- Email: [email protected]
Growth rates for pure birth Markov chains
Statistics and Probability Letters 78 (2008) 1534–1540 www.elsevier.com/locate/stapro
Growth rates for pure birth Markov chains K.B. Athreya Departments of Mathematics and Statistics, Iowa State University, Ames, IA 50011, United States Received 15 December 2006; received in revised form 2 November 2007; accepted 3 January 2008 Available online 15 January 2008
Abstract A pure birth Markov chain is a continuous time Markov chain {Z (t) : t ≥ 0} with state space S ≡ {0, 1, 2, . . .} such that for each i ≥ 0 the chain stays in state i for a random length of time that is exponentially distributed with mean λi−1 and then jumps to (i + 1). Suppose b(·) is a function from (0, ∞) → (0, ∞) that is nondecreasing and ↑ ∞. This paper addresses the two questions: (1) Given {λi }i≥0 what is the growth rate of Z (t)? (2) Given b(·) does there exist {λi } such that Z (t) grows at rate b(t)? c 2008 Elsevier B.V. All rights reserved.
MSC: primary 60J27; secondary 92D25
1. Introduction A pure birth Markov chain is a continuous time Markov chain {Z (t) : t ≥ 0} with state space S ≡ {0, 1, 2, . . .} such that ∀ i ≥ 0, the chain stays in state i for a random length of time that is exponentially distributed with mean λi−1 and then jumps to (i + 1) (see Karlin and Taylor (1975)). Let b(·) : [0, ∞) → [0, ∞) be nondecreasing ↑ ∞. This paper addresses the two questions: (1) Given {λi }i≥0 what is the growth rate of Z (t)? (2) Given b(·) does there exist {λi } such that Z (t) grows at rate b(t)? It is known that: (i) if λi ≡ λ then {Z (t) : t ≥ 0} is a Poisson process with rate λ and hence as t → ∞, Z (t) t →λ w.p. 1, (ii) if λi ≡ iλ and if Z (0) 6= 0 then {Z (t) : t ≥ 0} is the so-called Yule process (and a special case of Markov branching process) and hence that Z (t)e−λt converges w.p. 1 to a random variable Y that is exponentially distributed with mean one (see Athreya and Ney (2000)), (iii) if λi ≡ iλ + β with β > 0, {Z (t) : t ≥ 0} is a Yule process with constant immigration (at rate β) and Z (t)e−λt converges to a strictly positive random variable. This raises the question √ of what other rates are possible? For example, can Z (t) grow like t p for some 0 < p < ∞, or like log t or like e t ? Also, does the exponential growth hold if λi is only asymptotically linear, i.e., λi ∼ = iλ + β for i large? This paper presents some results for these questions. Theorem 2.1 gives the necessary and sufficient condition for exponential growth while Theorem 2.2 does that for polynomial growth. Theorem 2.3 gives sufficient conditions for a general growth rate b(·). 2. Main results Assume throughout that either λ0 > 0 or Z (0) 6= 0 or both. E-mail address: [email protected]. c 2008 Elsevier B.V. All rights reserved. 0167-7152/$ - see front matter doi:10.1016/j.spl.2008.01.016
K.B. Athreya / Statistics and Probability Letters 78 (2008) 1534–1540
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Theorem 2.1. Let {λi }i≥1 satisfy ∞ X 1 = ∞, λ i=1 i
∞ X 1 < ∞. 2 λ i=1 i
(2.1)
Then, for some 0 < c < ∞, lim Z (t)e−ct = ξ
exists w.p. 1
n→∞
(2.2)
with P(0 < ξ < ∞) = 1 iff n X 1 1 − n→∞ λi ci i=1 lim
exists and is finite.
