EURANDOM - Sept. 27, 2012
Asymptotic expected number of
passages of a random walk
through an interval
Offer Kella
The Hebrew University of Jerusalem
joint work with
Wolfgang Stadje
University of Osnabrück
EURANDOM - Sept. 27, 2012
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{Sn | n ≥ 0} - some stochastic process.
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{Sn | n ≥ 0} - some stochastic process.
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A - measurable set it its state space.
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{Sn | n ≥ 0} - some stochastic process.
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A - measurable set it its state space.
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In = 1{Sn ∈A} .
EURANDOM - Sept. 27, 2012
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{Sn | n ≥ 0} - some stochastic process.
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A - measurable set it its state space.
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In = 1{Sn ∈A} .
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Passage of lenth k through A:
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{Sn | n ≥ 0} - some stochastic process.
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A - measurable set it its state space.
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In = 1{Sn ∈A} .
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Passage of lenth k through A:
Sn 6∈ A, Sn+1 ∈ A, . . . , Sn+k ∈ A, Sn+k+1 6∈ A
EURANDOM - Sept. 27, 2012
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{Sn | n ≥ 0} - some stochastic process.
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A - measurable set it its state space.
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In = 1{Sn ∈A} .
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Passage of lenth k through A:
Sn 6∈ A, Sn+1 ∈ A, . . . , Sn+k ∈ A, Sn+k+1 6∈ A
or
(In , In+1 , . . . , In+k , In+k+1 ) = (0, 1, . . . , 1, 0) .
EURANDOM - Sept. 27, 2012
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{Sn | n ≥ 0} - some stochastic process.
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A - measurable set it its state space.
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In = 1{Sn ∈A} .
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Passage of lenth k through A:
Sn 6∈ A, Sn+1 ∈ A, . . . , Sn+k ∈ A, Sn+k+1 6∈ A
or
(In , In+1 , . . . , In+k , In+k+1 ) = (0, 1, . . . , 1, 0) .
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Vi - length of the ith completed passage (if exists).
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EURANDOM - Sept. 27, 2012
Questions:
1. If {Sn } is stationary and ergodic, what is the average passage
lengths?
EURANDOM - Sept. 27, 2012
Questions:
1. If {Sn } is stationary and ergodic, what is the average passage
lengths?
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2. If Sn = ni=1 Xi is a random walk with EXn > 0
EURANDOM - Sept. 27, 2012
Questions:
1. If {Sn } is stationary and ergodic, what is the average passage
lengths?
P
2. If Sn = ni=1 Xi is a random walk with EXn > 0
and Ax = (x, x + h] or Ax = (x, ∞),
EURANDOM - Sept. 27, 2012
Questions:
1. If {Sn } is stationary and ergodic, what is the average passage
lengths?
P
2. If Sn = ni=1 Xi is a random walk with EXn > 0
and Ax = (x, x + h] or Ax = (x, ∞),
does the following limit exist?
lim E (number of passages through Ax )
x→∞
EURANDOM - Sept. 27, 2012
Questions:
1. If {Sn } is stationary and ergodic, what is the average passage
lengths?
P
2. If Sn = ni=1 Xi is a random walk with EXn > 0
and Ax = (x, x + h] or Ax = (x, ∞),
does the following limit exist?
lim E (number of passages through Ax )
x→∞
If yes, then what is it?
EURANDOM - Sept. 27, 2012
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Assume that {Sn } is stationary and ergodic.
EURANDOM - Sept. 27, 2012
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Assume that {Sn } is stationary and ergodic.
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Recall Ii = 1{Si ∈A} .
EURANDOM - Sept. 27, 2012
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Assume that {Sn } is stationary and ergodic.
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Recall Ii = 1{Si ∈A} .
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Denote pij = P(I0 = i, I1 = j), pi = P(I0 = i) and assume
p10 > 0.
EURANDOM - Sept. 27, 2012
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Assume that {Sn } is stationary and ergodic.
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Recall Ii = 1{Si ∈A} .
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Denote pij = P(I0 = i, I1 = j), pi = P(I0 = i) and assume
p10 > 0.
