Open Access
Issue
Wuhan Univ. J. Nat. Sci.
Volume 31, Number 3, June 2026
Page(s) 291 - 298
DOI https://doi.org/10.1051/wujns/2026313291
Published online 24 June 2026

© Wuhan University 2026

Licence Creative CommonsThis is an Open Access article distributed under the terms of the Creative Commons Attribution License (https://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

0 Introduction

Branching process is a mathematical model used to describe the reproduction of species, which has received much attention. As the reproduction of species is subject to factors such as natural environment and social environment, many mathematicians have improved the classical branching process model. Smith and Wilkison first proposed branching processes in random environments and studied the extinction probability of branching processes in independent and identically distributed (i.i.d.) random environments and some properties of branching processes in Markovian environments[1-3]. Athreya and Karlin[4-5] discussed the extinction probability and some limiting behaviors of branching processes in stationary ergodic random environments. Holzheimer[6] introduced the controlled branching model in a random environment and studied its extinction probability. Yanev et al[7] investigated the sufficient condition for certain extinction and non-certain extinction of controlled branching processes in i.i.d. random environments; Bi et al[8] proposed a criterion for certain extinction of controlled branching processes in stationary ergodic random environments. Li et al[9] studied the Markov property and some limiting behaviors of controlled branching processes in random environments. Branching process with migration in random environments was first introduced by Key[10]. Li et al[11] discussed the conditional expectations and Markov property of branching processes with migration in random environments; Ren et al[12] studied the limiting behaviors of branching processes with migration in random environments. Li[13] investigated the asymptotic behavior for critical branching processes with migration. Wang et al[14] studied the limit theory for a supercritical branching process; Ren et al[15-16] studied the Markov property and limiting behaviors of the branching process affected by viral infectivity in a random environment, and the Markov property, conditional probability generating function and sufficient condition for certain extinction of bisexual branching processes influenced by viral infectivity in random environments. There have been many research results on the branching process in random environments, which can be seen in Refs. [17-21]. On the basis of Refs [12,15], the Markov property, conditional probability generating function, conditional expectation and sufficient condition for certain extinction of branching processes with migration and affected by viral infectivity in random environments are considered in this paper. Also, the existing correlation theory of branching process in random environments is generalized.

The remainder of this paper is organized as follows. In Section 1, some notations, definitions are introduced. Sections 2-4 are devoted to presenting the main results, including the Markov property, probability generating functions, conditional expectation and extinction probability. Section 5 is the conclusion.

1 Preliminaries

The model we shall be concerned with in this paper may be described as follows.

Let (Ω ,F,P)Mathematical equation be a given probability space, (Θ,Σ)Mathematical equation be a measurable space, and N={0,1,2,},Mathematical equationN+={1,2,3,},Mathematical equation

N m = { - m , - m + 1 , , 0,1 , } , m N Mathematical equation

Let ξ={ξ0,ξ1,ξ2,}Mathematical equation be a sequence of i.i.d. random variables, mapping from (Ω ,F,P)Mathematical equation to ΘMathematical equation and Let {Ini,nN,iN+}Mathematical equation and {Xni,nN,iN+}Mathematical equation denote two random variable sequences from (Ω ,F,P)Mathematical equation to N.Mathematical equation

For any θΘ,Mathematical equation we assume {pk(θ),kN}Mathematical equation and {αk(θ),kNm}Mathematical equation are two sequences of probability distributions, satisfying k=0kpk(θ)<,0p0(θ)+p1(θ)<1,k=-mkαk(θ)Mathematical equation

<   ,   0 α 0 ( θ ) < 1 Mathematical equation and there exists k{-1,-2,,-m}Mathematical equation such that αk(θ)>0.Mathematical equation

Definition 1   Suppose {Zn,nN}Mathematical equation and {Yn,nN}Mathematical equation are two sequences of random variables mapping from (Ω,F,P)Mathematical equation to NMathematical equation and (Ω,F,P)Mathematical equation to NmMathematical equation, respectively and satisfy

(i) Z0=1,Zn+1={i=1Zn+YnXniIni,Zn+Yn>00,Zn+Yn0,nNMathematical equation;

(ii) P(Xni=k|ξ)=pk(ξn), k,nN , iN+,P(Yn=k|ξ)=αk(ξn), nN, kNm,P(Ini=x|ξ)=bx(ξn)(1-b(ξn))1-x, nN , iN+ ,x=0,1;Mathematical equation

(iii) For given ξMathematical equation, {Xni,nN,iN+}Mathematical equation and {Ini,nN,iN+}Mathematical equation

are both independent sequences; {Xni,nN,iN+}Mathematical equation and

{ I n i , n N , i N + } Mathematical equation are mutually independent. For fixed nMathematical equation, {(Xni,Ini),iN+}Mathematical equation are identically distributed. For fixed nMathematical equation, {Yn,nN}Mathematical equation and {Zn,nN},{(Xni,Ini),iN+},{Xni,nN,Mathematical equation

i N + } Mathematical equation are independent respectively.