Theorem 2.2. Let {λi }i≥1 satisfy ! !2 n ∞ n X X X 1 1 = ∞ and lim λi = 0. 2 n→∞ λ λ i i=1 i=1 i i=1
(2.3)
(2.4)
Then, for some 0 < c < ∞, 0 < q < ∞ lim
t→∞
Z (t) =c tq
(2.5)
in probability
iff lim
n→∞
n X
! λi n − p = c−1/ p
where p =
i=1
1 . q
(2.6)
Also under (2.1) and (2.6), the convergence in (2.5) holds w.p. 1. Theorem 2.3. Let b : (0, ∞) → (0, ∞) satisfy (i) b(t) ↑ ∞ as t ↑ ∞. (ii) b(·) is differentiable and b0 (t) ↑ in t. R∞ (iii) 1 (b0 (b−11 (t)))2 dt < ∞. Let λi ≡ b0 (b−1 (i)), i ≥ 1. (a) Then, a pure birth Markov chain {Z (t) : t ≥ 0} with rates {λi }i≥0 satisfies t − a(Z ˜ (t)) → Y˜
w.p. 1 as t ↑ ∞
(2.7)
where a(t) ˜ ≡ b−1 (t) and Y˜ is a proper random variable with an absolutely continuous distribution. (b) If, further, limt↑∞ b(t+(t)) = 1 for any (t) → 0 as t → ∞, then b(t) Z (t) b(t − Y˜ )
→1
w.p. 1 as t ↑ ∞.
Theorem 2.4. Let b : (0, ∞) → (0, ∞) satisfy:
(2.8)
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K.B. Athreya / Statistics and Probability Letters 78 (2008) 1534–1540
(i) b(t) ↑ ∞ as t ↑ ∞. (ii) If a(t) ≡ b−1 (t) ≡ inf{x : b(x) ≥ t} then λn ≡ (a(n) − a(n − 1))−1 > 0
∀n ≥ 1
and n P
1 λ2 j=1 j a 2 (n)
→ 0.
(a) Then, a pure birth Markov Chain {Z (t) : t ≥ 0} with rates {λi }i≥1 satisfies lim
t→∞
(b) If further,
a(Z (t)) = 1 in probability. t
b(a(n)) n
→ 1 as n → ∞ and
b (t (1 + (t))) =1 t→∞ b(t) for any (t) → 0 as t → ∞ then Z (t) = 1 in probability. lim t→∞ b(t) lim
Remark. Theorems 2.3 and 2.4 give fairly general conditions under which Z (t) grows like b(t) w.p. 1 or in probability. 3. Proofs Proposition 3.1. Let {λi }i≥1 satisfy ∞ ∞ X X 1 1 = ∞, < ∞. (3.1) 2 λ i=1 i i=1 λi Pn 1 Let a(n) = i=1 λi . Let {Z (t) : t ≥ 0} be a pure birth process with rates {λi }i≥1 . Then, as t ↑ ∞, (t − a(Z (t))) converges w.p. 1 to a random variable Y such that EY = 0, EY 2 < ∞, and Y has an absolutely continuous distribution.
Proof. Let T0 = 0 < T1 < T2 < T3 < · · · be the jump times (birth times) of the pure birth process {Z (t) : t ≥ 0}. Then, L i = Ti − Ti−1 , i = 1, 2, . . . are independent, exponentially distributed random variables with E L i = λ1i . Thus ( ) n X 1 Yn ≡ Tn − a(n) = Li − λi i=1 n≥1
is a martingale. Further, since
P∞
1 i=1 λ2 i
< ∞ and EYn2 = V (Yn ) =
Pn
i=1
V (L i ) =
Pn
1 i=1 λ2 i
it follows that
is a L 2 bounded martingale. Hence Yn converges w.p. 1 and in L 2 to a random variable Y with EY = 0, {Yn }n≥1 P ∞ 1 2 EY = i=1 2 < ∞ (see Athreya and Lahiri (2006), Theorem 13.3.9). Also λi
P(L i > x) = e−λi x ∀x > 0 for i ≥ 1, ∞ ∞ X 1 1 X ⇒ P(L i > x) ≤ 2 <∞ x i=1 λi2 i=1 ⇒ Li → 0
w.p. 1 as i → ∞.
(3.2)
For any t > 0, TZ (t) ≤ t < TZ (t) + L Z (t) .
(3.3)
K.B. Athreya / Statistics and Probability Letters 78 (2008) 1534–1540
Also, for any s > 0, n Y E e−sTn =
λi . (λi + s)
i=1
Since lim Tn ≡ T∞ exists w.p. 1, ∞ Y λi E e−sT∞ = , (λ + s) i i=1 P∞ 1 Also, since i=1 λi = ∞, E(e−sT∞ ) = 0
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∀s > 0.