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Note: p10 = EI0 (1 − I1 )
EURANDOM - Sept. 27, 2012
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Assume that {Sn } is stationary and ergodic.
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Recall Ii = 1{Si ∈A} .
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Denote pij = P(I0 = i, I1 = j), pi = P(I0 = i) and assume
p10 > 0.
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Note: p10 = EI0 (1 − I1 ) = EI0 − EI0 I1
EURANDOM - Sept. 27, 2012
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Assume that {Sn } is stationary and ergodic.
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Recall Ii = 1{Si ∈A} .
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Denote pij = P(I0 = i, I1 = j), pi = P(I0 = i) and assume
p10 > 0.
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Note: p10 = EI0 (1 − I1 ) = EI0 − EI0 I1 = EI1 − EI0 I1
EURANDOM - Sept. 27, 2012
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Assume that {Sn } is stationary and ergodic.
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Recall Ii = 1{Si ∈A} .
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Denote pij = P(I0 = i, I1 = j), pi = P(I0 = i) and assume
p10 > 0.
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Note: p10 = EI0 (1 − I1 ) = EI0 − EI0 I1 = EI1 − EI0 I1 = p01 .
EURANDOM - Sept. 27, 2012
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Assume that {Sn } is stationary and ergodic.
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Recall Ii = 1{Si ∈A} .
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Denote pij = P(I0 = i, I1 = j), pi = P(I0 = i) and assume
p10 > 0.
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Note: p10 = EI0 (1 − I1 ) = EI0 − EI0 I1 = EI1 − EI0 I1 = p01 .
(Level crossing)
EURANDOM - Sept. 27, 2012
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Assume that {Sn } is stationary and ergodic.
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Recall Ii = 1{Si ∈A} .
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Denote pij = P(I0 = i, I1 = j), pi = P(I0 = i) and assume
p10 > 0.
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Note: p10 = EI0 (1 − I1 ) = EI0 − EI0 I1 = EI1 − EI0 I1 = p01 .
(Level crossing)
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Lk - number of completed passages by time k.
EURANDOM - Sept. 27, 2012
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Assume that {Sn } is stationary and ergodic.
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Recall Ii = 1{Si ∈A} .
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Denote pij = P(I0 = i, I1 = j), pi = P(I0 = i) and assume
p10 > 0.
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Note: p10 = EI0 (1 − I1 ) = EI0 − EI0 I1 = EI1 − EI0 I1 = p01 .
(Level crossing)
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Lk - number of completed passages by time k.
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Lk = ki=0 Ii (1 − Ii+1 ).
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EURANDOM - Sept. 27, 2012
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Assume that {Sn } is stationary and ergodic.
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Recall Ii = 1{Si ∈A} .
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Denote pij = P(I0 = i, I1 = j), pi = P(I0 = i) and assume
p10 > 0.
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Note: p10 = EI0 (1 − I1 ) = EI0 − EI0 I1 = EI1 − EI0 I1 = p01 .
(Level crossing)
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Lk - number of completed passages by time k.
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Lk = ki=0 Ii (1 − Ii+1 ).
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Hence, Lk /k → EI0 (1 − I1 ) = p10 = p01 > 0.
EURANDOM - Sept. 27, 2012
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Assume that {Sn } is stationary and ergodic.
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Recall Ii = 1{Si ∈A} .
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Denote pij = P(I0 = i, I1 = j), pi = P(I0 = i) and assume
p10 > 0.
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Note: p10 = EI0 (1 − I1 ) = EI0 − EI0 I1 = EI1 − EI0 I1 = p01 .
(Level crossing)
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Lk - number of completed passages by time k.
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Lk = ki=0 Ii (1 − Ii+1 ).
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Hence, Lk /k → EI0 (1 − I1 ) = p10 = p01 > 0.
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In particular, Lk → ∞.
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Nn = sup{k| Lk ≤ n}.
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Nn = sup{k| Lk ≤ n}.
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Nn /n → µ iff Lk /k → λ where

 ∞ λ=0
1/λ 0 < λ < ∞
µ=

0
λ=∞
EURANDOM - Sept. 27, 2012
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Nn = sup{k| Lk ≤ n}.
