Hypergeometric solution to a gambler's ruin problem with a nonzero halting probability

0.17 MB·English

by  Ken Yamamoto

Hypergeometric solution to a gambler’s ruin problem with a nonzero
halting probability
Ken Yamamoto
Department of Physics, Faculty of Science and Engineering, Chuo University, Kasuga, Bunkyo,
Tokyo 112-8551, Japan
3 E-mail: [email protected]
1
0 Abstract
2
This paper treats of a kind of a gambler’s ruin problem, which seeks the probability that a
n
a random walker first hits the origin at a certain time. In addition to a usual random walk which
J hopseitherrightwardsorleftwards,thepresentpaperintroducesthe‘halt’thatthewalkerdoesnot

7 hopwith a certainprobability. The solutionto the problemcanbe obtained exactlyusing a Gauss
hypergeometric function. The moment generating function of the duration is also calculated, and
]
h a calculation technique of the moments is developed. The author derives the long-time behavior
p of the ruin probability, which exhibits power-law behavior if the walker hops to the right and left
- with equal probability.
h
t
a PACS numbers: 05.40.Fb, 02.50.-r
m
Keywords: first-passage problem; gambler’s ruin; hypergeometric function; asymptotic analysis
[

2
1 Introduction
v
4
1 The gambler’s ruin problem is a classical subject of probability theory. Consider a random walker on
3
4 a one-dimensional lattice hopping to the right with probability p→ and left with p←(= 1 p→) in a
−
. single time step. The problem seeks statistical properties of duration T , the time at which the walker
1 x
1 from x first hits the origin. (Of course, Tx is a random variable.) Not only a model of bankruptcy [1]
2 and gambling [2], but queuing theory [3] and genetic algorithm [4] are concerned with the gambler’s
1
ruin problem.
:
v
A standard textbook of probability theory [5] gives detailed instructions on this problem. The
i
X central quantity is the probability P˜(x,t) that the walker at position x first hits the origin after time
r t. As thesolution of theequation P˜(x,t+1) = p→P˜(x+1,t)+p←P˜(x 1,t) withinitial and boundary
a −
conditions P˜(0,0) = 1 and P˜(x,0) = P˜(0,t) = 0 (x,t 1),
≥
x t
p(t−x)/2p(t+x)/2 t and x are of same parity,
P˜(x,t) = t t+x → ← (1)
 (cid:18) 2 (cid:19)
0 t and x are of the opposite parity.
(t−x)/2 (t+x)/2
The coefficient before p→ p← is the number of different paths from x hitting the origin first at
time t, and it is connected with the reflection principle of a random walk [6]. Obviously, P˜(x,t) = 0
holds when t < x, because t+tx = 0.
2
Instatisticalphysics,asimilarproblemiscalledthefirst passage problem[7]. Thisisamoregeneral
(cid:0) (cid:1)
problem than the ruin problem, in that the state space of a random walker (or a diffusion particle)
is not necessarily a one-dimensional lattice but higher-dimensional spaces and networks. Many fields
1
p0
p← p→
O
xx−−11 x x+1
halt
duration Tx
halt
time
Figure 1: An illustration of the problem. The random walker hops to the right with probability p→,
andto theleft withp←; hoppingdoesnot occurwith probability p0. (p→+p←+p0 = 1.) Theproblem
focuses on statistical properties of the duration T , which is the time the walker first hits the origin.
x
in statistical physics, including reaction-rate theory [8], neuron dynamics [9], and economic analysis
[10], have been formulated and analyzed based on first-passage properties.
The present paper analyzes an extended form of the classical ruin problem; a random walker
hops to the right with probability p→, to the left with p←, and it does not hop with probability
p0 = 1 p→ p←. Figure 1 schematically shows the problem. The only difference from the original
− −
ruin problem is that the halting probability p is introduced. The difference seems very small, but
0
the results become quite distinct. In fact, the solution (1) of non halting case (p = 0) is superseded
0
by a formula involving a Gauss hypergeometric function in halting case p = 0. Moreover, moment
0
6
analysis and asymptotic (long-time) behavior are developed.