∀s > 0
⇒ P(T∞ < ∞) = 0.
(3.4)
Now (3.2)–(3.4) show that as t ↑ ∞, w.p. 1 Z (t) ↑ ∞ and t − TZ (t) → 0. Since Yn = Tn − a(n) converges w.p. 1 to Y as n → ∞, it follows that t − a(Z (t)) → Y
w.p. 1. P∞ Since Y = L 1 − + i=2 (L i − λ1i ) = Y1 + (Y − Y1 ), say, and Y1 and Y − Y1 are independent and Y1 has an absolutely continuous distribution, so does Y . 1 λ1
Corollary 3.1. Under (3.1), a (Z (t)) →1 t
w.p. 1, as t → ∞.
Proposition 3.2. Let {λi }i≥1 satisfy (2.4). Then, as t → ∞, a (Z (t)) → 1 in probability t Pn 1 where a(n) = i=1 λi , n ≥ 1.
(3.5)
Proof. Let {Tn }n≥1 be as in Proposition 3.1. Let an = a(n), n ≥ 1. Then, Tn E =1 an and ! n X 1 Tn 1 = V . 2 an an2 i=1 λi By (2.4) V aTnn → 0 and hence
Tn →1 an
in probability.
(3.6)
Next, P(L n > an ) = P(λn L n > λn an ) ≤ e−λn an . Also by (2.4), (λn an )−2 → 0. So λn an → ∞ and hence Ln →0 an
in probability.
(3.7)
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K.B. Athreya / Statistics and Probability Letters 78 (2008) 1534–1540
Since ∀ i, λi > 0, ∀ n ≥ 1, Tn < ∞ w.p. 1 and hence Z (t) ↑ ∞ w.p. 1 as t ↑ ∞. Also since TZ (t) ≤ t < TZ (t)+1 = TZ (t) + L Z (t) , it follows from (3.6) and (3.7) that as t → ∞, t →1 a Z (t)
in probability
which implies (3.5).
Proof of Theorem 2.1. By Proposition 3.1, under (2.1) lim (t − a (Z (t))) ≡ Y
t→∞
exists in R w.p. 1.
Now (2.2) holds iff lim (log Z (t) − ct) = log ξ
t→∞
exists in R w.p. 1.
Thus under (2.1) and (2.2) holds iff log ξ 1 log Z (t) − a (Z (t)) = +Y lim t→∞ c c
exists in R.
Since Z (t) goes up by one at each step this holds iff (2.3) holds.
Proof of Theorem 2.2. By Proposition 3.2, under (2.4) a (Z (t)) →1 t Thus (2.5) holds iff a (Z (t)) (Z (t))1/q
in probability.
→ c−1/ p
as t → ∞.
Since Z (t) goes up one at each step this holds iff (2.6) holds.
−1 Proof of Theorem 2.3. Let a(t) ˜ ≡ b−1 (t), t > 0 and w(t) ≡ a˜ 0 (t) . Then w(·) is monotone and so the hypothesis of Theorem 2.3 imply those of Proposition 3.1. So w.p. 1, t − a (Z (t)) → Y w.p. 1. Again since w(·) is monotone the hypothesis of Theorem 2.3 imply that lim (a(n) ˜ − a(n)) ≡ α1
n→∞
exists.
Thus t − a˜ (Z (t)) → Y + α1 ≡ Y˜ w.p. 1. But a(t) ˜ = b−1 (t). Let (t) = a˜ (Z (t)) − t + Y˜ . Then Z (t) = b t − Y˜ + (t) . By hypothesis in (b) of Theorem 2.3, it follows that lim
t↑∞
Z (t) b(t − Y˜ )
= 1.
Again, Y˜ has an absolutely continuous distribution.
Proof of Theorem 2.4. The hypotheses of Theorem 2.4 imply that (2.4) holds. So by Proposition 3.2 a(Z (t)) →1 t
in probability
proving (a). Letting (t) = 1 −
a(Z (t)) t
one sees that (b) follows.