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Nn /n → µ iff Lk /k → λ where

 ∞ λ=0
1/λ 0 < λ < ∞
µ=

0
λ=∞
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Thus Nn /n → 1/p10 > 0
EURANDOM - Sept. 27, 2012
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Nn = sup{k| Lk ≤ n}.
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Nn /n → µ iff Lk /k → λ where

 ∞ λ=0
1/λ 0 < λ < ∞
µ=

0
λ=∞
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Thus Nn /n → 1/p10 > 0, Nn → ∞.
EURANDOM - Sept. 27, 2012
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Now,
LNn +k
LNn +k Nn + k
1
=
→ p10 ·
=1
n
Nn + k
n
p10
EURANDOM - Sept. 27, 2012
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Now,
LNn +k
LNn +k Nn + k
1
=
→ p10 ·
=1
n
Nn + k
n
p10
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LNn
X
i=1
Vi ≤
n
X
i=1
LNn +1
Vi <
X
i=1
Vi
EURANDOM - Sept. 27, 2012
Hence, from
LNn +1
LNn
n
X
LNn +1 1
LNn 1 X
1X
Vi <
Vi
Vi ≤
n LNn
n
n LNn +1
i=1
i=1
i=1
we have that
n
1X
Vi → v
n
i=1
⇐⇒
Lk
1 X
Vi → v
Lk
i=1
EURANDOM - Sept. 27, 2012
Next,
Lk
X
i=1
Vi ≤
k
X
i=1
Ii ≤
LX
k +1
i=1
Vi
EURANDOM - Sept. 27, 2012
Next,
Lk
X
i=1
Vi ≤
k
X
i=1
Ii ≤
LX
k +1
Vi
i=1
thus,
Lk
k
1X
1X
lim
Vi = lim
Ii = EI0 = p1
k→∞ k
k→∞ k
i=1
i=1
EURANDOM - Sept. 27, 2012
Next,
Lk
X
Vi ≤
i=1
k
X
Ii ≤
i=1
LX
k +1
Vi
i=1
thus,
Lk
k
1X
1X
lim
Vi = lim
Ii = EI0 = p1
k→∞ k
k→∞ k
i=1
and so
Lk
1 X
Vi =
Lk
i=1
i=1
1
k
PLk
i=1 Vi
Lk
k
→
p1
p10
EURANDOM - Sept. 27, 2012
Proposition:
{Sn } is stationary and ergodic, and A is such that
P(S0 ∈ A, S1 6∈ A) > 0, then, a.s.
n
1X
1
Vi →
n
P(S1 6∈ A|S0 ∈ A)
i=1
EURANDOM - Sept. 27, 2012
In fact, stationary and ergodic is sufficient, but all that we really
used was only that
k
k
1X
1X
1{Si ∈A} ≥
1{S0 ∈A,Si+1 6∈A}
k
k
i=0
both converge to a positive limit.
i=0
EURANDOM - Sept. 27, 2012
In fact, stationary and ergodic is sufficient, but all that we really
used was only that
k
k
1X
1X
1{Si ∈A} ≥
1{S0 ∈A,Si+1 6∈A}
k
k
i=0
i=0
both converge to a positive limit.
e.g. {Sn } is a positive (Harris) recurrent Markov process
EURANDOM - Sept. 27, 2012
In fact, stationary and ergodic is sufficient, but all that we really
used was only that
k
k
1X
1X
1{Si ∈A} ≥
1{S0 ∈A,Si+1 6∈A}
k
k
i=0
i=0
both converge to a positive limit.
e.g. {Sn } is a positive (Harris) recurrent Markov process
or (delayed) regenerative with finite mean inter-regeneration epoch.
EURANDOM - Sept. 27, 2012
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a− = − min(a, 0), a+ = max(a, 0).
EURANDOM - Sept. 27, 2012
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a− = − min(a, 0), a+ = max(a, 0).
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X , X1 , X2 , . . . i.i.d.
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a− = − min(a, 0), a+ = max(a, 0).
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X , X1 , X2 , . . . i.i.d.
