Superposition of Time-Changed Poisson Processes and Their Hitting Times

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i i i

(1.1) Y (t) := jNj(t)= Nk(t) j=1 j=1 k j X X X= i

(1.2) Z(t) := g(j)Nj(t) j=1 X and i f (1.3) W (t) := jNj(H (t)), j=1 X where Nj(t),t> 0, 1 ≤ j ≤ i are independent, homogeneous Poisson processes of rate λ, g is an integer-valued, bounded, monotonically increasing function, Hf (t) are subordinators related to a Bernˇstein function f, independent from Nj.

Recently, Sengar et. al (see [10]) have studied the process W (t), that is a time-changed Poisson process of order i. Furthermore, in this paper the authors have studied the process Y (t), time-changed with the inverse of Hf . The process above take jumps of amplitude bigger than i and in the case where g is an integer-valued function the maximal jump has amplitude equal to g(i).

Processes of the form (1.1) can arise in the study of queues in banks or post offices where different services are offered to clients. Since the types of services require a different amount of time to be carried out, the queues formed in front of each window is different. Therefore jNj(t) represents the random number of clients in the j-th queue while Y (t) is the total random number of clients in the post office or bank agency. The process Z(t) can be interpreted in the same way.

In this paper we study the hitting times for all processes, that is we study the distribution of the random variables

(1.4) Tk = inf{s : X(s)= k}, k ≥ 1. A similar analysis was made in the paper by Orsingher and Polito (2012) (see [7]) for the iterated Poisson process, for the time-changed Poisson process by Orsingher and Toaldo (2015) (see [9]) and also by Garra et. al (2016) (see [2]) for the space fractional Poisson process.

In some cases it is possible to evaluate the hitting probabilities

P[Tk < ∞] and, in particular for the process Y (t),t> 0 we show that k if k ≤ i P i [Tk < ∞]=   1 if k > i.

 SUPERPOSITION OF TIME-CHANGED POISSON PROCESSES AND THEIR HITTING TIMES 3

The same analysis is performed for the processes Z(t) and W (t). For the case where Hf (t) is itself a Poisson process we extend the results obtained in [7].

One of our main concerns in the analysis of the hitting time Tk deﬁned in (1.4), where X is one of the processes above. The random variables Tk can be interpreted as the time necessary for the total queue to obtain for the ﬁrst time the length k. We will show for W that P[Tk < ∞] < 1 for all k because all processes examined make upward jumps of arbitrary size with positive probability.

We will sometimes use the following notations. Let N (i) follow the Poisson distribution of order i with rate parameter λ> 0, then the probability mass function (pmf ) is given by λx1+...+xi P[N (i) = n]= e−iλ , n =0, 1,..., x1! ...xi! x1,x2,...,xi≥0 i X Pj=1 jxj =n where the summation is taken over all positive integers x1, x2,...,xi such that x1 +2x2 + ... + ixi = n.

2. Poisson process of order i The process defined in (1.1) during an infinitesimal interval of time of length dt can i(i+1) perform jumps of size up to 2 and thus substantially differs from the classical Poisson process. This can be shown by observing that the probability generating function of Y (t) reads i i − − j − i j E Y (t) E Pj=1 jNj (t) E i jNj(t) λt(1 u ) iλt+λt Pj=1 u (2.1) [u ]= [u ]= [Πj=1u ]= e = e . j=1 Y For the increment dY (t) we have therefore

dY (t) −iλdt+λdt Pi uj (2.2) E[u ]= e j=1 i = (1 − iλdt) 1+ λdt uj + ... . j=1 ! X Therefore Y (t) has the following distribution

P[dY (t)= j]= λdt for 1 ≤ j ≤ i (2.3) P[dY (t)=0]=1 − iλdt. Since Y (t) makes the jumps of length j with uniform law in all time intervals of length dt it can skip the first i − 1 levels even during the infinite time horizon space. A similar behaviour was noted in many other cases like the iterated Poisson process N1(N2(t)) where Nk,k =1, 2 are independent homogeneous Poisson process (see [7]), for the space- fractional Poisson process and its generalizations. In the present case, however, the first k (k < i) states are reached with positive probability. 4 SUPERPOSITION OF TIME-CHANGED POISSON PROCESSES AND THEIR HITTING TIMES