Then {Zn,nN}Mathematical equation is called a branching process with migration and affected by viral infectivity in random environment ξ.Mathematical equation

The above Zn,YnMathematical equation represents the total number of individuals in n-th generation and the number of individuals migrated in n-th generation, respectively; XniMathematical equation represents the number of individuals in (n+1)-th generation produced by the i-th individual in n-th generation. We suppose Ini=0Mathematical equation when the i-th individual in n-th generation was infected and not cured, that is, it produces no offspring; Ini=1Mathematical equation when the i-th individual in n-th generation didn't have the virus or was cured of it, that is, it can reproduce normally.

2 Markov Property

Since the branching process is a special case of Markov chain in random environments, the branching process in random environments can be studied with the help of the theory of Markov chain. Therefore, we first discuss the Markov property of branching processes with migration and affected by viral infectivity in random environments.

Definition 2   Suppose X={Xn,nN}Mathematical equation and ξ={ξn, nN}Mathematical equation are random sequences from (Ω ,F,P)Mathematical equation to NMathematical equation and ΘMathematical equation, respectively. For any x,nNMathematical equation, if

P ( X 0 = x 0 | ξ ) = P ( X 0 = x 0 | ξ 0 ) , Mathematical equation(1)

P ( X n + 1 = x | X 0 n , ξ ) = P ( ξ n , X n , x ) , Mathematical equation(2)

then XMathematical equation is called a Markov chain in random environment ξMathematical equation. For ease of exposition, we introduce some notations. Set A={(rl,xl):l=1i+jrlxl=k, rlN; xl=0,1; l=1,2,,i+j,iNm,jN},Fn(ξ)=σ(Z0,Z1,,Zn;ξ), nN,P*i(θ;j,k)=P(Zn+1=k|Yn=i,Zn=j,θ)=(rl,xl)AP(Xn1In1=x1r1,,Xn(i+j)In(i+j)=xi+jri+j|Yn=i,Zn=j,θ)={(rl,xl)Al=1i+jprl(θ)bxl(θ)(1-b(θ))1-xl,i+j>0;δ0k,i+j0.Mathematical equation

Theorem 1   { Z n , n N } Mathematical equation is a Markov chain in random environment ξMathematical equation with the one-step transition probabilities P(ξn;j,k)=i=-mP*i(ξn;j,k)αi(ξn).Mathematical equation

Proof   By Definition 2, to prove {Zn,nN}Mathematical equation is a Markov chain in random environment ξMathematical equation, it suffices to prove that (1) and (2) hold.

By Z0=1Mathematical equation we have (1) holds. Next we prove (2) and it suffices to show, for any k,nNMathematical equation, it follows that

P ( Z n + 1 = k | F n ( ξ ) ) = P ( ξ n ; Z n , k ) . Mathematical equation(3)

Since P(Zn+1=k|Fn(ξ))Mathematical equation and P(ξn;Zn,k)Mathematical equation are measurable with respect to Fn(ξ)Mathematical equation, it suffices to show, for any ΔFn(ξ)Mathematical equation, it follows that

Δ P ( Z n + 1 = k | F n ( ξ ) ) d P = Δ P ( ξ n ; Z n , k ) d P Mathematical equation(4)

Taking Δ={Z0=k0,,Zn-1=kn-1,Zn=j;ξ0B0,,Mathematical equation

ξ l B l } F n ( ξ ) Mathematical equation (where k0,k1,,kn-1,jN;B0,,BlΣMathematical equation),

we proceed to the proof of (4). We suppose lnMathematical equation (if l<nMathematical equation, let l'=nMathematical equation and Bl+1==Bl'=ΘΣMathematical equation).

Writing C={Z0=k0,,Zn-1=kn-1,Zn=j},D={ξ0B0,Mathematical equation

, ξ l B l } Mathematical equation, we get

Δ P ( ξ n ; Z n , k ) d P = Δ P ( Z n + 1 = k | Z n , ξ n ) d P = Δ i = - m P ( Z n + 1 = k , Y n = i | Z n , ξ n ) d P Mathematical equation