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EX − < ∞, EX > 0.
EURANDOM - Sept. 27, 2012
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a− = − min(a, 0), a+ = max(a, 0).
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X , X1 , X2 , . . . i.i.d.
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EX − < ∞, EX > 0.
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S0 = 0, Sn = ni=1 Xi , n ≥ 0.
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EURANDOM - Sept. 27, 2012
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a− = − min(a, 0), a+ = max(a, 0).
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X , X1 , X2 , . . . i.i.d.
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EX − < ∞, EX > 0.
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S0 = 0, Sn = ni=1 Xi , n ≥ 0.
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N x - number of passages of {Sn } through (x, x + h].
EURANDOM - Sept. 27, 2012
Example 1:
Simple random walk:
P(X = 1) = 1 − P(X = −1) = p = 1 − q > 1/2
EURANDOM - Sept. 27, 2012
Example 1:
Simple random walk:
P(X = 1) = 1 − P(X = −1) = p = 1 − q > 1/2
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P(ever reaching i + 1|start from i) = 1
EURANDOM - Sept. 27, 2012
Example 1:
Simple random walk:
P(X = 1) = 1 − P(X = −1) = p = 1 − q > 1/2
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P(ever reaching i + 1|start from i) = 1
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P(ever reaching i − 1|start from i) = q/p = r .
EURANDOM - Sept. 27, 2012
Gambler’s ruin:
1 − ri
1 − rN
ri − rN
P(reach 0 before N|start from i) =
1 − rN
P(reach N before 0|start from i) =
EURANDOM - Sept. 27, 2012
In particular:
1−r
1 − r h+1
r − r h+1
P(reach 0 before h + 1|start from 1) =
1 − r h+1
1 − rh
P(reach h + 1 before 0|start from h) =
1 − r h+1
r h − r h+1
P(reach 0 before h + 1|start from h) =
1 − r h+1
P(reach h + 1 before 0|start from 1) =
EURANDOM - Sept. 27, 2012
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A = 1, . . . , h.
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A = 1, . . . , h.
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ah - expected number of passages of A starting from 0.
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A = 1, . . . , h.
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ah - expected number of passages of A starting from 0.
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bh - expected number of passages of A starting from h + 1.
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A = 1, . . . , h.
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ah - expected number of passages of A starting from 0.
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bh - expected number of passages of A starting from h + 1.
ah = 1 +
r − r h+1
1−r
ah +
bh
h+1
1−r
1 − r h+1
EURANDOM - Sept. 27, 2012
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A = 1, . . . , h.
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ah - expected number of passages of A starting from 0.
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bh - expected number of passages of A starting from h + 1.
ah = 1 +
r − r h+1
1−r
ah +
bh
h+1
1−r
1 − r h+1
so
ah =
1 − r h+1
+ bh
1−r
EURANDOM - Sept. 27, 2012
ah =
1 − r h+1
+ bh
1−r
EURANDOM - Sept. 27, 2012
ah =
bh = r
1 − r h+1
+ bh
1−r
r h − r h+1
1 − rh
1+
a
+
bh
h
1 − r h+1
1 − r h+1
EURANDOM - Sept. 27, 2012
ah =
bh = r
1 − r h+1
+ bh
1−r
r h − r h+1
1 − rh
1+
a
+
bh
h
1 − r h+1
1 − r h+1
bh
r − r h+1 bh
1−r h
=1+
+
r ah
h+1
r
r
1−r
1 − r h+1
EURANDOM - Sept. 27, 2012
ah =
bh = r
1 − r h+1