In order to make more explicit the particular nature of Y (t), we will show that the hitting time i

Tk = inf {t : Y (t)= k} = inf t : jNj(t)= k ( j=1 ) X has the following form k , 1 ≤ k ≤ i − 1 (2.4) P[Tk < ∞]= i  , k ≥ i, 1 that it is initially increasing and then reaches a stationary form where all states are hit sooner or later with probability one. Remark 2.1. For i = 1, formula (2.4) conﬁrms that for the homogeneous Poisson processes all states are visited with probability one during an inﬁnite time interval. From (2.1) we can infer that

(λs)x1+...+xi (2.5) P[Y (s)= n]= e−iλs . x1! ...xi! x1,x2,...,xi≥0 i X Pj=1 jxj =n Result (2.5) can also be written down directly by applying convolution arguments. The hitting time of a stochastic process Y (t), t ≥ 0 be the ﬁrst time at which the given process hits a level k, denoted as Tk, and is deﬁned as

Tk = inf{t : Y (t)= k}.

(i) Theorem 2.1. Let N (t) be the Poisson process of order i and Tk be the ﬁrst hitting time of N (i)(t), then k i , 1 ≤ k ≤ i − 1 P[Tk < ∞]= . (1, k ≥ i

Proof. Let P[Tk ∈ ds], s> 0 be the distribution function of Tk, then

k (i) (i) P[Tk ∈ ds]= P[N (s)= k − h, N (s + ds)= k] h X=1 k = P[N (i)(s)= k − h, N (i)(s)+ N (i)(ds)= k] h X=1 k = P[N (i)(s)= k − h]P[N (i)(ds)= h] h X=1 k (2.6) = P[N (i)(s)= k − h]λds, h X=1 SUPERPOSITION OF TIME-CHANGED POISSON PROCESSES AND THEIR HITTING TIMES 5 where the last equality follows from

(i) E[uN (ds)] = exp[−iλds(1 − EuX )] =1 − iλds(1 − EuX )+ o(ds) i 1 =1 − iλds + iλds uj i j=1 ! X i =1 − iλds + λds uj . j=1 ! X Hence, λds, 1 ≤ l ≤ i P[N (i)(ds)= l]= (1 − iλds, l =0. Using the above result in equation (2.6), we get k (i) P[Tk ∈ ds]=λ P[N (s)= k − h]ds h X=1 =λ[P(N (i)(s)= k − 1)ds + P(N (i)(s)= k − 2)ds + ... + P(N (i)(s)=0)ds] k−1 =λ P(N (i)(s)= n)ds. n=0 X We now compute the hitting time probability of level k of N (i)(t). When k < i, then

∞ ∞ k−1 (k) P[Tk < ∞]= P[Tk ∈ ds]=λ P(N (s)= n)ds 0 0 n=0 Z Z X

k−1 ∞ (λs)x1+...+xi =λ  e−iλs  ds 0 x1! ...xi! Z n=0 x1,x2,...,xi≥0 X  i X   Pj=1 jxj =n    k−1   ∞ (λs)x1+...+xi =λ e−iλs ds 0 x1! ...xi! n=0 x1,x2,...,xi≥0 Z X i X Pj=1 jxj=n k−1 λx1+...+xi ∞ =λ e−iλssx1+...+xi ds x1! ...xi! 0 n=0 x1,x2,...,xi≥0 Z X i X Pj=1 jxj=n k−1 x1+...+xi λ Γ(x1 + ... + xi + 1) =λ x +...+x +1 x1! ...xi! (iλ) 1 i n=0 x1,x2,...,xi≥0 X i X Pj=1 jxj=n 6 SUPERPOSITION OF TIME-CHANGED POISSON PROCESSES AND THEIR HITTING TIMES

k−1 Γ(x1 + ... + xi + 1) = i−(x1+...+xi+1). x1! ...xi! n=0 x1,x2,...,xi≥0 X i X Pj=1 jxj=n

i Set xj = nj, 1 ≤ j ≤ i and n = x + j=1(j − 1)nj i.e. n1 + n2 + ... + ni = x in the internal summation. P

