Cette solution est disponible au format PDF

D1D2D3
D4
D1
Prob(D2D1) = 1
3,Prob(D3D1) = 5 9,Prob(D4D1) = 2 3;
D2
Prob(D1D2) = 2
3,Prob(D3D2) = 1 3,Prob(D4D2) = 1 2;
D3
Prob(D1D3) = 4
9,Prob(D2D3) = 2 3,Prob(D4D3) = 1 3;
D4
Prob(D1D4) = 1
3,Prob(D2D4) = 1 2,Prob(D3D4) = 2 3.
p=2
3>1 2
p=2
3

p=2
3
q=1
3
Xn= +1 Xn=−1
Sn=X1+· · · +XnSn
n
Snn p (Sn)n∈N
b
a a
Na= min{n>1 : Sn=a}N−b= min{n>1 : Sn=−b}
Na
N−b
b b
Prob( ) = Prob(Na< N−b).
Na< N−b
Prob(Na< N−b)
ui= Prob(Na< N−b|Sn0=i)n0
i
i∈ {−b, −b+ 1, ..., a −1, a}

uin0i
u0
(ui)−b6i6a
n0i
Xn0= +1 p Sn0−1=i−1
i−1
Xn0=−1q Sn0−1=i+ 1
i+ 1
−b<i<a
ui=pProb(Na< N−b|Sn0−1=i−1) + qProb(Na< N−b|Sn0−1=i+ 1).
ui=pui−1+qui+1
ua= 1 Sn0=a u−b= 0 Sn0=−b
qr2−r+p= 0
1ρ=p/q p =2
3q=1 3ρ6= 1
1ρ ui=λρi+µ
ua= 1 u−b= 0 λρa+µ= 1 λρ−b+µ= 0
λ=1
ρa−ρ−b=ρb ρa+b−1, µ =−1 ρa+b−1.
u0=λ+µ=ρb−1
ρa+b−1=qa(pb−qb) pa+b−qa+b.
p=2
3q=1 3
Prob( ) = 2b−1
2a+b−1.
a+∞
Prob( ) = 0.
b+∞
Prob( ) = 1
2a.
p=q=1
2
r2−2r+ 1 = 0
ui=λi+µ
ua= 1 u−b= 0 λa +µ= 1 −λb +µ= 0 λ=1
a+b
µ=b
a+bu0=µ
Prob( ) = b
a+b.

a+∞
Prob( ) = 0
b+∞
Prob( ) = 1.