= Δ i = - m P ( Z n + 1 = k | Y n = i , Z n = j , ξ n ) P ( Y n = i | Z n = j , ξ n ) d P = Δ i = - m P { l = 1 i + j X n l I n l = k | Y n = i , Z n = j , ξ n ) α i ( ξ n ) d P = Δ i = - m ( r l , x l ) A l = 1 i + j p r l ( ξ n ) b x l ( ξ n ) ( 1 - b ( ξ n ) ) 1 - x l α i ( ξ n ) d P = Δ i = - m P * ( ξ n ; j , k ) α i ( ξ n ) d P . Mathematical equation

Using the independence assumed in Definition 1 (iii) and the properties of conditional expectation, it is deduced that

Δ P ( Z n + 1 = k | F n ( ξ ) ) d P = Δ E { I { Z n + 1 = k } | F n ( ξ ) } d P = D E { I { Z n + 1 = k } I C | F n ( ξ ) } d P = Δ = D i = - m E { I { Z n + 1 = k } I C | Y n = i , F n ( ξ ) } P ( Y n = i | F n ( ξ ) ) d P = Δ = D i = - m E { I { l = 1 i + j X n l I n l = k } I C | Y n = i , F n ( ξ ) } α i ( ξ n ) d P = D i = - m P ( l = 1 i + j X n l I n l = k | Y n = i , F n ( ξ ) ) I C α i ( ξ n ) d P     = D i = - m P * i ( ξ n ; j , k ) α i ( ξ n ) d P . Mathematical equation

Therefore, (4) holds, that is, {Zn,nN}Mathematical equation is a Markov chain in random environment ξMathematical equation with the one-step transition probabilities claimed above.

3 Conditional Probability Generating Function and Conditional Expectation

Since the extinction probability and martingale convergence of branching process in random environments are often investigated from the conditional probability generating function and conditional expectation of the process, in what follows, we consider the conditional probability generating function and conditional expectation of branching process with migration and affected by viral infectivity in random environments.

For convenience, we give some notations. Writing

f ξ n ( s ) = E ( s X n i | ξ ) = k = 0 P k ( ξ n ) s k   ,   g ξ n ( s ) = E ( s Y n | ξ ) = k = - m α k ( ξ n ) s k   , Π ξ n ( s ) = E ( s Z n | ξ ) = k = 0 P k , n ( ξ n ) s k   ,   s [ 0,1 ]   , P k , n ( ξ n ) = P ( Z n = k | ξ )   , k , n N . Mathematical equation

Theorem 2   For any nN , s[0,1] ,Mathematical equationit follows that

Π ξ n + 1 ( s ) = Π ξ n ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) g ξ n ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) + k = 0 m - 1 P k , n ( ξ n ) r = - m - k - 1 α r ( ξ n ) [ 1 - ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) k + r ] . Mathematical equation(5)

Proof   From Definition 1, the definition of Πξn+1(s)Mathematical equation and properties of conditional expectation, we obtain

Π ξ n + 1 ( s ) = E ( s Z n + 1 | ξ ) = k = 0 E ( s i = 1 k + Y n X n i I n i | Z n = k , ξ ) P ( Z n = k | ξ )                                                                                                            + k = m P k , n ( ξ n ) r = - m E ( s i = 1 k + r X n i I n i | Y n = r , Z n = k , ξ ) α r ( ξ n ) . = k = 0 m - 1 P k , n ( ξ n ) { r = - k E ( s i = 1 k + r X n i I n i , Y n = r | Z n = k , ξ )                                                                                                          + r = - m - k - 1 α r ( ξ n ) }      + r = - m - k - 1 P ( Y n = r | Z n = k , ξ ) }                                                                                                                + k = m P k , n ( ξ n ) r = - m E ( s i = 1 k + r X n i I n i | Y n = r , Z n = k , ξ ) α r ( ξ n ) .             + k = m P k , n ( ξ n ) r = - m E ( s i = 1 k + r X n i I n i , Y n = r | Z n = k , ξ ) Mathematical equation

Since for nN,iN+,Xni,IniMathematical equation are independent, so we have

E ( s X n i I n i | ξ ) = E ( s X n i 0 | ξ ) P ( I n i = 0 | ξ ) + E ( s X n i 1 | ξ ) P ( I n i = 1 | ξ )      = 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) . Mathematical equation

Hence,

Π ξ n + 1 ( s ) = k = 0 m - 1 P k , n ( ξ n ) r = - k α r ( ξ n ) [ 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ] k + r + k = m P k , n ( ξ n ) r = - m α r ( ξ n ) [ 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ] k + r + k = 0 m - 1 P k , n ( ξ n ) r = - m - k - 1 α r ( ξ n ) = Π ξ n ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) g ξ n ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) + k = 0 m - 1 P k , n ( ξ n ) r = - m - k - 1 α r ( ξ n ) [ 1 - ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) k + r ] . Mathematical equation

This completes the proof.