+ bh
1−r
r h − r h+1
1 − rh
1+
a
+
bh
h
1 − r h+1
1 − r h+1
bh
r − r h+1 bh
1−r h
=1+
+
r ah
h+1
r
r
1−r
1 − r h+1
1 − r h+1
bh
=
+ r h ah
r
1−r
EURANDOM - Sept. 27, 2012
bh = ah −
bh =
1 − r h+1
1−r
r
(1 − r h+1 ) + r h+1 ah
1−r
EURANDOM - Sept. 27, 2012
bh = ah −
bh =
1 − r h+1
1−r
r
(1 − r h+1 ) + r h+1 ah
1−r
Subtracting gives
(1 − r h+1 )ah =
1+r
(1 − r h+1 )
1−r
EURANDOM - Sept. 27, 2012
bh = ah −
bh =
1 − r h+1
1−r
r
(1 − r h+1 ) + r h+1 ah
1−r
Subtracting gives
(1 − r h+1 )ah =
or
ah =
1+r
1−r
1+r
(1 − r h+1 )
1−r
EURANDOM - Sept. 27, 2012
bh = ah −
bh =
1 − r h+1
1−r
r
(1 − r h+1 ) + r h+1 ah
1−r
Subtracting gives
(1 − r h+1 )ah =
or
ah =
1+r
(1 − r h+1 )
1−r
1 + q/p
1+r
=
1 − q/p
1−r
EURANDOM - Sept. 27, 2012
bh = ah −
bh =
1 − r h+1
1−r
r
(1 − r h+1 ) + r h+1 ah
1−r
Subtracting gives
(1 − r h+1 )ah =
or
ah =
1+r
(1 − r h+1 )
1−r
1 + q/p
p+q
1+r
=
=
1 − q/p
p−q
1−r
EURANDOM - Sept. 27, 2012
bh = ah −
bh =
1 − r h+1
1−r
r
(1 − r h+1 ) + r h+1 ah
1−r
Subtracting gives
(1 − r h+1 )ah =
or
ah =
1+r
(1 − r h+1 )
1−r
1 + q/p
p+q
1
1+r
=
=
=
1 − q/p
p−q
p−q
1−r
EURANDOM - Sept. 27, 2012
For h = ∞:
ah = 1 + rah
EURANDOM - Sept. 27, 2012
For h = ∞:
ah = 1 + rah
or
ah =
1
1−r
EURANDOM - Sept. 27, 2012
For h = ∞:
ah = 1 + rah
or
ah =
p
1
=
1−r
p−q
EURANDOM - Sept. 27, 2012
The renewal process case X > 0:
EURANDOM - Sept. 27, 2012
The renewal process case X > 0:
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νx = sup{n| Sn ≤ x}.
EURANDOM - Sept. 27, 2012
The renewal process case X > 0:
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νx = sup{n| Sn ≤ x}.
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γ(x) = Sνx +1 − x.
EURANDOM - Sept. 27, 2012
The renewal process case X > 0:
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νx = sup{n| Sn ≤ x}.
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γ(x) = Sνx +1 − x.
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One passage through (x, x + h] iff γ(x) ≤ h.
EURANDOM - Sept. 27, 2012
The renewal process case X > 0:
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νx = sup{n| Sn ≤ x}.
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γ(x) = Sνx +1 − x.
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One passage through (x, x + h] iff γ(x) ≤ h.
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No passage through (x, x + h] iff γ(x) > h.
EURANDOM - Sept. 27, 2012
The renewal process case X > 0:
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νx = sup{n| Sn ≤ x}.
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γ(x) = Sνx +1 − x.
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One passage through (x, x + h] iff γ(x) ≤ h.
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No passage through (x, x + h] iff γ(x) > h.
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N x = 1{γ(x)≤h} .
EURANDOM - Sept. 27, 2012
The renewal process case X > 0:
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νx = sup{n| Sn ≤ x}.
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γ(x) = Sνx +1 − x.
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One passage through (x, x + h] iff γ(x) ≤ h.
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No passage through (x, x + h] iff γ(x) > h.
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N x = 1{γ(x)≤h} .
EURANDOM - Sept. 27, 2012
EN x = P(γ(x) ≤ h)
EURANDOM - Sept. 27, 2012
1
EN = P(γ(x) ≤ h) →
EX
x
Z
h
P(X > s)ds
0
EURANDOM - Sept. 27, 2012
1
EN = P(γ(x) ≤ h) →
EX
x
Z
h
P(X > s)ds =
0
EX ∧ h
EX
EURANDOM - Sept. 27, 2012
1
EN = P(γ(x) ≤ h) →
EX
x
For the case h = ∞, clearly
limit.