2 Hypergeometric solution
As in the classical problem described in the previous section, let P(x,t) be the probability that the
walker starting from x has duration Tx = t. P(x,t) satisfies the equation P(x,t+1) = p→P(x+1,t)+
p←P(x 1,t)+p0P(x,t) with initial and boundary conditions P(0,0) = 1 and P(x,0) = P(0,t) = 0
−
(x,t 1). However, itishardtosolvethisequation directly, andwetakeananotherwaybyemploying
≥
a classical result (1) effectively.
To calculate P(x,t), we classify the walker’s paths according to the number of hopping. We count
the paths consisting of j hops and t j halts. First, the different patterns of putting t j halts into
t steps are given by t−1 in total, w−here ‘t 1’ (not t) comes from the fact that a halti−ng step never
t−j −
comes to the last t-th step. Next, if we focus on only the hopping steps (and ignore the halting steps),
(cid:0) (cid:1)
the paths are reduced to those of classical ruin problem; the probability that the walker from x hits
the origin after j hops is given by P˜(x,j). Thus the total occurring probability of a path with j hops
and t j halts is
−
t−1 pt−jP˜(x,j).
t j 0
(cid:18) − (cid:19)
2
Summing up all j, we get
t
P(x,t) = t−1 pt−jP˜(x,j)
t j 0
j=x(cid:18) − (cid:19)
X
t
= p(→j−x)/2p(←j+x)/2pt0−jx j+x !(tj−−x1)!!(t j)!. (2)
Xj=x 2 2 −
j+xeven
(cid:16) (cid:17) (cid:16) (cid:17)
In the third equality, the range of summation is changed; the terms corresponding to j > t have no
contribution because 1 = 0. In the last equality we change the summation variable as j = x+2k.
(t−j)!
Thisresultcanbealsoobtained bysolvingdirectly theequation P(x,t+1) = p→P(x+1,t)+p←P(x
−
1,t)+p P(x,t) with conditions P(0,0) = 1, and P(x,0) = P(0,t) = 0 (x 1, t 1).
0
≥ ≥
We further proceed with calculation; we rewrite factorials using gamma functions, which is a
preparatorystep toward ahypergeometricfunction. Employingsomeformulasof thegamma function,
one obtains
1 1 Γ(x−t + 1 +k)Γ(x−t +k)
= 2 2 2 22k. (3)
(t x 2k)! Γ(t+1 x) Γ(x−t + 1)Γ(x−t)
− − − 2 2 2
(See Appendix A for the calculation in detail.) The probability P(x,t) is expressed as
(t 1)!
P(x,t) = px←pt0−xxΓ(x+1)Γ−(t x+1)
−
Γ(x+1) ∞ Γ(x−t +k+ 1)Γ(x−t +k) 1 4p→p← k
2 2 2
Γ(x−t + 1)Γ(x−t) Γ(x+k+1) k! p2
2 2 2 k=0 (cid:18) 0 (cid:19)
X
= px←pt0−x(x (t1)−!(1t)! x)!F x−2 t,x−2t+1;x+1; 4p→p2p← , (4)
− − (cid:18) 0 (cid:19)
where F(α,β;γ;z) is the Gauss hypergeometric function. This is an explicit form of the solution of
our problem. This solution cannot be deduced from the non-halting solution (1). The hypergeometric
functioninEq. (4)isagenuinehypergeometricfunction,inthesensethatitcannotbeexpressedusing
simplerfunctions. (Ithasbeenstudiedthatsomehypergeometricfunctionshavetractableexpressions,
e.g., F(1/2,1;3/2; z2) = z−1arctanz. [11])
−
We comment here on the convergence of the sum in Eq. (4)—at first sight, it seems to diverge
when 4p→p←/p20 > 1. By using the Pochhammer symbol (α)k := α(α+1)···(α+k−1) instead of
gamma functions,
P(x,t) = px←pt0−x(x (t1)−!(1t)! x)! ∞ (x−2(t+x1+)k1()x−2t)k k1! 4p→p2p← k.
