Sojourn time in <span class="mathjax-formula">$\mathbb {Z}^{+}$</span> for the Bernoulli random walk on <span class="mathjax-formula">$\mathbb {Z}$</span>

DOI:10.1051/ps/2010013 www.esaim-ps.org

SOJOURN TIME IN Z

FOR THE BERNOULLI RANDOM WALK ON Z

Aim´ e LACHAL

Abstract. Let (Sk)_k1 be the classical Bernoulli random walk on the integer line with jump param- eters p ∈ (0,1) and q = 1−p. The probability distribution of the sojourn time of the walk in the set of non-negative integers up to a fixed time is well-known, but its expression is not simple. By modifying slightly this sojourn time through a particular counting process of the zeros of the walk as done by Chung & Feller [Proc. Nat. Acad. Sci. USA35(1949) 605–608], simpler representations may be obtained for its probability distribution. In the aforementioned article, only the symmetric case (p=q= 1/2) is considered. This is the discrete counterpart to the famous Paul L´evy’s arcsine law for Brownian motion.

In the present paper, we write out a representation for this probability distribution in the general case together with others related to the random walk subject to a possible conditioning. The main tool is the use of generating functions.

Mathematics Subject Classification. 60G50, 60J22, 60J10, 60E10.

Received October 5, 2009. Revised March 29, 2010.

1. Introduction

Let (Xk)_k1 be a sequence of Bernoulli random variables with parameters p = P{Xk = 1} ∈ (0,1) and q= 1−p=P{Xk=−1}, and (Sk)_k0 be the random walk defined on the set of integersZ={. . . ,−1,0,1, . . .}

as S_k=S₀+_k

j=1X_j, k1 with initial locationS₀. For brevity, we writeP_i=P{. . .|S₀=i}andP₀=P.

The probability distribution of the sojourn time of the walk (S_k)_k0in Z⁺=Z∩[0,+∞) up to a fixed step n1,N_n =_n

j=01l_Z+(S_j) = #{j∈ {0, . . . , n}:S_j 0}, is well-known. A representation for this probability distribution can be derived with the aid of Sparre Andersen’s theorem (see [9,10] and, e.g., [11] Chap. IV, Sect. 20). This latter can be stated as the remarkable relationship, settingN₀= 0,

P{N_n=k}=P{N_k=k}P{N_n−k= 0}for 0kn,

Keywords and phrases. Random walk, sojourn time, generating function.

1 Institut National des Sciences Appliqu´ees de Lyon, Pˆole de Math´ematiques/Institut Camille Jordan, Bˆatiment L´eonard de Vinci, 20 avenue Albert Einstein, 69621 Villeurbanne Cedex, France.[email protected]; http://maths.insa-lyon.fr.

Article published by EDP Sciences c EDP Sciences, SMAI 2012

where the probabilities P{N_k = 0} and P{N_k = k}, k ∈ N, are implicitly known through their generating functions:

∞ k=0

P{Nk = 0}z^k= exp ∞

k=1

P{Sk <0}z^k k

, ∞

k=0

P{Nk=k}z^k= exp ∞

k=1

P{Sk 0}z^k k

·

Nevertheless, the result is not so simple. Rescaling the random walk and passing to the limit, we get Paul L´evy’s famous arcsine law for Brownian motion.

By modifying slightly the counting process of the positive terms of the random walk as done by Chung &

Feller (see [4] and, e.g., [5], Chap. III, Sect. 4 and [8], Chap. 8, Sect. 11), an alternative sojourn time of the walk (S_k)_k∈Nin Z⁺ up toncan be defined asT_n=_n

j=1δ_j with δj=

1 if (S_j>0) or (S_j = 0 andS_j−1>0), 0 if (S_j<0) or (S_j = 0 andS_j−1<0).

We putT0= 0. We obviously have 0Tn n. InTn,n1, one counts each stepj such thatSj >0 and only those steps such that Sj = 0 which correspond to a downstep: Sj−1 = 1. This convention is described in [4]

(and, e.g., in [5] and [8]) in the symmetric case p=q = 1/2 whenn is an even integer and, as written in [4]

–“The elegance of the results to be announced depends on this convention” (sic) –, it produces a remarkable result. Indeed, in this case, the sojourn time is even and its probability distribution takes the simple following form: for even integersksuch that 0kn, as in Sparre Andersen’s theorem,

P{T_n=k}=P{T_k=k}P{T_n−k= 0}= 1 2ⁿ

k k/2

(n−k) (n−k)/2

.

In this paper, we derive explicit expressions for the probability distribution of Tn in any case, that is for any p∈(0,1) and any integer (even or odd)n1. The main results are displayed in Theorems4.2and 5.3. We also compute the distribution of T_n under various constraints at the last step: S_n = 0, S_n > 0 or S_n < 0.

