**Proof:**

According to Lundberg in [18]:

$$P(N(t+h) = 0 | N(t) = 0) = \exp\left\{-\int\_{t}^{t+h} \lambda\_0(u) du\right\} \tag{43}$$

where *λ*0ð Þ*t* denotes the intensity function associated with the time-dependent (or nonstationary) *PCP*. If we make *n* ¼ 0 in (40), then we obtain

$$P\_0(t, t+h) = \exp\left\{-\int\_t^{t+h} \lambda\_0(v; a) dv\right\} = \frac{P\_0(t+h)}{P\_0(t)}\tag{44}$$

Thus,

$$P\_0(t+h) = P\_0(t) \cdot P\_0(t, t+h) \qquad \qquad \text{for} \quad t, h \ge 0.$$

The expression obtained in (41) may be interpreted as if no event occurred, then the *NHP* has independent increments.

Lemma 1.9: Let *N t*ð Þ be an *NHP* with transition intensities given by (21). Then this CP satisfies

$$\sum\_{j=0}^{m} \frac{\lambda\_j'(t; a)}{\lambda\_j(t; a)} = \lambda\_0(t; a) - \lambda\_{m+1}(t; a) \qquad \text{ for all} \quad m \ge 0. \tag{45}$$

*Some Results on the Non-Homogeneous Hofmann Process DOI: http://dx.doi.org/10.5772/intechopen.106422*

## **Proof:**

From (25), we have

$$\frac{\lambda\_j'(t;a)}{\lambda\_j(t;a)} = \lambda\_j(t;a) - \lambda\_{j+1}(t;a) \qquad \text{ for all} \quad j \ge 0. \tag{46}$$

Thus, (44) turns out the *m*th partial sum of a telescoping series and from here

$$\sum\_{j=0}^{m} \frac{\lambda\_j'(t; a)}{\lambda\_j(t; a)} = \lambda\_0(t; a) - \lambda\_{m+1}(t; a) \qquad \qquad \text{for all} \quad m \ge 0.$$

Now, using the above lemma, we will prove the following proposition:

Proposition 1.10: Let *N t*ð Þ be an NHP with marginal pmf given by (5), then *Pn*ð Þ*t* satisfies that

i. Process with time-dependent increments

$$\lim\_{h \to 0} \frac{P\_{n,n+1}(t, t+h)}{h} = \lambda\_n(t; a)$$

ii. The probability that no event occurs in ð � *t*, *t* þ *h* is

$$P\_0(t, t+h) = \mathbf{1} - h\lambda\_0(t; a) + o(h) \tag{47}$$

iii. The probability that one event occurs in ð � *t*, *t* þ *h* is

$$P\_1(t, t+h) = h\lambda\_0(t; a) - o(h) \tag{48}$$

iv. *Faddy's conjecture*<sup>2</sup> : If the transition intensities be an increasing sequence with *n*, i.e,

$$
\lambda\_0(t; a) < \lambda\_1(t; a) < \dots < \lambda\_n(t; a), \qquad \text{for any fixed } t \tag{49}
$$

then *Var N t* ½ � ð Þ >½ � *N t*ð Þ , this last inequality is reversed for a decreasing sequence.

### **Proof:**

i. As the *NHP* is an *MPP* then, according to Lundberg in [18], for 0 ≤*u* <*v*, *i* ≤*j*, *N t*ð Þ satisfies:

$$\underbrace{P(N(v)=j \mid N(u)=i)}\_{P\_{ij}(u,v)} = \binom{j}{i} \left(\frac{u}{v}\right)^i \left(1 - \frac{u}{v}\right)^{j-i} \frac{P\_j(v)}{P\_i(u)}\tag{50}$$

Replacing the expression *Pn*ð Þ*t* given in (12), when *κ* 6¼ 0, we obtain in (49) that the transition probabilities for the *NHP* are:

<sup>2</sup> See [21].

*Applied Probability Theory - New Perspectives, Recent Advances and Trends*

$$\begin{split} P\_{ij}(u,v) &= \binom{j}{i} \left(\frac{u}{v}\right)^i \left(1 - \frac{u}{v}\right)^{j-i} \frac{P\_j(v)}{P\_i(u)} \\ &= \binom{j}{i} \binom{u}{w} \left(\frac{v-u}{v}\right)^{j-i} \left[\frac{(-1)^j v^j P\_0^{(j)}(v)}{j!}\right] \\ &= \frac{(u-v)^{j-i}}{(j-i)!} \frac{P\_0^{(j)}(v)}{P\_0^{(i)}(u)} \\ &= \prod\_{m=1}^{j-i} \left[\frac{v-u}{m} \lambda\_{m+i-1}(u;a)\right] \exp\left\{-\int\_u^v \lambda\_j(w;a) dw\right\}. \end{split} \tag{51}$$

We complete the proof of the theorem by the following steps: Rewrite the product in (50) by replacing all instances of *i* ¼ *n*, *j* ¼ *n* þ 1, *u* ¼ *t* and *v* ¼ *t* þ *h*, and we make the limit as *h* approaches zero. Then the transition intensities given by (21) represent the instantaneous transitions probabilities of the *NHP*.