− − k=0 k (cid:18) 0 (cid:19)
X
If x > t, P(x,t) = 0 holds automatically because 1/(t x)! = 0. On the other hand, if x t, either
(x−t) or (x−t+1) becomes zero for k > t−x . (More −precisely, the former becomes zero w≤hen x and
2 k 2 k ⌊ 2 ⌋
t are of same parity, and the latter becomes zero otherwise.) The sum consists of a finite number of
terms in reality, so one does not need to worry about the convergence.
The exact solution (4) is difficult to understand intuitively. We show numerical evaluation of
P(x,t) in Fig. 2. The initial position of the walker is fixed as x = 50. We separate graphs according
to the parameter ∆p := p← p→; the panels (a), (b), and (c) respectively correspond to ∆p = 0.2,
−
0.1, and 0. (∆p is a key parameter for the average duration, as discussed in the following section.)
P(x,t) is a unimodal function of t for each parameter value. Power-law behavior P(x,t) t−3/2 is
∝
suggested in large t when ∆p = 0 (see (d)), which is further discussed in Sec. 4.
3
0.009 0.004
0.008
0.007 p =0.1 0.003
→
,t=50)00..000056 ,t=50) 0.002 p→=0.1
0.004
x x
P(0.003 P(
0.001
0.002 p =0.3 p =0.3
0.001 → →
0 0
0 100 200 300 400 500 0 200 400 600 800 1000
t t
(a) (b)
0.00012 10 3
−
0.00010 p =0.3
10−4 →
,t0) 0.00008 p→=0.3 ,t0) 10−5 slope
=5 0.00006 =5 −23
x x 10 6
P( 0.00004 p→=0.1 P( − p→=0.1
10 7
0.00002 −
0 10 8
0 2000 4000 6000 8000 10000 −102 103 104 105 106
t t
(c) (d)
Figure2: Numericalresults of P(x,t)as afunction of t (x = 50). Thegraphs areseparated according
to the value of ∆p = p← p→: (a) ∆p = 0.2, (b) ∆p = 0.1, and (c) ∆p = 0. The two curves in
−
each panel represent p→ = 0.1 and 0.3. Long-time behavior of P(x,t) in ∆p = 0 is shown in (d) on
logarithmic scales, which suggests a power law P(x,t) t−3/2.
∝
3 Analysis of moment
In order to see properties of the random variable T , moment analysis is developed in this section.
x
First, we calculate the moment generating function of T , defined as
x
∞
M (s) = estP(x,t). (5)
x
t=0
X
Calculation process is summarized in Appendix B, and the result is
x
e−s p0 e−s p0 2 p←
M (s) = − − . (6)
x
 2p→ −s 2p→ − p→
(cid:18) (cid:19)
 
M (s) contains no special functions, so this is simpler than P(x,t) of Eq. (4). Remarkably, M (s)
x x
is described by elementary functions, though its definition (5) is an infinite sum of which each term
contains a genuine hypergeometric function Remarkably, M (s) contains no special functions, though
x
the infinite summation in Eq. (5) involves a genuine hypergeometric function via P(x,t).
4
Repeatedly differentiating M (s) and putting s = 0, we have the first and second moments of the
x
duration T as
x
x
T = ,
x
h i p← p→
−
x2 p←+p→ 1
T2 = + x.
h xi (p← p→)2 (p← p→)3 − p← p→
− (cid:26) − − (cid:27)
The average and the other moments remain finite when p← > p→, and they diverge to infinity as
p← p→ 0. The variance of Tx is given by
− ց
p←+p→ 1
V(T ) := T2 T 2 = x.
x h xi−h xi (p← p→)3 − p← p→
(cid:26) − − (cid:27)
We can give a plain explanation for the average T . The walker moves to the left by a length
x
h i
∆p = p← p→ on average in a single hop. In other words, ∆p is a mean velocity of the walker toward
−
the origin. Hence, it takes x/∆p time steps on average to reach the origin.