The constraint S_n = 0 (with S₀ = 0) corresponds to the bridge of the random walk. The related results are respectively included in Theorems6.2and7.3. The main tool for this study is the use of generating functions, excursions theory associated with random walks together with clever algebra. The intermediate results are contained in Theorems4.1,5.1,6.1and7.1.

Finally, by rescaling suitably the random walk, we retrieve the distribution of the sojourn time in (0,+∞) for the Brownian motion with a possible drift. This includes of course Paul L´evy’s arcsine law for Brownian motion without drift.

Although this problem is old and classical, we are surprised not to have found any related reference in the literature.

2. Settings and mathematical background 2.1. Some preliminary identities

LetN=Z⁺={0,1,2, . . .}be the usual set of non-negative integers,N^∗=Z^+∗=N\ {0}={1,2,3, . . .} that of positive integers,Z^−∗ ={. . . ,−3,−2,−1}that of negative integers and Z^∗ =Z\ {0}. LetE ={0,2,4, . . .}

denote the set of even non-negative integers, E^∗ =E \ {0}={2,4,6, . . .} the set of even positive integers and O={1,3,5, . . .}the set of odd positive integers. Set, for suitable realz,

A(z) = 1−4z² and A(z) = 1−4pqz².

Set also fori∈ E

a_i= 1 i+ 2

i i/2

= 1

4(i+ 1)

i+ 2 (i+ 2)/2

and b_i= (i+ 2)a_i= i

i/2

.

Thea_i’s are closely related to the famous Catalan numbers: a_i =C_i/2/2. We shall make use of the following elementary identities.

Proposition 2.1. For any z such that|z|<1/2, A(z) = 1−4z²=−

i∈E

1 i−1

i i/2

zⁱ= 1−4

i∈E^∗

ai−2zⁱ,

(2.1) 1

A(z)= 1

√1−4z² =

i∈E

i i/2

zⁱ=

i∈E

b_izⁱ. We have the following convolution relationships.

Proposition 2.2. For any even integer i0,

j∈E:ji

a_ja_i−j= 1

2a_i+2 and

j∈E:ji

a_jb_i−j =1

4b_i+2. (2.2)

Proof. The generating function of the left-hand side of the first equality in (2.2) can be evaluated as

i∈E ^j∈E:_ji

ajai−j

zⁱ=

i∈E

aizⁱ ₂

=

1−A(z) 4z²

₂

= 1

8z⁴(1−2z²−A(z)) = 1 2

i∈E

ai+2zⁱ.

By identification of the coefficients of the foregoing generating functions, we immediately obtained the first equality in (2.2). Analogously, for the second equality in (2.2),

i∈E ^j∈E:_ji

ajbi−j

zⁱ=

i∈E

aizⁱ

i∈E

bizⁱ

=1−A(z)

4z²A(z) = 1

4z²A(z)− 1 4z² =1

4

i∈E

bi+2zⁱ

and the second equality in (2.2) holds.

We shall also use the identity below.

Proposition 2.3. For any xandy such that |x|<1/2 and|y|<1/2, 1

A(x) +A(y) =

i,j∈E

a_i+jxⁱy^j. (2.3)

Proof. Let us write 1

A(x) +A(y) = A(x)−A(y)

4(y²−x²) .On one hand, A(x)−A(y) = 4

k∈E

a_k(y^k+2−x^k+2).

On the other hand,

y^k+2−x^k+2

y²−x² =

i,j∈E:

i+j=k

xⁱy^j.

Therefore,

1

A(x) +A(y) =

k∈E

a_k

i,j∈E:

i+j=k

xⁱy^j=

i,j∈E

a_i+jxⁱy^j.

2.2. Some well-known identities on random walks

Here, we recall several well-known formulas in the theory of random walks. We refer,e.g., to [11]. We have, forj∈Zandk∈Nsuch that|j|kandk−j∈ E,

P{S_k=j}= k

(j+k)/2

p^(j+k)/2q^(k−j)/2. (2.4)

Using the representation _k

(k+1)/2

=b_k+1/2, we have in particular

P{Sk = 0}=

bk(pq)^k/2 ifkis even, 0 ifkis odd,

P{Sk = 1}=P−1{Sk = 0}= ₁

2bk+1p^(k+1)/2q^(k−1)/2 ifk is odd,

0 ifk is even,

(2.5)

P{S_k=−1}=P₁{S_k = 0}= ₁

2bk+1p^(k−1)/2q^(k+1)/2 ifk is odd,

0 ifk is even.