∞ k−1 Γ(n1 + n2 + ... + ni + 1) −(n1+n2+...+ni+1) P[Tk ∈ ds]= i 0 n1!n2! ...ni! Z n=0 n1,n2,...,ni≥0 X i X Pj=1 nj =x k− 1 1 x! 1 = n +n +...+n i n1!n2! ...ni! k 1 2 i n=0 n1,n2,...,ni≥0 X i X Pj=1 nj =x k− 1 1 1 1 1 x = + + ... + i i i i n=0 Xk− 1 1 = i n=0 X k = < 1 for k < i. i When k ≥ i, then

k−1 (i) (i) P[Tk ∈ ds]= P(N (s)= k − h, N (s + ds)= k) h k−i X= k−1 = P(N (i)(s)= k − h, N (i)(s)+ N (i)(ds)= k) h k−i X= k−1 = P(N (i)(s)= k − h)P(N (i)(ds)= h) h k−i X= k−1 = P(N (i)(s)= k − h)λds h k−i X= i =λ P[N (i)(s)= m]ds. m=1 X Now we have that

∞ ∞ i (i) P[Tk ∈ ds]=λ P(N (s)= m)ds 0 0 m=1 Z Z X SUPERPOSITION OF TIME-CHANGED POISSON PROCESSES AND THEIR HITTING TIMES 7

i ∞ (λs)x1+...+xi =λ  e−iλs  ds 0 x1! ...xi! Z m=1 x1,x2,...,xi≥0 X  i X   Pj=1 jxj=n    i  ∞  (λs)x1+...+xi =λ e−iλs ds 0 x1! ...xi! m=1 x1,x2,...,xi≥0 Z X i X Pj=1 jxj =n i λx1+...+xi ∞ =λ e−iλssx1+...+xi ds x1! ...xi! 0 m=1 x1,x2,...,xi≥0 Z X i X Pj=1 jxj =n i x1+...+xi λ Γ(x1 + ... + xi + 1) =λ x +...+x +1 x1! ...xi! (kλ) 1 i m=1 x1,x2,...,xi≥0 X i X Pj=1 jxj =n i Γ(x + ... + xi + 1) = 1 (k)−(x1+...+xi+1). x1! ...xi! m=1 x1,x2,...,xi≥0 X i X Pj=1 jxj =n

i Set xj = nj, 1 ≤ j ≤ i and m = x + j=1(j − 1)nj i.e. n1 + n2 + ... + ni = x. ∞ i P Γ(n1 + n2 + ... + ni + 1) −(n1+n2+...+ni+1) P[Tk ∈ ds]= (k) 0 n1!n2! ...ni! Z m=1 n1,n2,...,ni≥0 X i X Pj=1 nj =x i 1 x! 1 = n +n +...+n i n1!n2! ...ni! i 1 2 i m=1 n1,n2,...,ni≥0 X i X Pj=1 nj =x i 1 1 1 1 x = + + ... + i i i i m=1 Xi 1 i = = =1. i i m=1 X Hence, k i , 1 ≤ k ≤ i − 1 P[Tk < ∞]= . (1, k ≥ i

Remark 2.2. The above theorem states that the Poisson process of order i hits every level k when k is bigger than i and with probability k/i it will hit level k if k is less than i. This is consistent with the fact that arrivals are in packets of size i. 8 SUPERPOSITION OF TIME-CHANGED POISSON PROCESSES AND THEIR HITTING TIMES

Since the maximal size of jumps of process Y (t) is i, all states larger or equal i can be obtained with probability one, while the process can avoid with positive probability N(t) i i the ﬁrst i − 1 states. We have N1(t)+2N2(t)+ ... + iNi(t) = j=1 Xj, where Xj are independent integer-valued random variables uniformly distributed on the set {1, 2,...,i}. P 3. Weighted sum of Poisson processes In this section we study the behaviour of the process i