K.B. Athreya / Statistics and Probability Letters 78 (2008) 1534–1540
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4. Some examples 1 1 ct ci+β + i 2 , c > 0}i≥1 satisfies (2.1) and Z (t) grows like e . limi→∞ λi i −θ = α3 , 0 < α3 < ∞, θ < 1. Then (2.4) and (2.6) hold. In
1. (Exponential growth) Clearly, { λ1i ≡ 2. (Polynomial growth) If 1 grows like t q , q = 1−θ .
this case, Z (t)
p
3. (Exponential to a power) b(t) ≡ αet , α > 0, 0 < p ≤ 1 satisfies the conditions of Theorem 2.3. Here p λi ≡ w(i) = p 1/ p i(log αi )( p−1)/ p , i ≥ 1 and, hence, Z (t) grows like et . t
4. (Iterated exponential) b(t) = ee ⇒ a(t) = b−1 (t) = log log t ⇒ w(t) = conditions of Proposition 3.1 so t − log log Z (t) → Y w.p. 1
1 t log t
so w(i) ≡ i log i, i ≥ 2 satisfies
log Z (t) → e−Y w.p. 1 et t ⇒ Z (t) grows like ee . ⇒
Pn i 5. (Logarithmic.) Let w(i) = e−i , i ≥ 1. So Tn = i=1 e L i where {L i } are iid exp(1) random variables Pn Tn ⇒ en = i=1 e(i−n) L i and hence has the same distribution as Xn ≡
n X
e− j L j .
j=1
P −j Since E L j ≡ 1, X n converges w.p. 1 to X ∞ ≡ ∞ j=1 e L j < ∞ w.p. 1. Thus ! ∞ X log Tn d d −j log Tn − n −→ log −→ 1. e Lj ⇒ n j=1 Also P
⇒
log Tn >1+ n
∞ X
Tn =P = P Tn > e > en en Tn 1 1 1 ≤ E ≤ n n −1 e e (1 − e ) en
n(1+)
P(log Tn > n(1 + )) < ∞.
i=1
By the Borel–Cantelli, w.p. 1, lognTn < 1 + for large n. Next, log Tn P < 1 − = P(log Tn < n(1 − )) n = P(Tn < en(1−) ) Tn −n < e = P en ≤ P(L n < e−n ) = P(L 1 < e−n ) ∼ = c · e−n , 0 < c < ∞, for n large ∞ X log Tn < 1 − < ∞. ⇒ P n i=1 So by the Borel–Cantelli, w.p. 1 follows that log t → 1 w.p. 1. Z (t)
log Tn n
< 1 − for large n. So
log Tn n
→ 1 w.p. 1. Since TZ (t) ≤ t < TZ (t)+1 it
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K.B. Athreya / Statistics and Probability Letters 78 (2008) 1534–1540
5. Remark The referee raised the following question on an earlier version of Theorem 2.1 of the paper. Does there exist {λi } such that λi ∼ + β for some cP> 0, β ∈ R, (2.3) does not hold and still (2.2) holds? If λi ∼ = = ci + β for some c > 0, Pci ∞ 1 ∞ 1 < ∞ and so β ∈ R, then i=1 = ∞ and (2.1) holds. By the present version of Theorem 2.1, (2.2) holds i=1 2 λi λi
iff (2.3) holds. i log i Thus the answer to the referee’s question is in the negative. Here is an example. Let λ1 = 1 and λi = (log i+1) , P n 1 1 1 1 ∼ i ≥ 2. Then = + , i ≥ 2. So λi = i but (2.3) does not hold. However, (3.1) does hold. Now a(n) = , λi
i
i=1 λi
i log i
n ≥ 1. It can be shown that lim (a(n) − log n − log log n) = b
n→∞
exists in R.
So from Proposition 3.1, it follows that t − log Z (t) − log(log Z (t))
converges w.p. 1,
i.e., limt→∞ e−t Z (t) log Z (t) ≡ ξ exists w.p. 1. This implies Z (t) grows slower than et . It can be shown that if λi ∼ = ci + β for some c > 0, β ∈ R, then for any c1 < c < c2 , w.p. 1 lim Z (t)e−c1 t = ∞
t→∞
lim Z (t)e−c2 t = 0.
t→∞
Acknowledgement The author would like to thank the referee for raising the above question. References Athreya, K.B., Ney, P., 2000. Branching Processes. Dover. Athreya, K.B., Lahiri, S.N., 2006. Measure Theory and Probability Theory. Springer, New York. Karlin, S., Taylor, H.M., 1975. A First Course in Stochastic Processes. Academic Press, New York.