We define several generating functions. Forj ∈Z, let H_j be the generating function of theP{S_k =j}, k∈N, and, fori∈Z,H_i^F be the generating function of thePi{Sk∈F}, k∈N:

H_j(z) =^∞

k=0

P{S_k =j}z^k and H_i^F(z) =^∞

k=0

P_i{S_k∈F}z^k=

j∈F−i

H_j(z) (2.6)

whereF−i is the set of the numbers of the formj−i, j∈F. We explicitly have

Hj(z) =

⎧⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎩ 1 A(z)

1−A(z) 2qz

_j

= 1

A(z)

2pz 1 +A(z)

_j

ifj0, 1

A(z)

1−A(z) 2pz

_|j|

= 1

A(z)

2qz 1 +A(z)

_|j|

ifj0.

(2.7)

We need to introduce the first hitting time of a levela∈Z for the random walk: τ_a = min{n1 :S_n =a}.

The probability distribution ofτa fora∈Z^∗ can be expressed by means of the probabilitiesP{Sk=a},k∈N, as

P{τ_a=k}= |a|

k P{S_k=a}=|a|

k

k (k+a)/2

p^(k+a)/2q^(k−a)/2 fork|a|. (2.8) In some particular cases, we have fork∈ E^∗:

P{τ0=k}= 1 k−1

k k/2

(pq)^k/2

and fork∈ O:

P{τ₁=k}= 1 k

k (k+ 1)/2

p^(k+1)/2q^(k−1)/2, P{τ₋₁=k}= 1

k k

(k+ 1)/2

p^(k−1)/2q^(k+1)/2. We sum up these formulas as follows:

• fork∈ E^∗,

P{τ₀=k}= 4a_k−2(pq)^k/2; (2.9)

• fork∈ O,

P{τ₁=k}=P₋₁{τ₀=k}= 1

2qP{τ₀=k+ 1}= 2p a_k−1(pq)^(k−1)/2,

(2.10) P{τ−1=k}=P1{τ0=k}= 1

2pP{τ0=k+ 1}= 2q ak−1(pq)^(k−1)/2.

Remark 2.4. The convolution identities (2.2) can be interpreted as Darling–Siegert-type equations (see, e.g., [3]) which are due to the Markov property of the random walk. More precisely, the second identity of (2.2) is the analytic form of the probabilistic equality

P{Sn= 0}=

j∈E:jn

P{τ0=j}P{Sn−j= 0}

which is obtained by remarking that a trajectory starting at zero and terminating at zero at timennecessarily passes through zero at a time equal to or less thannand thenτ₀n. The first identity of (2.2) is the analytic form of the probabilistic equality

P{τ₂=n}=

jn−1j∈O:

P{τ₁=j}P{τ₁=n−j}

which is obtained by observing that a trajectory starting at zero and passing at level two at timenfor the first time necessarily crosses level one at a time less than n: τ1< n.

The corresponding generating functionsKa defined as Ka(z) =E(z^τa) =

∞ k=1

P{τa=k}z^k

are explicitly expressed by

Ka(z) =

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎩

1−A(z) 2qz

_a

ifa1, 1−A(z)

2pz _|a|

ifa−1, 1−A(z) ifa= 0.

(2.11)

We state a last elementary result. Noticing, by the Markov property, thatP{τ₀=j, S₁>0}=pP₁{τ₀=j−1}

andP{τ0=j, S1<0}=qP1{τ0=j−1},we get, by (2.10),P{τ0=j, S1>0}=P{τ0=j, S1<0}= ¹₂P{τ0= j} and then

E(z^τa1l_{S1>0}) =E(z^τa1l_{S1<0}) = 1

2(1−A(z)). (2.12)

2.3. Purpose of the article

The aim of this paper is to compute the probabilities

r^F_k,n=P{T_n=k, S_n ∈F},0kn,

in the four cases F = Z, F = {0}, F = Z^+∗ and F = Z^−∗. The corresponding results are displayed in Theorems5.1,6.1and7.1. The adopted way consists of first calculating the generating functionG_Fof ther_k,n^F ’s:

G^F(x, y) =

k,n∈N:

kn

r^F_k,nx^ky^n−k=

n∈N

E

x^Tny^n−Tn1l_{Sn∈F}

with the help of the excursions theory associated with random walks, and next of inverting this function.

Notice that r^F_0,0 = 1l_F(0) and for n 1, the events {Tn = 0} and {Tn = n} respectively coincide with {S₁0, . . . , S_n0}={τ₁> n}and{S₁0, . . . , S_n 0}={τ₋₁> n}. Therefore, forn1,

r_0,n^F =P{τ1> n, Sn∈F} and r^F_n,n=P{τ−1> n, Sn∈F}. (2.13) The particular probabilitiesr_0,n^F andr^F_n,n,n∈N, are generated by the partial functions ofG^F, namely

G^F(x,0) =

n∈N

r^F_n,nx^k and G^F(0, y) =

n∈N

r^F_0,nyⁿ. (2.14)

3. Generating function

Theorem 3.1. The generating function G^F can be written as

G^F(x, y) = [1 +A(x)]G^F(x,0) + [1 +A(y)]G^F(0, y)−21l_F(0)

A(x) +A(y) (3.1)

whereA(z) = 1−4pqz². The quantitiesG^F(x,0)andG^F(0, y)can be expressed as G^F(x,0) =H₀^F(x)−1−A(x)

2px H₋₁^F (x) and G^F(0, y) =H₀^F(y)−1−A(y)

2qy H₁^F(x) (3.2) whereH₀^F,H₁^F andH₋₁^F are defined by (2.6).