(3.1) Z(t)= g(j)Nj(t), j=1 X where Nj, j = 1,...,n are independent, homogeneous Poisson processes and g is an integer-valued, bounded and increasing function. We now give the characterisation of the weighted sum of Poisson processes in the following theorem. Theorem 3.1. The process Z deﬁned in (3.1) can be represented as a compound Poisson processes, that is

i N(t) d i (3.2) Z(t)= g(j)Nj(t) = g(Xj), j=1 j=1 X X i where the random variables Xj are independent, discrete and uniform on the set {1, 2,...,i} and N(t) is Poisson with parameter iλ. Proof. The probability generating function (pgf ) of Z(t) reads

Z(t) Pi g(j)N (t) E[u ]=E[u j=1 j ] i = E[ug(j)Nj (t)] j=1 Y i = exp[−λt + λtug(j)] j=1 Y i = exp[−iλt + λt ug(j)]. j=1 X Now we will ﬁnd the pgf of R.H.S. ∞ N t N t P ( ) g(Xi) P ( ) g(Xi) Eu j=1 j = E[u j=1 j |N(t)= n]P(N(t)= n) n=0 X∞ Pn g(Xi) = E[u j=1 j ]P(N(t)= n) n=0 X∞ n = Eug(Xj )P(N(t)= n) n=0 j=1 X Y SUPERPOSITION OF TIME-CHANGED POISSON PROCESSES AND THEIR HITTING TIMES 9

∞ (iλt)n = (Eug(X))ne−iλt n! n=0 X = exp(−iλt) exp iλtE[ug(X)] g(X) = exp −iλt(1 − E[u ]) i 1 = exp −iλt 1 − ug(j) i " j=1 !# X i = exp −iλt + λt ug(j) , " j=1 # X and this completes the proof. Remark 3.1. For g(j)= j, we retrieve the pgf of Y (t). Furthermore (λt)x1+...+xi P[Z(t)= n]= e−iλt . x1! ...xi! x1,...,xi≥0 i X Pj=1 g(j)xj =n

We pass now to the analysis of the hitting times of Tk,k ≥ 1 of the process Z(t). In the following theorem, we compute the distribution of Tk.

Theorem 3.2. The hitting time Tk,k ≥ 1 has distribution

k x1+...+xi −iλt (λt) (3.3) P[Tk ∈ ds]= λds e , s> 0. x1! ...xi! h=1 x1,...,xi≥0 g(Xh)≤k i X Pj=1 g(j)xj =k−g(h) Proof. Recall that

Z(t)= g(1)N1(t)+ g(2)N2(t)+ ... + g(i)Ni(t). We begin with i

P[Tk ∈ ds]= P[Z(s)= k − g(h),Z(s + ds)= k] h=1 gX(h)≤i k = P[Z(s)= k − g(h),Z(s)+ dZ(s)= k] h=1 g(Xh)≤k k = P[Z(s)= k − g(h)]P[dZ(s)) = g(h)] h=1 g(Xh)≤k k = P[Z(s)= k − g(h)]λds h=1 g(Xh)≤k 10 SUPERPOSITION OF TIME-CHANGED POISSON PROCESSES AND THEIR HITTING TIMES

k (λs)x1+...+xi =λds e−iλs , x1! ...xi! h=1 x1,...,xi≥0 g(Xh)≤k i X Pj=1 g(j)xj =k−g(h) which completes the proof.

We now ﬁnd the probability that Z hits an arbitrary level k. Theorem 3.3. The probability of hitting any level k is given by #(h :1 ≤ g(h) ≤ k) , 1 ≤ k