ii. Certainly, the function given by (9) is continuous for *t*≥0 and also analytic, due to *P*ð Þ *<sup>n</sup>* <sup>0</sup> ð Þ*t* , exists for all *n*≥1. Then it is possible to express *P*0ð Þ *t* þ *h* through a Taylor series as follows:

$$P\_0(t+h) = \sum\_{m=0}^{\infty} \frac{h^m}{m!} P\_0^{(m)}(t). \tag{52}$$

By substituting the expression for the *m*th derivative of *P*0ð Þ*t* obtained given by (27) in (51), we have:

$$P\_0(t+h) = P\_0(t) + \sum\_{m=1}^{\infty} \frac{h^m}{m!} \left[ (-1)^m \left( \prod\_{j=0}^{m-1} \lambda\_j(t; a) \right) P\_0(t) \right]. \tag{53}$$

Notice that *P*0ð Þ *t* þ *h* satisfies (41), then (52) is similar to:4

$$P\_0(t) \cdot P\_0(t, t+h) = P\_0(t) \left[ 1 + \sum\_{m=1}^{\infty} (-1)^m \frac{h^m}{m!} \left( \prod\_{j=0}^{m-1} \lambda\_j(t; a) \right) \right] \tag{54}$$

Let *n* ¼ *m* � 1 then:

$$\begin{split} P\_0(t, t+h) &= \mathbf{1} + \sum\_{n=0}^{\infty} \left( -\mathbf{1} \right)^{n+1} \frac{h^{n+1}}{(n+1)!} \left( \prod\_{j=0}^{n} \lambda\_j(t; a) \right) \\ &= \mathbf{1} - h \sum\_{n=0}^{\infty} \frac{(-h)^n}{(n+1)!} \left( \prod\_{j=0}^{n} \lambda\_j(t; a) \right) \end{split} \tag{55}$$

From the expansion of the first terms of (54), we get:

$$P\_0(t, t+h) = \mathbf{1} - h\lambda\_0(t; a) + o(h) \tag{56}$$

*Some Results on the Non-Homogeneous Hofmann Process DOI: http://dx.doi.org/10.5772/intechopen.106422*

where

$$\rho(h) = \sum\_{n=1}^{\infty} \frac{(-h)^{n+1}}{(n+1)!} \prod\_{j=0}^{n} \lambda\_j(t; a).$$

The last function satisfies that lim *h*!0 *o h*ð Þ*=h* ¼ 0 ([21, 22]).

iii. From (55) and the fact *P*0ð Þ¼ *t*, *t* þ *h PNt* ð Þ ð Þ� þ *h N t*ðÞ¼ 0 , we obtain

$$P(N(t+h) - N(t) > 0) = 1 - P\_0(t, t+h). \tag{57}$$

Given that the *NHP N t*ð Þ is an *NHPBP* and assuming that we have in a small time interval, then there will be only two cases: there is a birth or not in that period. Thus,

$$P(N(t+h) - N(t) > 0) = P(N(t+h) - N(t) = 1) = P\_1(t, t+h).$$

Then, from (56), we obtain:

$$P\_1(t, t+h) = h\lambda\_0(t; a) - o(h),\tag{58}$$

provided that *h* is infinitesimal.

iv. According to Steutel et al. in ref. [16], a non-degenerate distribution f g *Pn*ð Þ*t* is log-convex if and only if *Pn*ð Þ*<sup>t</sup>* <sup>&</sup>gt; 0 for all *<sup>n</sup>*<sup>≥</sup> 0 and *Pn*þ1ð Þ*<sup>t</sup> Pn*ð Þ*t* n o is a nondecreasing sequence. By assumption

$$\frac{P\_n(t)}{P\_{n-1}(t)} < \frac{P\_{n+1}(t)}{P\_n(t)} \qquad\qquad\text{for some}\quad n \ge 1\tag{59}$$

By substituting (5) into (58)

$$\begin{aligned} \frac{t^n}{n!} \left[ \frac{(-1)^n P\_0^{(n)}(t)}{t^{n-1}} \right] &< \frac{t^{n+1}}{(n+1)!} \left[ (-1)^{n+1} P\_0^{(n+1)}(t) \right] \\\ \frac{t^{n-1}}{(n-1)!} \left[ (-1)^{n-1} P\_0^{(n-1)}(t) \right] &< \frac{t^n}{n!} \left[ (-1)^n P\_0^{(n)}(t) \right] \\\ \frac{1}{n} \left( -\frac{P\_0^{(n)}(t)}{P\_0^{(n-1)}(t)} \right) &< \quad \frac{1}{n+1} \left( -\frac{P\_0^{(n+1)}(t)}{P\_0^{(n)}(t)} \right) \\\ \frac{1}{n} \lambda\_{n-1}(t; a) &< \quad \frac{1}{n+1} \lambda\_n(t; a) \end{aligned}$$

we know 1< *<sup>n</sup>*þ<sup>1</sup> *<sup>n</sup>* for all *n*. Hence, we have the following:

$$
\lambda\_{n-1}(t;a) < \frac{n+1}{n} \lambda\_{n-1}(t;a) < \lambda\_n(t;a). \tag{60}
$$

,

Thus, we obtain that (48) is satisfied and, therefore, the conjecture holds.

The expression (48) allows to identify under- or over-dispersion of a *CP*, then we can classify the process according to the fixed criteria given in (16).

Corollary 1.10.1: If *a* 6¼ 0 and *N t*ð Þ is an *NHP*, then it does not have independent increments.