Proof. Let us introduce the successive passage times by 0 recursively defined by N₀ = 0 and for m ∈ N, N_m+1 = min{n N_m+ 1 : S_n = 0}. We decompose the random walk into excursions and this yields the following decomposition for the generating functionG^F:

G^F(x, y) =

n∈N

E

x^Tny^n−Tn1l_{Sn∈F}

= 1l_F(0) +E

m∈N Nm+1

n=Nm+1

x^Tny^n−Tn1l_{Sn∈F}

.

For N_m+ 1 n N_m+1, we have T_n =T_Nm + (n−N_m)1l_{em0} where e_m is the excursion of the random walk on the interval [N_m, N_m+1]. In particular, e₀ is the first excursion, it is defined on the interval [0, τ₀].

The fact thate0 0 (respectively e0 0) is equivalent to that S1 > 0 (respectivelyS1 <0). We have also n−Tn=Nm−TNm+ (n−Nm)1l_{em0} and then, by the independence of the excursions,

G^F(x, y) = 1lF(0) +E

m∈N

x^TNmy^Nm−TNm

Nm+1

n=Nm+1

x^n−Nm1l_{em0}+y^n−Nm1l_{em0}

1l_{Sn∈F}

= 1l_F(0) +E

m∈N

x^TNmy^Nm−TNm

Nm+1−Nm

n=1

xⁿ1l_{em0}+yⁿ1l_{em0}

1l_{Sn+Nm∈F}

= 1l_F(0) +E

m∈N

x^TNmy^Nm−TNm

E _τ0

n=1

xⁿ1l_{e00}+yⁿ1l_{e00}

1l_{Sn∈F}

.

Hence, we have obtained a representation of the form

G^F(x, y) = 1l_F(0) +G⁰(x, y)[B₊^F(x) +B₋^F(y)] (3.3) with

G⁰(x, y) =E

m∈N

x^TNmy^Nm−TNm

and

B^F₊(x) =E τ0

n=1

1l_{{S1>0,Sn∈F}}xⁿ

, B₋^F(y) =E τ0

n=1

1l_{{S1<0,Sn∈F}}yⁿ

.

Actually, G⁰ is nothing but G^{0} (i.e., G^F associated with F ={0}) as it is immediately seen by definition.

The quantity G⁰(x, y) can be easily evaluated as follows. Set um =E

x^TNmy^Nm−TNm

. We have u0= 0 and form∈N,um+1=umE

x^τ01l_{S1>0}+y^τ01l_{S1<0}

. Now, by (2.12), E

x^τ01l_{S1>0}+y^τ01l_{S1<0}

= 1

2[2−A(x)−A(y)].

As a byproduct, the series (

m∈Nu_m) is geometrical with ratio [2−A(x)−A(y)]/2 and its sum is given by 2/[A(x) +A(y)]. This yields the following result which will be stated in Theorem6.1:

G⁰(x, y) = 2 A(x) +A(y)· Thus, (3.3) can be written as

G^F(x, y) = 1lF(0) + 2B^F₊(x) + 2B₋^F(y)

A(x) +A(y) · (3.4)

We then deduce

G^F(x,0) = 1l_F(0) + 2B₊^F(x)

1 +A(x), G^F(0, y) = 1l_F(0) + 2B₋^F(y) 1 +A(y) from which we extract

2B^F₊(x) = [1 +A(x)][G^F(x,0)−1l_F(0)], 2B₋^F(y) = [1 +A(y)][G^F(0, y)−1l_F(0)]. (3.5) Finally, putting (3.5) into (3.4) gives (3.1).

Let us check (3.2). In view of (2.10), the probabilityr^F_n,n admits the following representations:

r_n,n^F =P{Sn∈F} −P{τ−1n, Sn∈F}=P{Sn∈F} −ⁿ

j=1

P{τ−1=j}P−1{Sn−j∈F}

=P{Sn∈F} − 1 2p

j∈O:jn

P{τ0=j+ 1}P−1{Sn−j∈F}. (3.6)

Therefore, we have by (2.14) and (3.6) G^F(x,0) =

n∈N

r_n,n^F xⁿ=

n∈N

P{Sn ∈F}xⁿ− 1 2p

n∈N^∗

j∈O:jn

P{τ0=j+ 1}P−1{Sn−j∈F}xⁿ.