∞ ∞ k P[Tk < ∞]= P[Tk ∈ ds]= P(Y (s)= k − g(h))λds Z0 Z0 h=1 g(Xh)≤k

k ∞ (λs)x1+...+xi =λ  e−iλs  ds 0 x1! ...xi! Z h=1 x1,x2,...,xi≥0 g(Xh)≤k  k X  Pj=1 g(j)xj =k−g(h)    k   ∞ (λs)x1+...+xi =λ e−iλs ds 0 x1! ...xi! h=1 x1,x2,...,xi≥0 Z g(Xh)≤k i X Pj=1 g(j)xj =k−g(h) k λx1+...+xi ∞ =λ e−iλssx1+...+xi ds x1! ...xi! 0 h=1 x1,x2,...,xi≥0 Z g(Xh)≤k i X Pj=1 g(j)xj =k−g(h) k x1+...+xi λ Γ(x1 + ... + xi + 1) =λ x +...+x +1 x1! ...xi! (iλ) 1 i h=1 x1,x2,...,xi≥0 g(Xh)≤k i X Pj=1 g(j)xj =k−(h) k Γ(x + ... + xi + 1) = 1 k−(x1+...+xi+1) x1! ...xi! h=1 x1,x2,...,xi≥0 g(Xh)≤k i X Pj=1 g(j)xj =k−g(h) k 1 = i h=1 g(Xh)≤k #(h :1 ≤ g(h) ≤ k) = . i SUPERPOSITION OF TIME-CHANGED POISSON PROCESSES AND THEIR HITTING TIMES 11

Now assume k ≥ g(i), we get distribution of Tk as k−g(1)

P[Tk ∈ ds]= P[Z(s)= k − h, Z(s + ds)= k] − h=Xk g(i) k−g(1) = P[Z(s)= k − h, Z(s)+ dZ(ds)= k] − h=Xk g(i) k−g(1) = P[Z(s)= k − h]P[dZ(s)= h] − h=Xk g(i) k−g(1) =λds P[Z(s)= k − h] − h=Xk g(i) g(i) (3.5) =λds P[Z(s)= h]ds. hX=g(1) We now obtain the hitting time probability as follows ∞ P[Tk < ∞]= P[Tk ∈ ds] Z0 ∞ g(i) =λ P[Z(s)= h]ds 0 Z hX=g(1)

g(i) ∞ (λs)x1+...+xi =λ  e−iλs  0 x1! ...xi! Z h=g(1) x1,...,xi≥0 X  i X  Pj=1 g(j)xj =h    g(i)   λx1+...+xi ∞ =λ e−iλssx1+...+xi ds x1! ...xi! 0 h=g(1) x1,...,xi≥0 Z X i X Pj=1 g(j)xj =h g(i) x + ... + xi! = 1 i−(x1+...+xi+1) x1! ...xi! h=g(1) x1,...,xi≥0. X i X Pj=1 g(j)xj =h i Set x = x1 + ... + xi, xj = nj, 1 ≤ j ≤ i and h = x + j=1[g(j) − 1]xj in the inner summation g(i) P g(i) (n + ... + ni)! 1 (3.6) = 1 i−(n1+...+ni+1) = =1. n1! ...ni! i h=g(1) n1,...,ni≥0 h=g(1) X i X X Pj=1 nj =x From (3.5) and (3.6), we complete the proof. 12 SUPERPOSITION OF TIME-CHANGED POISSON PROCESSES AND THEIR HITTING TIMES

4. Weighted sums of subordinated Poisson processes In this section, we study the weighted sums of subordinated Poisson process time- changed by the Lévy subordinator. We derive its characterisation in the form of the compound Poisson process and compute its hitting time probabilities. The study of the of subordinated Poisson process is done by Maheshwari and Vellaisamy (2019) (see [6]) and results about their hitting time probabilities can be found in Orsingher and Toaldo (2015) (see [9]). We now briefly introduce the Lévy subordinator. Let f be a Bernˇstein function with integral representation ∞ (4.1) f(x)= (1 − e−sx)ν(ds), Z0 where ν is the Lévy measure, that is a non-negative measure such that ∞ (4.2) min(s, 1)ν(ds) < ∞. Z0 An alternative definition of Bernˇstein function is that, for all n it happens that (−1)nf (n)(x) ≤ 0. We denote by Hf the subordinator related to the Bernˇstein function f or to the related Lévy measure ν. The Laplace transform of Hf reads