The moment generating function (6) looks too complicated to calculate higher moments by differ-
entiation. Alternatively, we provide another calculation method for moments of the duration T .
x
We start from the difference equation for P(x,t):
P(x,t+1) = p→P(x+1,t)+p←P(x 1,t)+p0P(x,t).
−
Multiply t and take summation for t to get
∞ ∞
tP(x,t+1) = tp→P(x+1,t)+tp←P(x 1,t)+tp0P(x,t)
{ − }
t=0 t=0
X X
= p→ Tx+1 +p← Tx−1 +p0 Tx .
h i h i h i
The left-hand side can be expressed as
∞ ∞ ∞
tP(x,t+1) = (t+1)P(x,t+1) P(x,t+1) = T 1.
x
− h i−
t=0 t=0 t=0
X X X
∞
The last equality uses P(x,t+1) = 1, which means that a random walker surely hits the origin.
t=0
(This normalization breaks if p← < p→—see the discussion of total probability of ruin in Sec. 5 in
P
more detail.) We have a difference equation
p→ Tx+1 (p→+p←) Tx +p← Tx−1 = 1.
h i− h i h i −
A particular solution is given by Tx = x/(p← p→), and the characteristic equation p→λ2 (p→+
h i − −
p←)λ+p← = 0 admits the two roots 1 and p←/p→. Thus, the general solution is
x
x p←
T = +A+B ,
x
h i p← p→ p→
− (cid:18) (cid:19)
which has two constants A and B. In a limiting case p→ 0, the problem is still well-defined, and
→
T should be finite, so B = 0. Also, it is obvious that T = 0, so A = 0. The appropriate solution
x 0
h i h i
is therefore
x
T = .
x
h i p← p→
−
Nextwecalculatethesecondmoment. Multiplyingt2 andtakingsummation,wederiveadifference
equation
hTx2i−2hTxi+1 = p→hTx2+1i+p←hTx2−1i+p0hTx2i,
5
where we use t2 = (t + 1)2 2(t + 1) + 1 for the calculation on the left-hand side. We assume a
−
particular solution in the form T2 = C x2+C x, and determine two constants as
h xi 2 1
p←+p→ 1 1
C = , C = .
1 (p← p→)3 − p← p→ 2 (p← p→)2
− − −
The second moment is
T2 = x2 + p←+p→ 1 x = x2 + p0(1−p0)+4p→p←x.
h xi (p← p→)2 (p← p→)3 − p← p→ (p← p→)2 (p← p→)3
− (cid:26) − − (cid:27) − −
The homogeneous-solution part vanishes for the same reason as the average.
In short, this ‘bottom-up’ method calculates higher moments inductively, and generally Tk be-
h xi
comes a polynomial of degree k. For instance, the third moment is calculated as
x3 p←+p→ 1
T3 = + 3x2
h xi (p← p→)3 (p← p→)3 − p← p→
− (cid:26) − − (cid:27)
2(p←+p→)2+4p←p→ 3(p←+p→) 1
+ + x.
(p← p→)5 − (p← p→)3 p← p→
(cid:26) − − − (cid:27)
4 Asymptotic behavior
We look at long-time behavior of P(x,t) in this section. In particular, we confirm the power-law
behavior P(x,t) t−3/2 mentioned in Sec. 2 when p→ = p← (see Fig. 2 (d) for reference).
∝
Asummation andfactorial, as well as ahypergeometric function, arenotsuitable tostudy limiting
behavior, so we first reexpress the probability P(x,t) into a tractable form. (It may also be possible
to estimate factorials directly by Stirling’s formula.) According to Ref. [5], the solution (1) of the
classic ruin problem has another expression
1
P˜(x,t) = 2tp→(t−x)/2p←(t+x)/2 cost−1πφ sinπφ sinπxφdφ. (7)
· ·
Z0
This form is convenient in that one does not need to care about the parities of x and t. In fact, the
integral holds the information of parities:
x (t 1)!