The first sum in the above equality isH₀^F(x) while the second can be computed as follows: by (2.10),

n∈N^∗

j∈O:jn

P{τ0=j+ 1}P−1{Sn−j∈F}xⁿ=

j∈O

P{τ0=j+ 1}x^j

n∈N∗: nj

P−1{Sn−j∈F}x^n−j

= 2p

j∈O

P{τ−1=j}x^j

n∈N

P−1{Sn∈F}xⁿ

= 2pK₋₁(x)H₋₁^F (x)

and the expression ofG^F(x,0) in (3.2) ensues. The derivation of that ofG^F(0, y) is quite similar. The Proof of

Theorem3.1is finished.

In [7], we propose a Proof of (3.1) based on a recursive relationship concerning the probabilitiesrk,n which is similar to the proof originally described in [4]. We also refer the reader to the book [6] for a combinatorial approach for tackling such problems.

4. The case F = {j } for j ∈ Z

We suppose thatF ={j}for a fixedj∈Z^∗. So, we are dealing with a random walk with a prescribed location after thenth step. The case where j= 0 will be considered in Section6. We set for simplicityr^{j}_k,n =r^j_k,n and G^{j}(x, y) =G^j(x, y),H_i^{j}(z) =H_i^j(z).

4.1. Generating function

In order to write the generating function G^j(x, y), in view of (3.1), we need to know thatH_i^j(z) =H_j−i(z) and to evaluate the functionsG^j(x,0) andG^j(0, y). We have

G^j(x,0) =H₀^j(x)−1−A(x)

2px H₋₁^j (x). But

H₋₁^j (x) =

⎧⎪

⎪⎨

⎪⎪

⎩

2px

1 +A(x)H₀^j(x) ifj∈Z^+∗, 2px

1−A(x)H₀^j(x) ifj∈Z^−∗.

Then, forj∈Z^+∗,

G^j(x,0) =H₀^j(x)

1−1−A(x) 2px

2px 1 +A(x)

= 2Kj(x) 1 +A(x) and, forj ∈Z^+∗,

G^j(x,0) =H₀^j(x)

1−1−A(x) 2px

2px 1−A(x)

= 0. Similarly, we have

G^j(0, y) =

⎧⎨

⎩

2Kj(y)

1 +A(y) ifj ∈Z^−∗, 0 ifj ∈Z^+∗. From this and (3.1), we immediately derive the functionG(x, y).

Theorem 4.1. The generating function Gis given by

G^j(x, y) =

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

2K_j(x)

A(x) +A(y) ifj∈Z^+∗

2Kj(y)

A(x) +A(y) ifj∈Z^−∗

(4.1)

whereA(z) = 1−4pqz².

4.2. Distribution of the sojourn time

We now invert the generating functionG^j given by (4.1) in order to derive the coefficientsr_k,n^j .

Theorem 4.2. The probability r_k,n^j =P{Tn=k, Sn =j} admits the following expression: for 0knsuch that n−k is even,

if j1:

r_k,n^j =

⎧⎪

⎪⎨

⎪⎪

⎩ 2j

jik,i∈N:

i−j∈E

an−i

i

i (i+j)/2

p^(n+j)/2q^(n−j)/2 ifjk andk−j is even,

0 ifj > k ork−j is odd;

(4.2)

if j−1:

r^j_k,n =

⎧⎪

⎪⎨

⎪⎪

⎩

2|j|

|j|ik,i∈N:

i+j∈E

an−i

i

i (i+j)/2

p^(n+j)/2q^(n−j)/2 ifjk−nandk−j is even,

0 ifj < k−nork−j is odd,

(4.3)

whereai= 1 i+ 2

i i/2

for i∈ E.

Proof. Assume first thatj1. We expandG^j(x, y) by using (2.3):

G^j(x, y) = 2 ∞ i=0

P{τj=i}xⁱ

l,m∈E

al+mx^ly^m= 2

i,l,m∈N:

l,m∈E

al+mP{τj =i}x^i+ly^m.

By performing the transformationsi+l=kandm=n−kin the last sum, we get G^j(x, y) = 2

i,k,n∈N:

k−i∈E,n−k∈E

a_n−iP{τ_j=i}x^ky^n−k=

k,n∈N:

n−k∈E i∈N:

k−i∈E

2a_n−iP{τ_j =i}

x^ky^n−k.

We finally obtain (4.2) by identifying the coefficients of the two expansions ofG^j and using (2.8). For the case where j −1, we invoke an argument of duality which is explained in Remark 4.3 below. The expression of r^j_k,n forj−1 can be deduced from that related to the case wherej1 by interchangingpandq,kandn−k,

j and−j, and this proves (4.3).

