Chapter 3. Renewal Processes
Outline
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Outline
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Distribution and Limiting Behavior of
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… X1
X2
N(t)
S1 S2 SN(t) SN(t)+1 t
Distribution and Limiting Behavior of
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Distribution and Limiting Behavior of
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1. pmf of N(t)
(fundamental relationship)
Example
Density of N(t) for Poisson Process Let N be a Poisson process with rate λ. Recall that the sum of i.i.d. Exponential random variables is an Erlang random variable, such that
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…X1
X2
N(t)
S1 S2 SN(t) SN(t)+1t
Exp
What is the pmf of N(t) ?
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2. Limiting Time Average
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Strong Law of Large Numbers
If X1, X2, … is an i.i.d. random sequence with sample mean Mn(X),
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w.p. = with probability
1 2lim 1 n
n
X X X
n
P
2. Limiting Time Average (con’t)
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Strong Law for Renewal Processes
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Average of the first N(t) inter-arrival times
rate as t -> ∞
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Central Limit Theorem
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3. Central Limit Theorem for
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#
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=
4. Renewal Function
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4. Renewal Function (con’t)
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4. Renewal Function (con’t)
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4. Renewal Function (con’t)
Proof method in p.100 of the textbook
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4. Renewal Function (con’t)
Big Problem!!!
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4. Renewal Function (con’t)
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4. Renewal Function (con’t)
Why? See next slide.
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Laplace Transform of f *g
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Laplace Transform of f *g (con’t)
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4. Renewal Function (con’t)
Elementary Renewal Theorem
Before proving elementary renewal theorem, we first introduce the stopping time (or stopping rule) and Wald’s equation.
Stopping Time (Rule)
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Yes
Let Xn, n=1, 2,… be independent and let Then, is N is a stopping time? No
Stopping Time (Rule)
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For a renewal process {N(t), t≥0}, N(t) is not a stopping time for interarrival sequence Xi’s. Why? [Homework]
Stopping Time (Rule)
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For a renewal process, N(t) + 1 is a stopping time for the interarrival sequence X1, X2, . . . It can be seen that the following events are equivalent: #
• Examples. To illustrate some examples of random times that are stopping rules and some that are not, consider a gambler playing roulette with a typical house edge, starting with $100: – Playing one, and only one, game corresponds to the stopping time τ = 1, and is
a stopping rule.
– Playing until he either runs out of money or has played 500 games is a stopping rule.
– Playing until he is the maximum amount ahead he will ever be is not a stopping rule and does not provide a stopping time, as it requires information about the future as well as the present and past.
– Playing until he doubles his money (borrowing if necessary if he goes into debt) is not a stopping rule, as there is a positive probability that he will never double his money. (Here it is assumed that there are limits that prevent the employment of a martingale system or variant thereof (such as each bet being triple the size of the last). Such limits could include betting limits but not limits to borrowing.)
– Playing until he either doubles his money or runs out of money is a stopping rule, even though there is potentially no limit to the number of games he plays, since the probability that he stops in a finite time is 1
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Stopping Time (Rule)
Wald's Equation
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E[N ] is finite
E[X] is finite
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#
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Example (Simple Random Walk)
Solution
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4. Renewal Function (con’t)
Elementary Renewal Theorem
A simple consequence ?
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Example
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4. Renewal Function (con’t)
Elementary Renewal Theorem
A simple consequence? Not always!!!
Corollary [Ross Corollary 3.3.3 ]
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Elementary Renewal Theorem
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inf a: greatest lower bound of a sup a: least upper bound of a
(1)
(Corollary 3.3.3)
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(2)
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Corollary
Elementary Renewal Theorem
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=
(Corollary 3.3.3)
Example
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Example (M/G/1 Loss)
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Solution
A renewal occurs every time that a customer actually enters the booth.
If N(t) denotes the number of customers who enter the booth by t, then {N(t) , t≥0} will be a renewal process having mean interarrival time of
Since the time to the next customer follows exponential with rate λ from the memoryless property. So,
(1) Since the rate at which customers enter the booth is just the long-run renewal rate, it is given by
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#
Solution (con’t)
(2) The loss rate of customers is obtained by
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(3) The utilization of the booth is
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Lattice
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More examples.
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True!!
Background of Blackwell's Theorem
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Blackwell's Theorem
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(Proof in Feller)
Example
Solution
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Directly Riemann Integrable
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Directly Riemann Integrable (con’t)
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Key Renewal Theorem
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Blackwell theorem
Regenerative Processes
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0
0
x1 x2 x3 x4
t
t
i.i.d.
S1 = X1
Renewal Theory
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i.i.d. cycles
total law of probability
Renewal Theory (con’t)
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Renewal Theory (con’t)
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Laplace Transform on both sides
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Inverse Laplace transform on both sides
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Renewal Theory (con’t)
solution
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#
Elementary Renewal Theorem
Example
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Proof
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Corollary
Regenerative Processes
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