1 − t and x are of same parity,
cost−1πφ sinπφ sinπxφdφ = 2t(t+x)!(t−x)! (8)
· ·  2 2
Z0 0 otherwise.
In Appendix C we prove this formula. Applying Eq. (7) to Eq. (2), we have the integral form of the
probability P(x,t):
t
P(x,t) = t−1 pt−jP˜(x,j)
t j 0
j=1(cid:18) − (cid:19)
X
= 2p→(1−x)/2p←(1+x)/2 1 t tt−1j pt0−j(2√p→p←cosπφ)j−1sinπφ·sinπxφdφ
Z0 Xj=1(cid:18) − (cid:19) 
1
= 2p→(1−x)/2p←(1+x)/2 (p0+2√p→p←cosπφ)t−1sinπφ sinπxφdφ
·
Z0
= 2p→(1−x)/2p←(1+x)/2(p0+2√p→p←)t−1 1 p0+2√p→p←cosπφ t−1sinπφ sinπxφdφ.
Z0 (cid:18) p0+2√p→p← (cid:19) ·
6
Here we estimate the integral
1 p0+2√p→p←cosπφ t−1
I := sinπφ sinπxφdφ
Z0 (cid:18) p0+2√p→p← (cid:19) ·
= πt p0+2√p→p←cos νt t−1sin ν sin xν 1 dν,
Z0 (cid:18) p0+2√p→p← (cid:19) t · t · πt
where the integral variable is changed as ν = πtφ in the second equality. For large t,
p0+2√p→p←cos νt t−1 1 √p→p← ν2 t−1 exp √p→p← ν2 .
(cid:18) p0+2√p→p← (cid:19) ≃ (cid:18) − p0+2√p→p← t2(cid:19) ≃ (cid:18)−p0+2√p→p← t (cid:19)
Hence the integrand is a rapidly decreasing function of ν, and we can extend the upper limit of
the integration to infinity. Together with the approximation sin(ν/t) ν/t, we can carry out the
≃
integration as
I x ∞exp √p→p← ν2 ν2dν = x p0+2√p→p← 3/2t−3/2.
≃ πt3 Z0 (cid:18)−p0+2√p→p← t (cid:19) 4√π (cid:18) √p→p← (cid:19)
Therefore, the asymptotic form is expressed as
P(x,t) x p→(1−x)/2p←(1+x)/2(p0+2√p→p←)t−1 p0+2√p→p← 3/2t−3/2.
≃ 2√π √p→p←
(cid:18) (cid:19)
By the arithmetic mean-geometric mean inequality, we note
p0+2√p→p← p0+p→+p← = 1,
≤
with equality holding if and only if p→ = p← (or ∆p = 0 by the notation in Sec. 2). Thus, the
long-time behavior of P(x,t) is completely different according to whether ∆p equals zero or not.
If ∆p = 0 (i.e., p→ = p←), p0 + 2√p→p← < 1, and hence P(x,t) asymptotically exhibits
• 6 6
exponential decay due to (p0+2√p→p←)t.
If ∆p = 0 (i.e., p→ = p← =:p), a power law with exponent 3/2
• −
x
P(x,t) t−3/2
≃ 2√πp
is concluded (recall Fig. 2 (d)).
5 Concluding remarks and discussions
In the present paper, we have solved a gambler’s ruin problem where a random walker hops to the
right with probability p→, left with p←, and does not hop with p0. We have calculated exactly the
probability P(x,t) that the walker starting from position x has duration T = t. The average and
x
higher moments of T are calculated in two ways: by the moment generating function, and by the
x
bottom-up calculation. The asymptotic form of P(x,t) is derived, and a power law P(x,t) t−3/2 is
∼
obtained when p→ = p←.