Remark 4.3. Let us introduce the dual random walk (S^∗_n)_n0 with S_k^∗ = _k

j=1X_j^∗ = −Sk. This sequence is the Bernoulli random walk with interchanged parameters q=P{X_j^∗= +1} and p = P{X_j^∗ = −1}. The corresponding sojourn time is defined asT_n^∗=_n

j=1δ^∗_j with δ_j^∗=

1 if (S_j^∗>0) or (S_j^∗= 0 andS_j−1^∗ >0), 0 if (S_j^∗<0) or (S_j^∗= 0 andS_j−1^∗ <0),

=

1 if (Sj<0) or (Sj= 0 and Sj−1<0), 0 if (Sj>0) or (Sj= 0 and Sj−1>0).

We see thatδ_j^∗= 1−δj and thenT_n^∗ =n−Tn which implies r_k,n^j =P{T_n^∗=n−k,S_n^∗ =−j}.As a result, the probabilityr_k,n^j can be deduced from the probabilityr^−j_n−k,n by interchangingpandq.

5. The case F = Z (random walk without conditioning)

In this part, we study the sojourn time without conditioning the extremity of the random walk. This corresponds to the case F =Z. We set for simplifying the notationsr^Z_k,n =rk,n and G^Z(x, y) =G(x, y). A possible expression for the rk,n’s can be obtained from Theorem4.2 by summing ther^j_k,n, j ∈ Z. Actually, r^j_k,n= 0 for|j|> n; so,

rk,n = n j=−n

r^j_k,n.

We propose another representation which can be deduced from the generating functionG.

5.1. Generating function

In order to derive the generating functionG(x, y), in view of (3.1), we need to evaluate the functionsH_i^Z(z), G(x,0) andG(0, y). On one hand, by definition,

H_i^Z(z) = ∞ k=0

Pi{Sk ∈Z}z^k= ∞ k=0

z^k = 1 1−z· On the other hand, by (3.2),

G(x,0) =H₀^Z(x)−1−A(x)

2px H₋₁^Z (x) = 1 1−x

1−1−A(x) 2px

= 2px−1 +A(x) 2px(1−x) ·

Similarly,

G(0, y) = 2qy−1 +A(y) 2qy(1−y) · We can then write outG(x, y).

Theorem 5.1. The generating function Gis given by G(x, y) = 1

A(x) +A(y)

2p−1 +A(x)

1−x +2q−1 +A(y) 1−y

(5.1) whereA(z) = 1−4pqz².

Proof. We have by (3.1)

G(x, y) =[1 +A(x)]G(x,0) + [1 +A(y)]G(0, y)−2 A(x) +A(y) ·

Let us compute the terms [1 +A(x)]G(x,0) and [1 +A(y)]G(0, y). We have

[1 +A(x)][2px−1 +A(x)] = 2px−1 + 2pxA(x) +A(x)²= 2px[1−2qx+A(x)]

and then

[1 +A(x)]G(x,0) = 1−2qx+A(x) 1−x · Similarly,

[1 +A(y)]G(0, y) = 1−2py+A(y) 1−y · Therefore, we get

[1 +A(x)]G(x,0) + [1 +A(y)]G(0, y)−2 =

1−2qx

1−x +1−2py 1−y −2

+A(x)

1−x+A(y) 1−y

=

1−2qx 1−x −2q

+

1−2py 1−y −2p

+A(x)

1−x+ A(y) 1−y

= 2p−1 +A(x)

1−x +2q−1 +A(y) 1−y

from which we obtain (5.1).

An interesting consequence of Theorem5.1concerns the “partial” generating function ˜Gof the probabilities rk,n limited to the even indiceskandn:

G(x, y) =˜

k,n∈E

r_k,nx^ky^n−k.

For this function, we have the simple result below.

Corollary 5.2. The generating function G˜ is given by

G(x, y) =˜ 4pq

[1−2p+A(x)][1−2q+A(y)] (5.2)

whereA(z) = 1−4pqz². In particular,

G(x, y) = ˜˜ G(x,0) ˜G(0, y). (5.3)

Proof. We first try to relate ˜GtoG. For this, we observe that 1

2[G(x, y) +G(−x,−y)] =

k,n∈N:

kn

1 + (−1)ⁿ

2 r_k,nx^ky^n−k=

k∈N,n∈E:

kn

r_k,nx^ky^n−k.

We know that, for all even integern, ifk is odd, thenr_k,n= 0. We therefore have checked that G˜(x, y) = 1

2[G(x, y) +G(−x,−y)]. Thus,

G(x, y) =˜ 1 A(x) +A(y)

2p−1 +A(x)

1−x² +2q−1 +A(y) 1−y²

· We have now

2p−1 +A(x) = (2p−1)²−A(x)²

2p−1−A(x) = 4pq 1−x² 1−2p+A(x) and similarly,

2q−1 +A(y) = 4pq 1−y² 1−2q+A(y)· This gives

G˜(x, y) = 4pq A(x) +A(y)

1

1−2p+A(x)+ 1 1−2q+A(y)

which in turn, with [1−2p+A(x)] + [1−2q+A(y)] = A(x) +A(y), yields (5.2). This formula supplies in particular

G(x,˜ 0) = 2q

1−2p+A(x) and G(0, y) =˜ 2p 1−2q+A(y)

and we immediately obtain (5.3).

