Sample Paths, Convergence, and Averages

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Sample Paths, Convergence, and Averages

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Sample Paths, Convergence, and Averages



Definition: {an : n = 1, 2, …} converges to b as n  ∞ means that  > 0, n0(), such that n > n0(), |an – b| < Example: an = 1/n converges to 0 as n  ∞ since  > 0, choosing n0() = 1/ ensures that n > n0(), an = 1/n < 1/n0() < 



Convergence Almost Surely(with Probability 1)

{Yn : n = 1, 2, …} converges almost surely to μ as n  ∞, if ε > 0, P{limn∞ |Yn – μ| > ε} = 0In other words, the probability mass of the set of sample paths that misbehave (Yn deviates from μ by more than ε) is 0 if we take n to be large enough Note: This does not mean that no sample path misbehaves, and only that the total probability mass of these “bad” paths is (of measure) zeroA sample path misbehaves if no matter what value of n we choose, it is still possible to have |Yn – μ| Consider the average of n coin flips. Most sample paths will converge to a value of ½. However, some sample paths wont, e.g., the sample path 1,1,1,1,1,1,… We have convergence almost surely, because even though there are an uncountable number of “bad” paths, that number becomes vanishingly small (of measure 0) as n goes to infinity



Convergence in Probability

Yn : n = 1, 2, …} converges in probability to μ as n  ∞, if  ε > 0, limn∞P{|Yn – μ| > ε} = 0(the limit applies to the probability, and not the random variables)In other words, the odds that an individual sample path behaves badly for Yn go to 0 as n  ∞However, this does not preclude the possibility that all (or a non-negligible number) sample paths occasionally behave badly as n  ∞Convergence almost surely implies convergence in probability, but the converse is not trueExample: See review problem S4.4



Strong & Weak Laws of Large Numbers

Weak Law: Let X1, X2, X3,… be i.i.d. random variables with mean E[X], Sn = Σ{i=1,…,n}Xi, and Yn = Sn/n Then Yn converges in probability to E[X]Strong Law: Let X1, X2, X3,… be i.i.d. random variables with mean E[X], Sn = Σ{i=1,…,n}Xi, and Yn = Sn/n Then Yn converges almost surely to E[X]Implications: Almost every sample paths of successive trials of Xn converges to a mean value of E[X]There can still be bad sample paths, but the odds of picking one tend to zero as n  ∞



Time & Ensemble Averages

Basic concept: Run a single experiment for 10 hours versus run 100 experiments of 6 minutes eachRecord average value of experiment variable over the 10 hours of the first experiment and compare it to the average of the experiment variable at the end of each of the 100 6 minutes experimentsTime averageEnsemble average where pi = limt∞P{N(t) = i}




An ergodic system is positive recurrent, aperiodic, and irreducibleIrreducible: We can get from any state to any other statePositive recurrent: Every system state is visited infinitely often, and the mean time between successive visits is finite (and the visit of each state is a renewal point – the system probabilistically restart itself)Aperiodic: The system state is not deterministically coupled to a particular time periodFor an ergodic system, the ensemble average exists and with probability 1 is equal to the ensemble average



Non-Ergodic Systems

Systems where the state evolution depends on some initial conditionsFlip a coin and with probability p the system will receive 1 job/sec, and with probability 1 – p it will receive 2 jobs/secSystem state at time t will be very different depending on the starting condition Periodic systemsAt the start of every 5mins interval, the system receives 2 jobs each taking 1min to processThe ensemble average does not exist in this case (different results depending on when the systems are sampled)



System Averages

Average number of jobs in system: NTime average: In every time interval during which the number of jobs is constant, record interval duration, ti, and number of jobs, niTime average of number of jobs is Ntime = (ΣiNiti)/(Σiti)Ensemble average: Run M experiments of duration t, where Mt = Σiti, and record nk(t), k = 1, 2, …, MEnsemble average of number of jobs is Nensemble = Σknk(t)/MAverage job time in system









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