f (4.3) E[e−µH (t)]= e−tf(µ). − α αµα 1 For example, if f(µ) = µ , 0 <α< 1, ν(dx) = Γ(1−α) , we have the stable subordinator of order α. We now prove the equality in distribution of W with its compound Poisson representation. Theorem 4.1. The following identity holds

i N(Hf (t),iλ) f d i (4.4) W (t)= jNj(H (t),λ) = Xj, j=1 j=1 X X where Xj, j =1, 2 ... are independent, uniformly distributed random variables (on the set of natural numbers 1 ≤ j ≤ i), Nj(•,λ) are independent homogeneous Poisson processes f as well as Nj(•, iλ). Of course H (t) are L´evy subordinators related to the Bernˇstein function f. Proof. The probability generating function of (4.4) reads

N Hf t ,iλ N Hf t ,iλ P ( ( ) ) Xi P ( ( ) ) Xi f E[u j=1 j ]=E[E[u j=1 j |H (t)]] f f 2 i =E[e−iλH (t)+λH (t)(u+u +...+u )] 2 i f =E[e−[iλ−λ(u+u +...+u )]H (t)] 2 i =e−tf[iλ−λ(u+u +...+u )]. SUPERPOSITION OF TIME-CHANGED POISSON PROCESSES AND THEIR HITTING TIMES 13

On the other hand ∞ Pi jN (Hf (t)) Pi jN (Hf (t)) f f E[u j=1 j ]= E[u j=1 j |H (t)= y]P[H (t) ∈ dy] 0 Z ∞ Pi jN (y) f = E[u j=1 j ]P[H (t) ∈ dy] 0 Z ∞ 2 i = e−iλy+λy(u+u +...+u )P[Hf (t) ∈ dy] 0 Z 2 i f =E[e−[iλ−λ(u+u +...+u )]H (t)] 2 i =e−tf[iλ−λ(u+u +...+u )] and this concludes the proof.

i f We now pass to the analysis of the hitting times of the process W (t)= j=1 jNj(H (t)). We observe that also in this case we can write for P Tk = inf{s < t : W (s)= k} that k

P[Tk ∈ ds]= P[W (s)= k − h, W (s + ds)= k] h X=1 k f f f f = P[W (s)= k − h, N1(H (s)+ dH (s)) + ... + iNi(H (s)+ dH (s)) = k] h X=1 k f f = P[W (s)= k − h, N1(dH (s))+ ... + iNi(dH (s)) = h] h X=1 k = P[W (s)= k − h]P[dW (s)= h], h X=1 where (4.5) 1 − (ds)f(iλ)+ o(h), h =0  P[dW (s)= h]=  .  x ... x  (−λ) 1+ + i 1 d(x1+...+xi) (−ds) x +...+x (x +...+x ) f(iλ) + o(ds), h =1, 2,...   x1!...xi! i 1 i dλ 1 i  x1,x2,...,xi≥0 i jx h   Pj=1P j=        and  ∞ P[W (s)= k − h]= P[W (s)= k − h|Hf (s)= y]P[Hf (s) ∈ dy] 0 Z ∞ f = P[N1(y)+ ... + iNi(y)= k − h]P[H (s) ∈ dy] Z0 14 SUPERPOSITION OF TIME-CHANGED POISSON PROCESSES AND THEIR HITTING TIMES

∞ (λy)x1+...+xi = e−iλy P[Hf (s) ∈ dy] 0 x1! ...xi! Z x1,...,xi≥0 i X Pj=1 jxj =k−h λx1+...+xi ∞ = e−iλyyx1+...+xi P[Hf (s) ∈ dy] x1! ...xi! 0 x1,...,xi≥0 Z i X Pj=1 jxj =k−h (−λ)x1+...+xi 1 ∞ dx1+...+xi (e−iλy) = P[Hf (s) ∈ dy] x1+...+xi x1+...+xi x1! ...xi! i 0 dλ x1,...,xi≥0 Z i X Pj=1 jxj =k−h (−λ)x1+...+xi dx1+...+xi ∞ = e−iλyP[Hf (s) ∈ dy] x1+...+xi x1+...+xi x1! ...xi!i dλ 0 x1,...,xi≥0 Z i X Pj=1 jxj =k−h