Corollary 1   For any nN , s[0,1] ,Mathematical equation it follows that

Π ξ 1 ( s ) = ( 1 - b ( ξ 0 ) + b ( ξ 0 ) f ξ 0 ( s ) ) g ξ 0 ( 1 - b ( ξ 0 ) + b ( ξ 0 ) f ξ 0 ( s ) )       + r = - m - 2 α r ( ξ 0 ) [ 1 - ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) r + 1 ] Mathematical equation

Π ξ n + 1 ( s ) = [ 1 - b ( ξ 0 ) + b ( ξ 0 ) f ξ 0 ( ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) ) ]   i = 0 n g ξ i ( 1 - b ( ξ i ) + b ( ξ i ) f ξ i ( ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) ) ) + k = 0 m - 1 P k , n ( ξ n ) r = - m - k - 1 α r ( ξ n ) [ 1 - ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) r + k ] + i = 0 n - 1 l = i + 1 n g ξ l ( 1 - b ( ξ l ) + b ( ξ l ) f ξ l ( ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) ) )   k = 0 m - 1 P k , i ( ξ i ) r = - m - k - 1 α r ( ξ i ) { 1 - [ 1 - b ( ξ i ) + b ( ξ i ) f ξ i ( ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) ) ) ] r + k } . Mathematical equation

Proof   From P1,0(ξ0)=P(Z0=1|ξ)=1Mathematical equation, we have Πξ0(s)=s.Mathematical equation Using the recursion of (5), we obtain

Π ξ 1 ( s ) = Π ξ 0 ( 1 - b ( ξ 0 ) + b ( ξ 0 ) f ξ 0 ( s ) ) g ξ 0 ( 1 - b ( ξ 0 )   + b ( ξ 0 ) f ξ 0 ( s ) ) + r = - m - 2 α r ( ξ 0 ) [ 1 - ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) r + 1 ]   = ( 1 - b ( ξ 0 ) + b ( ξ 0 ) f ξ 0 ( s ) ) g ξ 0 ( 1 - b ( ξ 0 ) + b ( ξ 0 ) f ξ 0 ( s ) )   + r = - m - 2 α r ( ξ 0 ) [ 1 - ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) r + 1 ]   . Mathematical equation

Applying the recursion of Theorem 2, we then have

Π ξ n + 1 ( s ) = Π ξ n ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) )   g ξ n ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) + k = 0 m - 1 P k , n ( ξ n ) r = - m - k - 1 α r ( ξ n ) [ 1 - ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) k + r ] = g ξ n ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) { Π ξ n - 1 ( 1 - b ( ξ n - 1 ) + b ( ξ n - 1 ) f ξ n - 1 ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) )                                        = [ 1 - b ( ξ 0 ) + b ( ξ 0 ) f ξ 0 ( ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) ) ]   g ξ n - 1 ( 1 - b ( ξ n - 1 ) + b ( ξ n - 1 ) f ξ n - 1 ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) )         i = 0 n g ξ i ( 1 - b ( ξ i ) + b ( ξ i ) f ξ i ( ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) ) ) + k = 0 m - 1 P k , n - 1 ( ξ n - 1 ) r = - m - k - 1 α r ( ξ n - 1 ) [ 1 - ( 1 - b ( ξ n - 1 ) + b ( ξ n - 1 ) f ξ n - 1 ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) ) k + r ] }                             + k = 0 m - 1 P k , n ( ξ n ) r = - m - k - 1 α r ( ξ n ) [ 1 - ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) k + r ] + k = 0 m - 1 P k , n ( ξ n ) r = - m - k - 1 α r ( ξ n ) [ 1 - ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) k + r ]     + i = 0 n - 1 l = i + 1 n g ξ l ( 1 - b ( ξ l ) + b ( ξ l ) f ξ l ( ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) ) ) = g ξ n ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) g ξ n - 1 ( 1 - b ( ξ n - 1 ) + b ( ξ n - 1 )   f ξ n - 1 ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) ) Π ξ n - 1 ( 1 - b ( ξ n - 1 ) + b ( ξ n - 1 )    + i = 0 n - 1 l = i + 1 n g ξ l ( 1 - b ( ξ l ) + b ( ξ l ) f ξ l ( ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) ) )   f ξ n - 1 ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) ) + g ξ n ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) )   k = 0 m - 1 P k , n - 1 ( ξ n - 1 ) r = - m - k - 1 α r ( ξ n - 1 ) [ 1 - ( 1 - b ( ξ n - 1 ) + b ( ξ n - 1 )   f ξ n - 1 ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) ) k + r ] + k = 0 m - 1 P k , n ( ξ n ) r = - m - k - 1 α r ( ξ n ) [ 1 - ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) k + r ] = Mathematical equation

This proof is completed.