Growth rates for pure birth Markov chains K.B. Athreya Departments of Mathematics and Statistics, Iowa State University, Ames, IA 50011, United States Received 15 December 2006; received in revised form 2 November 2007; accepted 3 January 2008 Available online 15 January 2008
Abstract A pure birth Markov chain is a continuous time Markov chain {Z (t) : t ≥ 0} with state space S ≡ {0, 1, 2, . . .} such that for each i ≥ 0 the chain stays in state i for a random length of time that is exponentially distributed with mean λi−1 and then jumps to (i + 1). Suppose b(·) is a function from (0, ∞) → (0, ∞) that is nondecreasing and ↑ ∞. This paper addresses the two questions: (1) Given {λi }i≥0 what is the growth rate of Z (t)? (2) Given b(·) does there exist {λi } such that Z (t) grows at rate b(t)? c 2008 Elsevier B.V. All rights reserved.
MSC: primary 60J27; secondary 92D25
1. Introduction A pure birth Markov chain is a continuous time Markov chain {Z (t) : t ≥ 0} with state space S ≡ {0, 1, 2, . . .} such that ∀ i ≥ 0, the chain stays in state i for a random length of time that is exponentially distributed with mean λi−1 and then jumps to (i + 1) (see Karlin and Taylor (1975)). Let b(·) : [0, ∞) → [0, ∞) be nondecreasing ↑ ∞. This paper addresses the two questions: (1) Given {λi }i≥0 what is the growth rate of Z (t)? (2) Given b(·) does there exist {λi } such that Z (t) grows at rate b(t)? It is known that: (i) if λi ≡ λ then {Z (t) : t ≥ 0} is a Poisson process with rate λ and hence as t → ∞, Z (t) t →λ w.p. 1, (ii) if λi ≡ iλ and if Z (0) 6= 0 then {Z (t) : t ≥ 0} is the so-called Yule process (and a special case of Markov branching process) and hence that Z (t)e−λt converges w.p. 1 to a random variable Y that is exponentially distributed with mean one (see Athreya and Ney (2000)), (iii) if λi ≡ iλ + β with β > 0, {Z (t) : t ≥ 0} is a Yule process with constant immigration (at rate β) and Z (t)e−λt converges to a strictly positive random variable. This raises the question √ of what other rates are possible? For example, can Z (t) grow like t p for some 0 < p < ∞, or like log t or like e t ? Also, does the exponential growth hold if λi is only asymptotically linear, i.e., λi ∼ = iλ + β for i large? This paper presents some results for these questions. Theorem 2.1 gives the necessary and sufficient condition for exponential growth while Theorem 2.2 does that for polynomial growth. Theorem 2.3 gives sufficient conditions for a general growth rate b(·). 2. Main results Assume throughout that either λ0 > 0 or Z (0) 6= 0 or both. E-mail address: [email protected]. c 2008 Elsevier B.V. All rights reserved. 0167-7152/$ - see front matter doi:10.1016/j.spl.2008.01.016
K.B. Athreya / Statistics and Probability Letters 78 (2008) 1534–1540
1535
Theorem 2.1. Let {λi }i≥1 satisfy ∞ X 1 = ∞, λ i=1 i
∞ X 1 < ∞. 2 λ i=1 i
(2.1)
Then, for some 0 < c < ∞, lim Z (t)e−ct = ξ
exists w.p. 1
n→∞
(2.2)
with P(0 < ξ < ∞) = 1 iff n X 1 1 − n→∞ λi ci i=1 lim
exists and is finite.