We make a brief discussion on the continuum limit, which is an appropriate scaling limit where
the step width and time interval of the walker tend to zero. Let the step width be δ and time interval
7
be ǫ, and position and time are scaled as ξ = δx and τ = ǫt. In the continuum limit we take δ,ǫ 0,
→
together with
δ δ2
(p→ p←) v, D
− ǫ → 2ǫ →
are both kept finite. v is the mean displacement per time unit, and D is the diffusioin coefficient
corresponding to a non-halting (or formally called a simple) random walk. The probability density
function
1 x t
ρ(ξ,τ) =lim P ,
ǫ δ ǫ
(cid:18) (cid:19)
is suitable in the continuum limit, rather than the probability distribution P(x,t). (‘lim’ represents
the continuum limit.) Carrying out calculation similar to that in Ref. [5],
ξ (ξ+vτ)2
ρ(ξ,τ) = exp
4π(1−p0)Dτ3 (cid:18)−4(1−p0)Dτ(cid:19)
p
isobtained. TheprobabilitydensityρiscalledtheinverseGaussiandistribution[12]. Inthecontinuum
limit, the halting effect p is reflected only upon the diffusion coefficient as (1 p )D. Comparing
0 0
−
P(x,t) of a discrete problem in Eq. (4) and above ρ(ξ,τ) of a continuum limit, we conclude that the
discrete random walk is far more difficult than the continuous diffusion, and that the results about
the discrete random walk in this paper cannot be attained from continuous diffuision problem.
Combining Eqs. (4), (5), and (6), we get
x
∞ estpx←pt0−x xt−11 F x−2 t,x−2t+1;x+1; 4p→p2p← = e−2sp−→p0 −s e−2sp−→p0 2− pp→← .
t=0 (cid:18) − (cid:19) (cid:18) 0 (cid:19) (cid:18) (cid:19)
X
 
We have not used the condition p→ +p← +p0 = 1 in the derivation of Eqs. (4) and (6), so p→, p←,
and p0 can take any value independently. Set p→ = z/4 and p← = p0 = 1,
x
∞ t 1 x t x t+1 2(e−s 1) 2(e−s 1) 2 4
est − F − , − ;x+1;z = − − .
x 1 2 2  z −s z − z
t=0 (cid:18) − (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
X
 
Furthermore, set s = ln2 (i.e., e−s = 2),
−
∞ x
1 t 1 x t x t+1 2 2√1 z
− F − , − ;x+1;z = − − .
2t x 1 2 2 z
t=0 (cid:18) − (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
X
These formulas are nontrivial, but the author cannot tell whether they are useful in practice. We
stress that an infinite series of which each term involves a genuine hypergeometric function is hardly
known.
We comment on the total probability of ruin. For the discrete problem, putting s = 0 the moment
8
generating function M (s),
x
∞
P(x,t) = M (0)
x
t=0
X
x
2
1 p0 1 p0 p←
= − −
 2p→ −s 2p→ − p→
(cid:18) (cid:19)
 x
p→+p← (p← p→)2
= −
2p→ − p 2p→ !
1 p← p→,
= p← x ≥
 p← < p→.
 p→
(cid:18) (cid:19)
If p← > p→, the gambler surely comes to ruin, but If p← < p→, the gambler can manage to avoid
running out of the bankroll with nonzero probability.
The equation P(x,t+1) = p→P(x+1,t)+p←P(x 1,t)+p0P(x,t) describes a traffic jam, where
−
P(x,t) stands for the probability distributionthat the jam consists of x cars at time t, and p→ and p←
are involved in the rates at which cars enter and leave the jam. The scaling behavior with exponent
3/2 hasbeenobserved inthelifetime distribution ofjams [13]. We thinkthat resultsobtained in this
−
papercan become theoretical bases; in particular, our discrete problem is comparable to amicroscopic
description of traffic, where discreteness is not negligible.
Acknowledgment
The author is grateful to Dr. Yoshihiro Yamazaki and Dr. Jun-ichi Wakita for their beneficial
comments.