EN x
Z
h
P(X > s)ds =
0
EX ∧ h
EX
= 1 for all x hence also in the
EURANDOM - Sept. 27, 2012
Theorem: Let 0 < h < ∞,
(a) If X has a nonarithmetic distribution,
lim EN x =
x→∞
E |X | ∧ h
.
EX
(b) If X has an arithmetic distribution with span α
then the same holds for h ∈ {kα| k ≥ 0}.
EURANDOM - Sept. 27, 2012
Theorem: Let h = ∞ and EX + < ∞. Then
lim EN x =
x→∞
EX +
.
EX
EURANDOM - Sept. 27, 2012
For the renewal case
E |X | ∧ h
EX ∧ h
=
EX
EX
EURANDOM - Sept. 27, 2012
For the renewal case
E |X | ∧ h
EX ∧ h
=
EX
EX
and
as obtained earlier.
EX +
EX
=
=1
EX
EX
EURANDOM - Sept. 27, 2012
For the random walk case:
EURANDOM - Sept. 27, 2012
For the random walk case:
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α = 1.
EURANDOM - Sept. 27, 2012
For the random walk case:
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α = 1.
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EX = p − q.
EURANDOM - Sept. 27, 2012
For the random walk case:
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α = 1.
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EX = p − q.
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|X | = 1
EURANDOM - Sept. 27, 2012
For the random walk case:
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α = 1.
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EX = p − q.
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|X | = 1 ⇒ |X | ∧ h = 1
EURANDOM - Sept. 27, 2012
For the random walk case:
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α = 1.
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EX = p − q.
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|X | = 1 ⇒ |X | ∧ h = 1 ⇒ E |X | ∧ h = 1
EURANDOM - Sept. 27, 2012
For the random walk case:
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α = 1.
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EX = p − q.
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|X | = 1 ⇒ |X | ∧ h = 1 ⇒ E |X | ∧ h = 1
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EX + = p.
EURANDOM - Sept. 27, 2012
For the random walk case:
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α = 1.
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EX = p − q.
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|X | = 1 ⇒ |X | ∧ h = 1 ⇒ E |X | ∧ h = 1
I
EX + = p.
Thus
E |X | ∧ h
1
=
EX
p−q
EURANDOM - Sept. 27, 2012
For the random walk case:
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α = 1.
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EX = p − q.
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|X | = 1 ⇒ |X | ∧ h = 1 ⇒ E |X | ∧ h = 1
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EX + = p.
Thus
E |X | ∧ h
1
=
EX
p−q
EX +
p
=
EX
p−q
as obtained earlier.
EURANDOM - Sept. 27, 2012
In fact, whenever P(|X | ≤ b) = 1 and h ≥ b, then
E |X | ∧ h
E |X |
=
EX
EX
independent of h.
EURANDOM - Sept. 27, 2012
THAT’S IT FOR NOW...
EURANDOM - Sept. 27, 2012
... or is there time to show the proof?
EURANDOM - Sept. 27, 2012
I
R(A) = E
P∞
n=0 1{Sn ∈A} .
EURANDOM - Sept. 27, 2012
P∞
I
R(A) = E
n=0 1{Sn ∈A} .
I
Nonarithmetic.
EURANDOM - Sept. 27, 2012
P∞
I
R(A) = E
n=0 1{Sn ∈A} .
I
Nonarithmetic.
I
R((x, x + h]) →
h
EX
as x → ∞.
EURANDOM - Sept. 27, 2012
P∞
I
R(A) = E
n=0 1{Sn ∈A} .
I
Nonarithmetic.
I
R((x, x + h]) →
I
R((−∞, x]) → 0 as x → −∞.
h
EX
as x → ∞.
EURANDOM - Sept. 27, 2012
P∞
I
R(A) = E
n=0 1{Sn ∈A} .
I
Nonarithmetic.
I
R((x, x + h]) →
I
R((−∞, x]) → 0 as x → −∞.
I
R((x, x + h]) ≤ ah + b for all x, h.
h
EX
as x → ∞.