Theorem 3   For any nNMathematical equation, it follows that

E ( Z n + 1 | ξ ) = E ( Z n | ξ ) b ( ξ n ) f ξ n ' ( 1 ) + b ( ξ n ) g ξ n ' ( 1 ) f ξ n ' ( 1 )   - b ( ξ n ) f ξ n ' ( 1 ) k = 0 m - 1 P k , n ( ξ n ) r = - m - k - 1 α r ( ξ n ) ( k + r ) . Mathematical equation

Proof   Taking the derivatives on both sides of (5) gives

Π ξ n + 1 ' ( s ) = Π ξ n ' ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) b ( ξ n ) f ξ n ' ( s ) g ξ n ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) + Π ξ n ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) g ξ n ' ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) b ( ξ n ) f ξ n ' ( s ) + k = 0 m - 1 P k , n ( ξ n ) r = - m - k - 1 α r ( ξ n )   [ - ( k + r ) ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( s ) ) k + r - 1 b ( ξ n ) f ξ n ' ( s ) ] . Mathematical equation

Moreover,

E ( Z n + 1 | ξ ) = Π ξ n + 1 ' ( 1 ) = Π ξ n ' ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( 1 ) ) b ( ξ n )   f ξ n ' ( 1 ) g ξ n ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( 1 ) ) + Π ξ n ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( 1 ) )   g ξ n ' ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( 1 ) ) b ( ξ n ) f ξ n ' ( 1 ) + k = 0 m - 1 P k , n ( ξ n )   r = - m - k - 1 α r ( ξ n ) [ - ( k + r ) ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( 1 ) ) k + r - 1 b ( ξ n ) f ξ n ' ( 1 ) ] . Mathematical equation

Due to Πξn(1)=fξn(1)=gξn(1)=1,Mathematical equation then for any nNMathematical equation, we can deduce that

E ( Z n + 1 | ξ ) = Π ξ n ' ( 1 ) b ( ξ n ) f ξ n ' ( 1 ) + b ( ξ n ) f ξ n ' ( 1 ) g ξ n ' ( 1 )   + k = 0 m - 1 P k , n ( ξ n ) r = - m - k - 1 α r ( ξ n ) [ - ( k + r ) b ( ξ n ) f ξ n ' ( 1 ) ] = E ( Z n | ξ ) b ( ξ n ) f ξ n ' ( 1 ) + b ( ξ n ) f ξ n ' ( 1 ) g ξ n ' ( 1 )   - b ( ξ n ) f ξ n ' ( 1 ) k = 0 m - 1 P k , n ( ξ n ) r = - m - k - 1 α r ( ξ n ) ( k + r ) ,   n N . Mathematical equation

The proof is completed.

Corollary 2   For any nNMathematical equation, it follows that

E ( Z n + 1 | ξ ) = i = 0 n b ( ξ i ) f ξ i ' ( 1 ) + i = 0 n g ξ i ' ( 1 ) l = i n b ( ξ l ) f ξ l ' ( 1 )   - i = 0 n l = i n b ( ξ l ) f ξ l ' ( 1 ) k = 0 m - 1 r = - m - k - 1 α r ( ξ i ) P k , i ( ξ i ) ( k + r ) . Mathematical equation

Proof   Applying the recursion of Theorem 3, we have

E ( Z n + 1 | ξ ) = E ( Z n | ξ ) b ( ξ n ) f ξ n ' ( 1 ) + b ( ξ n ) f ξ n ' ( 1 ) g ξ n ' ( 1 ) - b ( ξ n ) f ξ n ' ( 1 ) k = 0 m - 1 P k , n ( ξ n ) r = - m - k - 1 α r ( ξ n ) ( k + r ) = b ( ξ n ) f ξ n ' ( 1 ) { b ( ξ n - 1 ) f ξ n - 1 ' ( 1 ) E ( Z n - 1 | ξ ) + b ( ξ n - 1 ) f ξ n - 1 ' ( 1 ) g ξ n - 1 ' ( 1 ) - b ( ξ n - 1 ) f ξ n - 1 ' ( 1 ) k = 0 m - 1 P k , n - 1 ( ξ n - 1 ) r = - m - k - 1 α r ( ξ n - 1 ) ( k + r ) } + b ( ξ n ) f ξ n ' ( 1 ) g ξ n ' ( 1 ) - b ( ξ n ) f ξ n ' ( 1 ) k = 0 m - 1 P k , n ( ξ n ) r = - m - k - 1 α r ( ξ n ) ( k + r ) = b ( ξ n ) f ξ n ' ( 1 ) b ( ξ n - 1 ) f ξ n - 1 ' ( 1 ) { b ( ξ n - 2 ) f ξ n - 2 ' ( 1 ) E ( Z n - 2 | ξ ) + b ( ξ n - 2 ) f ξ n - 2 ' ( 1 ) g ξ n - 2 ' ( 1 ) - b ( ξ n - 2 ) f ξ n - 2 ' ( 1 ) k = 0 m - 1 P k , n - 2 ( ξ n - 2 )   r = - m - k - 1 α r ( ξ n - 2 ) ( k + r ) } + b ( ξ n ) f ξ n ' ( 1 ) b ( ξ n - 1 ) f ξ n - 1 ' ( 1 ) g ξ n - 1 ' ( 1 ) - b ( ξ n ) f ξ n ' ( 1 ) b ( ξ n - 1 ) f ξ n - 1 ' ( 1 ) k = 0 m - 1 P k , n - 1 ( ξ n - 1 ) r = - m - k - 1 α r ( ξ n - 1 ) ( k + r )   + b ( ξ n ) f ξ n ' ( 1 ) g ξ n ' ( 1 ) - b ( ξ n ) f ξ n ' ( 1 ) k = 0 m - 1 P k , n ( ξ n ) r = - m - k - 1 α r ( ξ n ) ( k + r ) Mathematical equation