Theorem 2.2. Let {λi }i≥1 satisfy ! !2 n ∞ n X X X 1 1 = ∞ and lim λi = 0. 2 n→∞ λ λ i i=1 i=1 i i=1
(2.3)
(2.4)
Then, for some 0 < c < ∞, 0 < q < ∞ lim
t→∞
Z (t) =c tq
(2.5)
in probability
iff lim
n→∞
n X
! λi n − p = c−1/ p
where p =
i=1
1 . q
(2.6)
Also under (2.1) and (2.6), the convergence in (2.5) holds w.p. 1. Theorem 2.3. Let b : (0, ∞) → (0, ∞) satisfy (i) b(t) ↑ ∞ as t ↑ ∞. (ii) b(·) is differentiable and b0 (t) ↑ in t. R∞ (iii) 1 (b0 (b−11 (t)))2 dt < ∞. Let λi ≡ b0 (b−1 (i)), i ≥ 1. (a) Then, a pure birth Markov chain {Z (t) : t ≥ 0} with rates {λi }i≥0 satisfies t − a(Z ˜ (t)) → Y˜
w.p. 1 as t ↑ ∞
(2.7)
where a(t) ˜ ≡ b−1 (t) and Y˜ is a proper random variable with an absolutely continuous distribution. (b) If, further, limt↑∞ b(t+(t)) = 1 for any (t) → 0 as t → ∞, then b(t) Z (t) b(t − Y˜ )
→1
w.p. 1 as t ↑ ∞.
Theorem 2.4. Let b : (0, ∞) → (0, ∞) satisfy:
(2.8)
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K.B. Athreya / Statistics and Probability Letters 78 (2008) 1534–1540
(i) b(t) ↑ ∞ as t ↑ ∞. (ii) If a(t) ≡ b−1 (t) ≡ inf{x : b(x) ≥ t} then λn ≡ (a(n) − a(n − 1))−1 > 0
∀n ≥ 1
and n P
1 λ2 j=1 j a 2 (n)
→ 0.
(a) Then, a pure birth Markov Chain {Z (t) : t ≥ 0} with rates {λi }i≥1 satisfies lim
t→∞
(b) If further,
a(Z (t)) = 1 in probability. t
b(a(n)) n
→ 1 as n → ∞ and
b (t (1 + (t))) =1 t→∞ b(t) for any (t) → 0 as t → ∞ then Z (t) = 1 in probability. lim t→∞ b(t) lim
Remark. Theorems 2.3 and 2.4 give fairly general conditions under which Z (t) grows like b(t) w.p. 1 or in probability. 3. Proofs Proposition 3.1. Let {λi }i≥1 satisfy ∞ ∞ X X 1 1 = ∞, < ∞. (3.1) 2 λ i=1 i i=1 λi Pn 1 Let a(n) = i=1 λi . Let {Z (t) : t ≥ 0} be a pure birth process with rates {λi }i≥1 . Then, as t ↑ ∞, (t − a(Z (t))) converges w.p. 1 to a random variable Y such that EY = 0, EY 2 < ∞, and Y has an absolutely continuous distribution.
Proof. Let T0 = 0 < T1 < T2 < T3 < · · · be the jump times (birth times) of the pure birth process {Z (t) : t ≥ 0}. Then, L i = Ti − Ti−1 , i = 1, 2, . . . are independent, exponentially distributed random variables with E L i = λ1i . Thus ( ) n X 1 Yn ≡ Tn − a(n) = Li − λi i=1 n≥1
is a martingale. Further, since
P∞
1 i=1 λ2 i
< ∞ and EYn2 = V (Yn ) =
Pn
i=1
V (L i ) =
Pn
1 i=1 λ2 i
it follows that
is a L 2 bounded martingale. Hence Yn converges w.p. 1 and in L 2 to a random variable Y with EY = 0, {Yn }n≥1 P ∞ 1 2 EY = i=1 2 < ∞ (see Athreya and Lahiri (2006), Theorem 13.3.9). Also λi
P(L i > x) = e−λi x ∀x > 0 for i ≥ 1, ∞ ∞ X 1 1 X ⇒ P(L i > x) ≤ 2 <∞ x i=1 λi2 i=1 ⇒ Li → 0
w.p. 1 as i → ∞.
(3.2)
For any t > 0, TZ (t) ≤ t < TZ (t) + L Z (t) .
(3.3)
K.B. Athreya / Statistics and Probability Letters 78 (2008) 1534–1540
Also, for any s > 0, n Y E e−sTn =
λi . (λi + s)
i=1
Since lim Tn ≡ T∞ exists w.p. 1, ∞ Y λi E e−sT∞ = , (λ + s) i i=1 P∞ 1 Also, since i=1 λi = ∞, E(e−sT∞ ) = 0
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∀s > 0.
∀s > 0
⇒ P(T∞ < ∞) = 0.