5.2. Distribution of the sojourn time

In this part, we invert the generating function Ggiven by (5.1) in order to derive the coefficientsrk,n. Theorem 5.3. The probability r_k,n=P{T_n =k} admits the following expression: for 0kn,

r_k,n= 1l_E(n−k)

2p

n−kini∈E:

a_i(pq)^i/2−4

n−kin−2i∈E: j∈E:

ji+k−n

a_ja_i−j

(pq)^i/2+1

+ 1l_E(k)

2q

kini∈E:

ai(pq)^i/2−4

kin−2i∈E: j∈E:

ji−k

ajai−j

(pq)^i/2+1

(5.4)

whereai= 1 i+ 2

i i/2

for i∈ E.

More specifically:

• For oddn

r_k,n=

⎧⎪

⎪⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎪⎪

⎩

2p

n−kin−1i∈E:

ai(pq)^i/2−4

n−kin−3i∈E: j∈E:

ji+k−n

ajai−j

(pq)^i/2+1 if kis odd,

2q

kin−1i∈E:

a_i(pq)^i/2−4

kin−3i∈E: j∈E:

ji−k

a_ja_i−j

(pq)^i/2+1 if kis even;

(5.5)

• For evenn

rk,n=

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎩

2p

n−kini∈E:

a_i(pq)^i/2−4

n−kin−2i∈E: j∈E:

ji+k−n

a_ja_i−j

(pq)^i/2+1

+2q

kini∈E:

ai(pq)^i/2−4

kin−2i∈E: j∈E:

ji−k

ajai−j

(pq)^i/2+1 if k is even, 0if kis odd.

(5.6)

Proof. Rewrite (5.1) as

G(x, y) = 2p 1−x

1

A(x) +A(y)−1−A(x) 1−x

1 A(x) +A(y) + 2q

1−y

1

A(x) +A(y)−1−A(y) 1−y

1

A(x) +A(y). (5.7) We expand the first term of (5.7). By (2.1) and (2.3),

2p 1−x

1

A(x) +A(y) = 2p

m∈N

x^m

j,l∈E

aj+l(pq)^(j+l)/2x^jy^l

= 2p

j,l∈E,m∈N

aj+l(pq)^(j+l)/2x^j+my^l

= 2p

k,n∈N

j,l∈E,m∈N:

j+m=k,l=n−k

aj+l(pq)^(j+l)/2

x^ky^n−k (5.8)

= 2p

k,n∈N:

kn,n−k∈E

mk,n−m∈Em∈N:

a_n−m(pq)^(n−m)/2

x^ky^n−k (5.9)

= 2p

k,n∈N:

kn,

1l_E(n−k)

n−kini∈E:

a_i(pq)^i/2

x^ky^n−k. (5.10)

Let us explain certain transformations made in the above calculations:

(1) In (5.8), we have introduced the indices k = j +m and n = j +l +m. Then j = k−m and l =n−k, the conditionsj, l ∈ E are equivalent to k−m 0, n−k 0, k−m ∈ E, n−k ∈ E, or mkn, n−k∈ E, n−m∈ E. This leads to (5.9);

(2) From (5.9) to (5.10), we have puti=n−m.

We expand the second term of (5.7). By (2.1) and (2.3), 1−A(x)

1−x

1

A(x) +A(y)= 4

m∈N

x^m

i∈E^∗

a_i−2(pq)^i/2xⁱ

j,l∈E

a_j+l(pq)^(j+l)/2x^jy^l

= 4

i∈E^∗,j,l∈E,m∈N

ai−2aj+l(pq)^(i+j+l)/2x^i+j+my^l

= 4

k,n∈N

i∈E∗,j,l∈E,m∈N:

i+j+m=k,l=n−k

ai−2aj+l(pq)^(i+j+l)/2

x^ky^n−k (5.11)

= 4

k,n∈N:

kn,n−k∈E

i∈E∗,m∈N:

i+mk,i+n−m∈E

a_i−2a_n−m−i(pq)^(n−m)/2

x^ky^n−k (5.12)

= 4

k,n∈N:

kn,n−k∈E m∈N:

n−m∈E i∈E∗:

ik−m

ai−2an−m−i

(pq)^(n−m)/2

x^ky^n−k (5.13)

= 4

k,n∈N:

kn,n−k∈E m∈N:

n−m∈E j∈E:

jk−m−2

ajan−m−j−2

(pq)^(n−m)/2

x^ky^n−k (5.14)