= = i = 0 n b ( ξ i ) f ξ i ' ( 1 ) + i = 0 n g ξ i ' ( 1 ) l = i n b ( ξ l ) f ξ l ' ( 1 ) - i = 0 n l = i n b ( ξ l ) f ξ l ' ( 1 ) k = 0 m - 1 r = - m - k - 1 α r ( ξ i ) P k , i ( ξ i ) ( k + r ) . Mathematical equation

This completes the proof.

4 Extinction Probability

Definition 3   Let qm(ξ)=P(Zn=0|Z0=m,ξ) , qm=P(Zn=0|Z0=m).Mathematical equation qm(ξ)Mathematical equation and qmMathematical equation are defined to be the conditional extinction probability and extinction probability of process {Zn,nN},Mathematical equation respectively. Process {Zn,nN}Mathematical equation is called to be certainly extinct, if there exists nN+Mathematical equation such that qm=1Mathematical equation. Otherwise, process {Zn,nN}Mathematical equation is called to be non-certainly extinct.

Lemma 1   For any state mN+Mathematical equation, Markov Chain {Zn,nN}Mathematical equation is transient, if for any nN, Mathematical equationHn(1-b(ξn)+b(ξn)fξn(0))>0Mathematical equation holds, where Hn(s)=E(sYn+Zn).Mathematical equation

Proof   For any mN+,Mathematical equation it follows that

P ( l = 1 Z n + l = m | Z n = m ) 1 - P ( Z n + 1 = 0 | Z n = m )   = 1 - P ( i = 1 Z n + Y n X n i I n i = 0 | Z n = m )   = 1 - k = 0 P ( Z n + Y n = k ) P ( i = 1 k X n i I n i = 0 | Z n = m )   = 1 - k = 0 P ( Z n + Y n = k ) E { ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( 0 ) ) k }   = 1 - H n ( 1 - b ( ξ n ) + b ( ξ n ) f ξ n ( 0 ) ) . Mathematical equation

By the condition Hn(1-b(ξn)+b(ξn)fξn(0))>0Mathematical equation, we can deduce that for any mN+Mathematical equation, P(l=1Zn+l=m|Zn=m)<1.Mathematical equation Namely {Zn,nN}Mathematical equation in state mN+Mathematical equation is transient. Therefore,for any kN+Mathematical equation, it holds that limnP(Zn=k|Z0=m)=0Mathematical equation,and limnP(Zn=0|Z0=m)+limnP(Zn=|Z0=m)=1.Mathematical equation

Theorem 4   If for any nN,Mathematical equation it follows that Hn(1-b(ξn)+b(ξn)fξn(0))>0Mathematical equation, where Hn(s)=E(sYn+Zn)Mathematical equation and there exists n0N+Mathematical equation such that, for any ln0Mathematical equation, E{b(ξn)fξn'(1)(l+Yn)}lMathematical equation, then for every mN+Mathematical equation, {Zn,nN}Mathematical equation is certainly extinct, i.e., for every mN+,qm=1.Mathematical equation

Proof   By Lemma 1, to prove qm=1,Mathematical equation it suffices to prove that for any mN+,Mathematical equation limnP(Zn=|Z0=m)=0.Mathematical equation From Ref. [19], we derive that for any n1N+Mathematical equation, it follows that

l i m n P ( Z n = | Z 0 = m ) n = 0 k = n 1 P ( Z n = k | Z 0 = m ) P ( m i n n > 0 Z n n 1 | Z 0 = k ) . Mathematical equation