(3.4)
Now (3.2)–(3.4) show that as t ↑ ∞, w.p. 1 Z (t) ↑ ∞ and t − TZ (t) → 0. Since Yn = Tn − a(n) converges w.p. 1 to Y as n → ∞, it follows that t − a(Z (t)) → Y
w.p. 1. P∞ Since Y = L 1 − + i=2 (L i − λ1i ) = Y1 + (Y − Y1 ), say, and Y1 and Y − Y1 are independent and Y1 has an absolutely continuous distribution, so does Y . 1 λ1
Corollary 3.1. Under (3.1), a (Z (t)) →1 t
w.p. 1, as t → ∞.
Proposition 3.2. Let {λi }i≥1 satisfy (2.4). Then, as t → ∞, a (Z (t)) → 1 in probability t Pn 1 where a(n) = i=1 λi , n ≥ 1.
(3.5)
Proof. Let {Tn }n≥1 be as in Proposition 3.1. Let an = a(n), n ≥ 1. Then, Tn E =1 an and ! n X 1 Tn 1 = V . 2 an an2 i=1 λi By (2.4) V aTnn → 0 and hence
Tn →1 an
in probability.
(3.6)
Next, P(L n > an ) = P(λn L n > λn an ) ≤ e−λn an . Also by (2.4), (λn an )−2 → 0. So λn an → ∞ and hence Ln →0 an
in probability.
(3.7)
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K.B. Athreya / Statistics and Probability Letters 78 (2008) 1534–1540
Since ∀ i, λi > 0, ∀ n ≥ 1, Tn < ∞ w.p. 1 and hence Z (t) ↑ ∞ w.p. 1 as t ↑ ∞. Also since TZ (t) ≤ t < TZ (t)+1 = TZ (t) + L Z (t) , it follows from (3.6) and (3.7) that as t → ∞, t →1 a Z (t)
in probability
which implies (3.5).
Proof of Theorem 2.1. By Proposition 3.1, under (2.1) lim (t − a (Z (t))) ≡ Y
t→∞
exists in R w.p. 1.
Now (2.2) holds iff lim (log Z (t) − ct) = log ξ
t→∞
exists in R w.p. 1.
Thus under (2.1) and (2.2) holds iff log ξ 1 log Z (t) − a (Z (t)) = +Y lim t→∞ c c
exists in R.
Since Z (t) goes up by one at each step this holds iff (2.3) holds.
Proof of Theorem 2.2. By Proposition 3.2, under (2.4) a (Z (t)) →1 t Thus (2.5) holds iff a (Z (t)) (Z (t))1/q
in probability.
→ c−1/ p
as t → ∞.
Since Z (t) goes up one at each step this holds iff (2.6) holds.
−1 Proof of Theorem 2.3. Let a(t) ˜ ≡ b−1 (t), t > 0 and w(t) ≡ a˜ 0 (t) . Then w(·) is monotone and so the hypothesis of Theorem 2.3 imply those of Proposition 3.1. So w.p. 1, t − a (Z (t)) → Y w.p. 1. Again since w(·) is monotone the hypothesis of Theorem 2.3 imply that lim (a(n) ˜ − a(n)) ≡ α1
n→∞
exists.
Thus t − a˜ (Z (t)) → Y + α1 ≡ Y˜ w.p. 1. But a(t) ˜ = b−1 (t). Let (t) = a˜ (Z (t)) − t + Y˜ . Then Z (t) = b t − Y˜ + (t) . By hypothesis in (b) of Theorem 2.3, it follows that lim
t↑∞
Z (t) b(t − Y˜ )
= 1.
Again, Y˜ has an absolutely continuous distribution.
Proof of Theorem 2.4. The hypotheses of Theorem 2.4 imply that (2.4) holds. So by Proposition 3.2 a(Z (t)) →1 t
in probability
proving (a). Letting (t) = 1 −
a(Z (t)) t
one sees that (b) follows.
