *N t*ð Þ in the period ½ � *t*, *t* þ 1 is

$$\mathbb{E}[N(t+1) - N(t)|N(t) = n] = \mathbb{E}[\Lambda | N(t) = n].$$

The moment generating function of the process will uniquely determine the distribution of the process, on comparing expression (34) with *P*0½ � ð Þ 1 � *z t* given for *a* ¼ 1 and as shown in **Table 1**, we find the particular cases: the *PCP* if Λ � *δγ* ð Þ*λ* (i.e. has a degenerate *cdf* at *λ* ¼ *γ*), the *NBCP* if Λ � Γð Þ *γ*, *δ* and the Geometric Counting Process if Λ � exp ð Þ*δ* .