= 4

k,n∈N:

kn,n−k∈E

m∈N\{0,1}:

n−m∈E j∈E:

jk−m

ajan−m−j

(pq)^(n−m)/2+1

x^ky^n−k (5.15)

= 4

k,n∈N:

kn

1l_E(n−k)

in−2i∈E: j∈E:

ji+k−n

a_ja_i−j

(pq)^i/2+1

x^ky^n−k. (5.16)

Let us explain the transformations used in the above computations:

(1) In (5.11), we have introduced the indicesk=i+j+mandn=i+j+l+m. Thenj=−i+k−m, l=n−k andi∈ E, the conditionsj, l∈ E are equivalent tok−i−m0, n−k0, k−i−m∈ E, n−k∈ E, ori+mkn, n−k∈ E, n−m∈ E. This leads to (5.12);

(2) From (5.13) to (5.14), we have puti=j+ 2;

(3) From (5.14) to (5.15), we have changedminto m−2;

(4) From (5.15) to (5.16), we have puti=n−m.

By invoking the argument of duality which is explained in Remark 4.3, the two last terms of (5.7) can be deduced from the two first ones by interchangingpandq, andxandy (that iskandn−k). This yields

2q 1−y

1

A(x) +A(y) = 2q

k,n∈N:

kn

1l_E(k)

kini∈E:

ai(pq)^i/2

x^ky^n−k, (5.17)

1−A(y) 1−y

1

A(x) +A(y) = 4

k,n∈N:

kn

1l_E(k)

in−2i∈E: j∈E:

ji−k

ajai−j

(pq)^i/2+1

x^ky^n−k. (5.18)

By identifying the coefficients of the series lying in the definition ofGand in (5.10), (5.16), (5.17) and (5.18),

we extract (5.4).

Remark 5.4. The probabilities

r0,n=P{Tn= 0}=P{S10, . . . , Sn0}=P{τ1> n}, rn,n=P{Tn=n}=P{S10, . . . , Sn 0}=P{τ−1> n}

are given by

r_0,n= 1−2p

in−1i∈E:

a_i(pq)^i/2 and r_n,n= 1−2q

in−1i∈E:

a_i(pq)^i/2. (5.19) The probabilities

r_1,n=P{T_n= 1}=P{S₁0, . . . , S_n−10, S_n>0}=P{τ₁=n}, rn−1,n=P{Tn=n−1}=P{S10, . . . , Sn−10, Sn<0}=P{τ−1=n}

are given byr_1,n=r_n−1,n= 0 ifnis even and, ifnis odd,

r1,n= 2an−1p^(n+1)/2q^(n−1)/2 and rn−1,n= 2an−1p^(n−1)/2q^(n+1)/2. (5.20) Formulas (5.20) can be easily deduced from (5.5) and (5.6). Formulas (5.19) could be deduced from Theorem5.3.

The computations are postponed to the report [7].

Proposition 5.5. For even n, k such that0kn, the following relationship holds:

r_k,n =r_k,kr_0,n−k. (5.21)

Proof. The probabilities r_k,n,k, n∈ E, are generated by the function ˜G which was introduced in Section5.2.

By (5.3), the quantity ˜G(x, y) can be factorized into the product of ˜G(x,0) and ˜G(0, y). Writing that G˜(x,0) ˜G(0, y) =

k,l∈E

(rk,kr0,l)x^ky^l=

k,n∈E:

kn

(rk,kr0,n−k)x^ky^n−k,

we conclude, by identification, that (5.21) holds true. Formula (5.21) could be obtained by using the explicit representation (5.19) of the probabilitiesrk,k andr0,n−k and the representation (5.6) ofrk,n. Nevertheless, the computations are very cumbersome. They are included in the report [7]. In [7], the reader can find also a proof

by induction.

Corollary 5.6. In the symmetric case (p = q = 1/2), the well-known following expression holds for even integers n, k such that0kn:

r_k,n = 1 2ⁿ

k k/2

(n−k) (n−k)/2

. Proof. In the case wherep= 1/2, we have the particular identities

P{τ0=j}=P{Sj−2= 0} −P{Sj= 0} and P{τ−1=j}=P{τ0=j+ 1}

which are due to the fact thatbj−2−bj/4 =aj−2 and to (2.10) respectively, and then rk,k =P{τ−1k+ 1}=

jk+1j∈O:

P{τ−1=j}=

jk+1j∈O:

P{τ0=j+ 1}=

jk+2j∈E:

P{τ0=j}

=

jk+2j∈E:

[P{Sj−2= 0} −P{Sj= 0}] =P{Sk= 0}= 1 2^k

k k/2

.

Analogously,

r0,n−k =P{τ1n−k+ 1}=P{Sn−k= 0}= 1 2^n−k

(n−k) (n−k)/2

.

We conclude with the help of (5.21).