EURANDOM - Sept. 27, 2012
EN x
=
E
P∞
n=0 1{Sn 6∈(x,x+h],Sn+1 ∈(x,x+h]}
EURANDOM - Sept. 27, 2012
EN x
P∞
=
E
=
P∞
n=0 1{Sn 6∈(x,x+h],Sn+1 ∈(x,x+h]}
n=0 P(Sn
∈ (−∞, x] ∪ (x + h, ∞), Sn+1 ∈ (x, x + h])
EURANDOM - Sept. 27, 2012
EN x
P∞
=
E
=
P∞
∈ (−∞, x] ∪ (x + h, ∞), Sn+1 ∈ (x, x + h])
=
P∞
∈ ((−∞, x] ∪ (x + h, ∞)) ∩ (x − X , x + h − X ])
n=0 1{Sn 6∈(x,x+h],Sn+1 ∈(x,x+h]}
n=0 P(Sn
n=0 P(Sn
EURANDOM - Sept. 27, 2012
EN x
P∞
=
E
=
P∞
∈ (−∞, x] ∪ (x + h, ∞), Sn+1 ∈ (x, x + h])
=
P∞
∈ ((−∞, x] ∪ (x + h, ∞)) ∩ (x − X , x + h − X ])
=
ER((x − X + , x − X + + X + ∧ h])
+ER((x + h + X − − X − ∧ h, x + h + X − ])
n=0 1{Sn 6∈(x,x+h],Sn+1 ∈(x,x+h]}
n=0 P(Sn
n=0 P(Sn
EURANDOM - Sept. 27, 2012
EN x
P∞
=
E
=
P∞
∈ (−∞, x] ∪ (x + h, ∞), Sn+1 ∈ (x, x + h])
=
P∞
∈ ((−∞, x] ∪ (x + h, ∞)) ∩ (x − X , x + h − X ])
=
ER((x − X + , x − X + + X + ∧ h])
+ER((x + h + X − − X − ∧ h, x + h + X − ])
→
EX + ∧h
EX
n=0 1{Sn 6∈(x,x+h],Sn+1 ∈(x,x+h]}
n=0 P(Sn
n=0 P(Sn
+
EX − ∧h
EX
EURANDOM - Sept. 27, 2012
EN x
P∞
=
E
=
P∞
∈ (−∞, x] ∪ (x + h, ∞), Sn+1 ∈ (x, x + h])
=
P∞
∈ ((−∞, x] ∪ (x + h, ∞)) ∩ (x − X , x + h − X ])
=
ER((x − X + , x − X + + X + ∧ h])
+ER((x + h + X − − X − ∧ h, x + h + X − ])
→
EX + ∧h
EX
n=0 1{Sn 6∈(x,x+h],Sn+1 ∈(x,x+h]}
n=0 P(Sn
n=0 P(Sn
+
EX − ∧h
EX
=
E |X |∧h
EX
EURANDOM - Sept. 27, 2012
EN x
P∞
=
E
=
P∞
∈ (−∞, x] ∪ (x + h, ∞), Sn+1 ∈ (x, x + h])
=
P∞
∈ ((−∞, x] ∪ (x + h, ∞)) ∩ (x − X , x + h − X ])
=
ER((x − X + , x − X + + X + ∧ h])
+ER((x + h + X − − X − ∧ h, x + h + X − ])
→
EX + ∧h
EX
n=0 1{Sn 6∈(x,x+h],Sn+1 ∈(x,x+h]}
n=0 P(Sn
n=0 P(Sn
+
EX − ∧h
EX
=
E |X |∧h
EX
Note:
R((y , y + X ± ∧ h]) ≤ R((y , y + h]) ≤ ah + b
EURANDOM - Sept. 27, 2012
For h = ∞ there is a similar argument only that in the end one has
ER((x − X + , x]) →
EX +
EX
EURANDOM - Sept. 27, 2012
For h = ∞ there is a similar argument only that in the end one has
ER((x − X + , x]) →
EX +
EX
which is allowed since R((x − X + , x]) ≤ aX + + b and EX + < ∞.
EURANDOM - Sept. 27, 2012
REALY THE END!