Let pk=P(minn>0Znn1|Z0=k).Mathematical equation Next we prove pk=0.Mathematical equation Take n1n0Mathematical equation and let

( Z n + Y n ) * = { 0 , Z n < n 1 , Z n + Y , n Z n n 1 . Mathematical equation

Constructing an auxiliary process

Z 0 * = k   ,   Z n + 1 * = i = 1 ( Z n * + Y n ) * X n i I n i   ,   n N . Mathematical equation

If k,ln1Mathematical equation, then P(Zn+1=l|Zn=k)=P(Zn+1*=l|Zn*=k).Mathematical equation Hence,

P ( m i n n > 0 Z n n 1 | Z 0 = k ) = P ( m i n n > 0 Z n * n 1 | Z 0 * = k ) = 1 - l i m n P ( Z n * = 0 | Z 0 * = k ) = 1 - q k * Mathematical equation(6)

For any kn1,MN+,Mathematical equation it follows that

P ( Z n * = 0 | Z 0 * = k ) = 1 - P ( Z n * M | Z 0 * = k ) - P ( 1 Z n * < M | Z 0 * = k ) .   Mathematical equation

By Lemma 1, for every ε>0Mathematical equation, when nMathematical equation is sufficiently large, for any M<Mathematical equation, it holds that

P ( 1 Z n * < M | Z 0 * = k ) ε 2 . Mathematical equation(7)

By Chebyshev's inequality, one can derive

P ( Z n * M | Z 0 * = k ) E ( Z n * | Z 0 * = k ) M . Mathematical equation

By the assumption that for every ln1Mathematical equation,E{b(ξn)fξn'(1)(l+Yn)}lMathematical equation holds, we have

E ( Z n + 1 * | Z n * = l ) = E { E ( Z n + 1 * | Z n * = l , ξ ) | Z n * = l } = E { E ( i = 1 ( Z n * + Y n ) * X n i I n i | Z n * = l , ξ ) | Z n * = l } = E { b ( ξ n ) f ξ n ' ( 1 ) E [ ( l + Y n ) | ξ ] } l . Mathematical equation

Consequently,

  E ( Z n * | Z 0 * = k ) = l E ( Z n * , Z n - 1 * = l | Z 0 * = k ) = l E ( Z n * | Z n - 1 * = l , Z 0 * = k ) P ( Z n - 1 * = l | Z 0 * = k ) = l E ( Z n * | Z n - 1 * = l ) P ( Z n - 1 * = l | Z 0 * = k ) l l P ( Z n - 1 * = l | Z 0 * = k ) = E ( Z n - 1 * | Z 0 * = k ) k . Mathematical equation(8)

By (7) and (8), we obtain that for every ε>0Mathematical equation, there exists sufficiently large M>0Mathematical equation such that

P ( Z n * M | Z 0 * = k ) < ε 2   ,   n 1 . Mathematical equation

Hence, for any ε>0Mathematical equation, P(Zn*=0|Z0*=k)1-ε,Mathematical equation whenever nMathematical equation and MMathematical equation are sufficiently large. Therefore, pk=0Mathematical equation, which implies limnP(Zn=|Z0=m)=0,Mathematical equation i.e., qm=1.Mathematical equation This completes the proof.

Theorem 5   Suppose 1) For any nNMathematical equation, it holds Hn(1-b(ξn)+b(ξn)fξn(0))>0Mathematical equation, where Hn(s)=E(sYn+Zn)Mathematical equation; 2) There exists a sequence of i.i.d. random variables {ζn,nN+}Mathematical equation such that supkn1k+Ynkζn , a.s.Mathematical equation Moreover, {ζn,nN+}Mathematical equation and each of the set {ξ,Yn,Xni,Ini,nN,iN+}Mathematical equation

are independent. Then if E{log(b(ξ0)fξ0'(1)ζ0)}0Mathematical equation, we can conclude that {Zn,nN}Mathematical equation is certainly extinct, i.e., qm=1,Mathematical equation for all mN+.Mathematical equation

Proof   To start with, consider the case E{b(ξ0)fξ0'(1)ζ0}1Mathematical equation. Using Jensen's inequality gives

E { l o g ( b ( ξ 0 ) f ξ 0 ' ( 1 ) ζ 0 ) } l o g { E ( b ( ξ 0 ) f ξ 0 ' ( 1 ) ζ 0 ) } 0 . Mathematical equation

On the other hand, for any kn1,Mathematical equation using the condition supkn1k+Ynkζn , a.s.Mathematical equation yields

E { b ( ξ 0 ) f ξ 0 ' ( 1 ) ( k + Y n ) } E { b ( ξ 0 ) f ξ 0 ' ( 1 ) k ζ 0 ) } k . Mathematical equation

By Theorem 4, we obtain qm=1Mathematical equation, i.e., {Zn,nN}Mathematical equation is certainly extinct.