A Calculation of Eq. (3)
Here we follow the calculation of Eq. (3) in detail. Employing the duplication formula Γ(2z) =
π−1/222z−1Γ(z)Γ(z+ 1), we rewrite
2
1 1 √π
= = . (9)
(t x 2k)! Γ(t x 2k+1) 2t−x−2kΓ(t−x k+ 1)Γ(t−x k+1)
− − − − 2 − 2 2 −
By Euler’s reflection formula Γ(z)Γ(1 z) = π/sinπz,
−
t x 1 1 π
Γ − k+ =
2 − 2 Γ(k t−x + 1)sinπ(t−x k+ 1)
(cid:18) (cid:19) − 2 2 2 − 2
1 ( 1)kπ
= −
Γ(k t−x + 1)sinπ(t−x + 1)
− 2 2 2 2
( 1)k t x 1 1 t x
= − Γ − + Γ − . (10)
Γ(k t−x + 1) 2 2 2 − 2
− 2 2 (cid:18) (cid:19) (cid:18) (cid:19)
Note that sinπ(t−x k+ 1) = ( 1)ksinπ(t−x + 1) because k is an integer. Similarly,
2 − 2 − 2 2
t x ( 1)k t x x t
Γ − k+1 = − Γ − +1 Γ − (11)
2 − Γ(k+ x−t) 2 2
(cid:18) (cid:19) 2 (cid:18) (cid:19) (cid:18) (cid:19)
9
is obtained. Substituting Eqs. (10) and (11) into Eq. (9), then using the duplication formula again as
t x 1 t x √π
Γ − + Γ − +1 = Γ(t x+1),
2 2 2 2t−x −
(cid:18) (cid:19) (cid:18) (cid:19)
we eventually come to the result
1 1 Γ(x−t + 1 +k)Γ(x−t +k)
= 2 2 2 22k.
(t x 2k)! Γ(t+1 x) Γ(x−t + 1)Γ(x−t)
− − − 2 2 2
B Calculation of the moment generating function (6)
We follow here the calculation of Eq. (6). Substituting Eq. (2) into Eq. (5),
∞ ∞
(t 1)!
Mx(s) = est pk→px←+kpt0−x−2kx(x+k)!k!(−t x 2k)!
t=0 k=0 − −
X X
∞ ∞
1 (t 1)!
= pk→px←+kp−0x−2kx(x+k)!k! (t x− 2k)!(p0es)t
k=0 t=0 − −
X X
∞ ∞
1 (τ +x+2k 1)!
= pk→px←+kp−0x−2kx(x+k)!k!(p0es)x+2k τ! − (p0es)τ, (12)
k=0 τ=0
X X
where a summation variable is changed as τ = t x 2k in the last equality. Applying the negative
− −
binomial expansion
∞
(τ +m 1)!
(1 y)−m = − yτ
− τ!(m 1)!
τ=0 −
X
with y = p es and m = x+2k, one obtains
0
∞
(τ +x+2k 1)!
− (p es)τ = (x+2k 1)!(1 p es)−x−2k,
0 0
τ! − −
τ=0
X
and simplifies Eq. (12) into
p← x ∞ (x+2k 1)! √p→p← 2k
M (s)= x − .
x e−s p (x+k)!k! e−s p
(cid:18) − 0(cid:19) k=0 (cid:18) − 0(cid:19)
X
We can transform as follows using the duplication formula Γ(2z) = π−1/222z−1Γ(z)Γ(z+ 1),
2
(x+2k 1)! = Γ(x+2k)
−
2x−1+2k x x+1
= Γ +k Γ +k
√π 2 2
(cid:18) (cid:19)
(cid:16) (cid:17)
2x−1 x x+1 Γ x +k Γ x+1 +k
= Γ Γ 2 2 22k
√π 2 2 Γ x Γ x+1
(cid:18) (cid:19) (cid:0) 2(cid:1) (cid:0) 2 (cid:1)
(cid:16) (cid:17)
Γ x +k Γ x+1 +k
= Γ(x) 2 2 22k(cid:0). (cid:1) (cid:0) (cid:1)
Γ x Γ x+1
(cid:0) 2(cid:1) (cid:0) 2 (cid:1)
(cid:0) (cid:1) (cid:0) (cid:1)
10