K.B. Athreya / Statistics and Probability Letters 78 (2008) 1534–1540
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4. Some examples 1 1 ct ci+β + i 2 , c > 0}i≥1 satisfies (2.1) and Z (t) grows like e . limi→∞ λi i −θ = α3 , 0 < α3 < ∞, θ < 1. Then (2.4) and (2.6) hold. In
1. (Exponential growth) Clearly, { λ1i ≡ 2. (Polynomial growth) If 1 grows like t q , q = 1−θ .
this case, Z (t)
p
3. (Exponential to a power) b(t) ≡ αet , α > 0, 0 < p ≤ 1 satisfies the conditions of Theorem 2.3. Here p λi ≡ w(i) = p 1/ p i(log αi )( p−1)/ p , i ≥ 1 and, hence, Z (t) grows like et . t
4. (Iterated exponential) b(t) = ee ⇒ a(t) = b−1 (t) = log log t ⇒ w(t) = conditions of Proposition 3.1 so t − log log Z (t) → Y w.p. 1
1 t log t
so w(i) ≡ i log i, i ≥ 2 satisfies
log Z (t) → e−Y w.p. 1 et t ⇒ Z (t) grows like ee . ⇒
Pn i 5. (Logarithmic.) Let w(i) = e−i , i ≥ 1. So Tn = i=1 e L i where {L i } are iid exp(1) random variables Pn Tn ⇒ en = i=1 e(i−n) L i and hence has the same distribution as Xn ≡
n X
e− j L j .
j=1
P −j Since E L j ≡ 1, X n converges w.p. 1 to X ∞ ≡ ∞ j=1 e L j < ∞ w.p. 1. Thus ! ∞ X log Tn d d −j log Tn − n −→ log −→ 1. e Lj ⇒ n j=1 Also P
⇒
log Tn >1+ n
∞ X
Tn =P = P Tn > e > en en Tn 1 1 1 ≤ E ≤ n n −1 e e (1 − e ) en
n(1+)
P(log Tn > n(1 + )) < ∞.
i=1
By the Borel–Cantelli, w.p. 1, lognTn < 1 + for large n. Next, log Tn P < 1 − = P(log Tn < n(1 − )) n = P(Tn < en(1−) ) Tn −n < e = P en ≤ P(L n < e−n ) = P(L 1 < e−n ) ∼ = c · e−n , 0 < c < ∞, for n large ∞ X log Tn < 1 − < ∞. ⇒ P n i=1 So by the Borel–Cantelli, w.p. 1 follows that log t → 1 w.p. 1. Z (t)
log Tn n
< 1 − for large n. So
log Tn n
→ 1 w.p. 1. Since TZ (t) ≤ t < TZ (t)+1 it
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K.B. Athreya / Statistics and Probability Letters 78 (2008) 1534–1540
5. Remark The referee raised the following question on an earlier version of Theorem 2.1 of the paper. Does there exist {λi } such that λi ∼ + β for some cP> 0, β ∈ R, (2.3) does not hold and still (2.2) holds? If λi ∼ = = ci + β for some c > 0, Pci ∞ 1 ∞ 1 < ∞ and so β ∈ R, then i=1 = ∞ and (2.1) holds. By the present version of Theorem 2.1, (2.2) holds i=1 2 λi λi
iff (2.3) holds. i log i Thus the answer to the referee’s question is in the negative. Here is an example. Let λ1 = 1 and λi = (log i+1) , P n 1 1 1 1 ∼ i ≥ 2. Then = + , i ≥ 2. So λi = i but (2.3) does not hold. However, (3.1) does hold. Now a(n) = , λi
i
i=1 λi
i log i
n ≥ 1. It can be shown that lim (a(n) − log n − log log n) = b
n→∞
exists in R.
So from Proposition 3.1, it follows that t − log Z (t) − log(log Z (t))
converges w.p. 1,
i.e., limt→∞ e−t Z (t) log Z (t) ≡ ξ exists w.p. 1. This implies Z (t) grows slower than et . It can be shown that if λi ∼ = ci + β for some c > 0, β ∈ R, then for any c1 < c < c2 , w.p. 1 lim Z (t)e−c1 t = ∞
t→∞
lim Z (t)e−c2 t = 0.
t→∞
Acknowledgement The author would like to thank the referee for raising the above question. References Athreya, K.B., Ney, P., 2000. Branching Processes. Dover. Athreya, K.B., Lahiri, S.N., 2006. Measure Theory and Probability Theory. Springer, New York. Karlin, S., Taylor, H.M., 1975. A First Course in Stochastic Processes. Academic Press, New York.