Now we turn to the case

E { b ( ξ 0 ) f ξ 0 ' ( 1 ) ζ 0 } > 1   a n d   E { l o g ( b ( ξ 0 ) f ξ 0 ' ( 1 ) ζ 0 ) } 0 . Mathematical equation

Let {Zn*,nN}Mathematical equation be the auxiliary process constructed in Theorem 4, then for {Zn*,nN}Mathematical equation, it follows that supkn1k+Ynkζn , a.s..Mathematical equation

Denote a sequence of random variable by

X n = l o g ( b ( ξ n ) f ξ n ' ( 1 ) ζ n )   ,   n N , Mathematical equation

then {Xn,nN}Mathematical equation is i.i.d.. Let S0=0 , Sn=l=0n-1Xl , nN+.Mathematical equation Since E{b(ξ0)fξ0'(1)ζ0}>1Mathematical equation, then P(X0=0)<1Mathematical equation. Meanwhile, E{log(b(ξ0)fξ0'(1)ζ0)}0,Mathematical equation then for any T>0Mathematical equation, there exists non-negative integer stopping time τTMathematical equation, where τT=inf{n>0:Sn-T} and P(τT<)=1, i.e., SτT-T, a.s..Mathematical equation

Suppose An=σ(ξ0,ξ1,,ξn-1;ζ0,ζ1,,ζn-1).Mathematical equation Below we prove that for any nN,m>n1Mathematical equation, it follows that

E ( Z n * | A n , Z 0 * = m ) m e S n   a . s . . Mathematical equation(9)

In fact,

E ( Z 1 * | A 1 , Z 0 * = m ) = E { X 01 I 01 + + X 0 ( Z 0 * + Y 0 ) * I 0 ( Z 0 * + Y 0 ) * | A 1 , Z 0 * = m } = E { X 01 I 01 + + X 0 ( m + Y 0 ) * I 0 ( m + Y 0 ) * | A 1 } = b ( ξ 0 ) f ξ 0 ' ( 1 ) E { ( m + Y 0 ) * | A 1 } b ( ξ 0 ) f ξ 0 ' ( 1 ) m E ( ζ 0 | A 1 ) = b ( ξ 0 ) f ξ 0 ' ( 1 ) m ζ 0 = m e S 1 . Mathematical equation

Namely, (9) holds for n=1.Mathematical equation Assuming (9) holds for n=lN+,Mathematical equation we will prove it holds for l+1Mathematical equation. By the independence of Yn,Zn*Mathematical equation and ζnMathematical equation, we obtain

E ( Z l + 1 * | A l + 1 , Z 0 * = m ) = E { X l 1 I l 1 + + X l ( Z l * + Y l ) * I l ( Z l * + Y l ) * | A l + 1 , Z 0 * = m } = b ( ξ l ) f ξ l ' ( 1 ) E { ( Z l * + Y l ) * | A l + 1 , Z 0 * = m } ζ l b ( ξ l ) f ξ l ' ( 1 ) E { Z l * | A l , Z 0 * = m } ζ l b ( ξ l ) f ξ l ' ( 1 ) m e S l = m e S l + 1 . Mathematical equation

Therefore, (9) holds for any nN+.Mathematical equation We proceed to prove that

E ( Z n * | Z 0 * = m ) m e - T   a . s . Mathematical equation

In fact, since {τT=n}AnMathematical equation, then

E ( Z τ T * | Z 0 * = m ) = E { E ( Z τ T * | τ T , Z 0 * = m ) } = n = 1 P ( τ T = n ) E { E ( Z τ T * | A n , Z 0 * = m ) | τ T = n } n = 1 P ( τ T = n ) E ( m e S n | τ T = n ) = m E ( e S τ T ) m e - T   . Mathematical equation

Combining Chebyshev's inequality, we have P(ZτT*1|Z0*=m)me-TMathematical equation. When TMathematical equation is sufficient large, it follows that

P ( Z τ T * = 0 | Z 0 * = m ) 1   ,   ( T + ) . Mathematical equation

then, {Zn,nN}Mathematical equation is of certain extinction.

5 Conclusion

In this paper, we mainly studied the property of Markov, probability generating function and conditional expectation of branching process with migration and affected by viral infectivity in random environments, and the sufficient conditions for certain extinction of processes. The theory of branching process in random environments is generalized. In future, we shall study the limit theory of the process such as large deviation, laws of the iterated logarithm, and some properties of branching processes affected by viral infectivity in independent non-identically random environment and stationary ergodic random environment.

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