MIT6_262S11_lec12

X t x5 t s1 s2 s3 e x2 1 t x d t t 0 e

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Unformatted text preview: Going to the limit t → ∞ N (t,ω )+1 ￿ i=1 2 Xi (ω ) 2t ￿ N￿ ) (t,ω 2 Xi (ω ) N (t, ω ) 1t lim Y (t, ω ) dt = lim t→∞ t→∞ t 0 t n=1 2N (t, ω ) ￿ E X2 = ￿ ￿ 2E [X ] ￿ This is infinite if E X 2 = ∞. Think of example where pX (￿) = 1 − ￿, pX (1/￿) = ￿. 5 Similar examples: Age Z (t) = t − SN (t) and duration, ￿ X (t) = SN (t)+1 − SN (t). ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿￿ ￿ ￿ ￿ ￿ ￿￿ ￿ ￿ Z (t) t S1 S2 S3 ￿ S4 ￿ ￿ E X2 1 t￿ X (τ ) dτ = t→∞ t 0 2E [X ] lim S5 S6 WP1. ✛ ✛ ￿ X (t) X5 ✲ ✲ t S1 S2 S3 ￿ ￿ ￿ E X2 1 t￿ X (τ ) dτ = t→∞ t 0 E [X ] lim S4 S5 S6 WP1. 6 Time-averages for renewal rewards Residual life, age, and duration are examples of as­ signing rewards to renewal processes. The reward R(t) at any time t is restricted to b e a function of the inter-renewal p eriod containing t. In simplest form, R(t) is restricted to b e a function ￿ R(Z (t), X (t)). The time-average for a sample path of R(t) is found by analogy to residual life. Start with the nth interrenewal interval. ￿ Sn(ω ) Rn(ω ) = Sn−1(ω ) R(t, ω ) dt Interval 1 goes from 0 to S1, with Z (t) = t. interval n, Z (t) = t − Sn−1, i.e., SN (t) = Sn−1. For 7 Rn = = = = ￿ Sn Sn−1 ￿ Sn Sn−1 ￿ Sn Sn−1 ￿ Xn 0 R(t) dt ￿ R(Z (t), X (t)) dt R(t − Sn−1, Xn) dt R(z, Xn) dz This is a function only of the rv Xn. Thus E [Rn] = ￿∞ x=0 ￿x z =0 R(z, x) dz dFX (x). Assuming that this expectation exists, ￿ 1t E [Rn] lim R(τ ) dτ = t→∞ t τ =0 X WP1 8 Example: Suppose we want to find the kth moment of the age. ￿ Then R(Z (t), X (t)) = Z k (t). Thus E [Rn] = = ￿∞ x=0 ￿x z =0 ￿ ∞ k+1 x 0 k+1 z k dz dFX (x) dFX (x) = ￿ 1 ￿ k+1￿ EX k ￿ ￿ E X k+1 1t lim R(τ ) dτ = t→∞ t τ =0 (k+1)X WP1 9 Stopping trials for stochastic processes It is often important to analyze the initial segment of a stochastic p...
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This note was uploaded on 01/13/2012 for the course ELECTRICAL 6.262 taught by Professor Staff during the Fall '11 term at MIT.

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