x1+...+xi x1+...+xi (−λ) d E −iλHf (s) = x +...+x x +...+x [e ] x1! ...xi!i 1 i dλ 1 i x1,...,xi≥0 i X Pj=1 jxj =k−h

x1+...+xi x1+...+xi (−λ) d −sf(iλ) = x +...+x x +...+x e . x1! ...xi!i 1 i dλ 1 i x1,...,xi≥0 i X Pj=1 jxj =k−h In conclusion the hitting time of the state k has probability

k x1+...+xi x1+...+xi P (−λ) d 1 [Tk < ∞]= x +...+x x +...+x x1! ...xi!i 1 i dλ 1 i f(iλ) h=1 x1,...,xi≥0 X i X Pj=1 jxj =k−h (−λ)x1+...+xi d(x1+...+xi) × (x +...+x ) f(iλ), x1! ...xi! dλ 1 i x1,...,xi≥0 i X Pj=1 jxj =h as shown by integrating with respect to s and substituting (4.5) in the expression of P[W (s) = k − h]. This result shows that the explicit distribution of the hitting time Tk crucially depends on the Bernˇstein function f as pointed out in [2] and [9]. We work a special case of the results obtained in here in the next section.

5. A Special case We here study the process

i λ β (5.1) U(t)= jNj (N (t)), j=1 X which is a special case of W (t), where the time change is performed by means of a λ homogeneous Poisson process of rate β independent from Nj , 1 ≤ j ≤ i. We prove in the SUPERPOSITION OF TIME-CHANGED POISSON PROCESSES AND THEIR HITTING TIMES 15 next section that U(t),t > 0 can be represented in the form of a compound Poisson as shown in the next Theorem. Theorem 5.1. The following relationship holds

i N iλ(N β (t)) λ β d i (5.2) U(t)= jNj (N (t)) = Xj, j=1 j=1 X X i where the Xj are independent random variables, uniformly distributed on the set {1,...,i} and N iλ(t) is a Poisson process of rate iλ independent from N β(t) and independent from i all the Xj random variables. Proof. The probability generating function of the right-hand side of (5.2) is ∞ ∞ Niλ Nβ t P ( ( )) Xi X k λ β E[u j=1 j ]= Eu P[N (r)= k]P[N (t)= r] k r=0 X=0 X ∞ r (βt) X = e−βt e−iλr+λirEu r! r=0 X ∞ r (βt) 1 i = e−βt e−iλr+λr(u +...+u ) r! r=0 X −[iλ−λ(u1+...+ui)] =e−βt+βt(e ). The probability generating function of the left-hand side of (5.2) writes

∞ i i jN λ N β t jN λ N β t β β E[uPj=1 j ( ( ))]= E[u j ( ( ))|N (t)= k]P[N (t)= k] k j=1 X=0 Y ∞ i k j (βt) = e−λk(1−u )e−βt k! k j=1 X=0 Y ∞ k 1 j (βt) = e−λki+λk(u +...+u )e−βt k! k X=0 −[iλ−λ(u1+...+ui)] =e−βt+βt(e ), and this concludes the proof. Remark 5.1. In view of (2.1), we can write down the distribution of process (5.1) as

∞ i P P λ P β [U(t)= m]= jNj (r)= m N (t)= r , m ≥ 1 r=1 " j=1 # X∞ X (βt)r (λr)x1+...+xi = e−βt e−iλr r! x1! ...xi! r=1 x1,x2,...,xi≥0 X i X Pj=1 jxj =m 16 SUPERPOSITION OF TIME-CHANGED POISSON PROCESSES AND THEIR HITTING TIMES

x1+...+xi λ β (5.3) = E e−iλN (t)(N β(t))x1+...+xi . x1! ...xi! x1,x2,...,xi≥0 i X h i Pj=1 jxj =m We now consider the hitting time distribution of the process U(t),t> 0. We can write that k

(5.4) P[Tk ∈ ds]= P[U(s)= k − h]P[dU(s)= k]. h X=1 Note that the increment of the iterated Poisson process N α(N β(t)) has distribution

βe−αds +1 − βds if r =0 P α β (5.5) [dN (N (t)) = r]=  α  e−α βds if r ≥ 1  r! as can be ascertained by observing that ∞ P[dN α(N β(t)) = r]= P[dN α(m)= r]P[N β(t)= m] m=1 X and that dN β(t) = 0 implies that N α(0) = 0. The increment of the process U(t) has distribution 1 − βdt + e−αiβdt if r =0 P (5.6) [dU(t)= r]=  P i α  βdt j=1 jNj (1) = r if r ≥ 1. hP i From (2.5) we have that  i x1+...+xi P α −iα α (5.7) jNj (1) = r = e . x1! ...xi! " j=1 # x1,x2,...,xi≥0 X i X Pj=1 jxj =r From (5.4) and formulas (5.3) and (5.6) we can write down the distribution of the hitting time, for k ≥ 1 (5.8) k i x1+...+xi λ β P E −iλN (s) β x1+...+xi P α [Tk ∈ ds]= e (N (s)) βds jNj (1) = h . x1! ...xi! h=1 x1,x2,...,xi≥0 " j=1 # X i X h i X Pj=1 jxj =k−h By integrating with respect to the time s we get the hitting probability of the state k as

k ∞ i λx1+...+xi P −iλr x1+...+xi P α (5.9) [Tk < ∞]= e r jNj (1) = h x1! ...xi! h=1 x1,x2,...,xi≥0 r=0 " j=1 # X i X X X Pj=1 jxj=k−h SUPERPOSITION OF TIME-CHANGED POISSON PROCESSES AND THEIR HITTING TIMES 17 because ∞ ∞ ∞ β E e−iλN (s)(N β(s))x1+...+xi ds = e−iλrrx1+...+xi P N β(s)= r ds Z0 Z0 r=0 h i ∞X 1 = e−iλrrx1+...+xi . β r=0 X We are able to evaluate (5.3) for k = 1 and obtain

∞ i x1+...+xi P λ −iλrP α [T1 < ∞]= e jNj (1) = 1 x1! ...xi! x1,x2,...,xi≥0 r=0 " j=1 # i X X X Pj=1 jxj =0 ∞ λx1+...+xi = e−iλr e−iλ x1! ...xi! r=0 x1,x2,...,xi≥0 X i X Pj=1 jxj=1 ∞ 1 =e−iλ e−iλyλ = λe−iλ < 1. 1 − e−iλ y=0 X This result coincides with equation (37) of Orsingher and Polito (2012) (see [7]) for a rate λα = iλ. For the hitting time T2 we have that (see formula (5.9))

∞ i ∞ i P −iλrP α −iλr P α [T2 < ∞]= e jNj (1) = 2 + e λr jNj (1) = 1 r=0 " j=1 # r=1 " j=1 # X∞ X ∞X X λ2 = e−iλr λe−iλ + e−iλ + e−iλrλrλe−iλ 2 r=0 r=1 X X∞ λ2 1 =e−iλ λ + + λ2e−iλ re−iλr 2 1 − e−iλ r=1 X λ2 e−iλ λ2e−2iλ = λ + + 2 1 − e−iλ (1 − e−iλ)2 λ = 1+ P[T < ∞]+(P[T < ∞])2 > P[T < ∞]. 2 1 1 1 Furthermore P[T2 < ∞] < 1 because λ2 e−iλ λ2e−2iλ λ + + < 1. 2 1 − e−iλ (1 − e−iλ)2 This inequality, after same manipulation becomes 2 2 λ λ 2 e−iλ λ + + e−2iλ −λ + < 1 − e−iλ 2 2 =1+ e−2λi − 2e−iλ 18 SUPERPOSITION OF TIME-CHANGED POISSON PROCESSES AND THEIR HITTING TIMES or λ2 λ2 e−iλ λ + +2 + e−2iλ −1 − λ + < 1. 2 2 The left-hand side of the above equation can be increased as λ2 λ2 e−iλ λ + +2+ − λ − 1 = 1+ λ2 e−iλ < 1 2 2 since 1 + λ2 < eiλ.

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