1 n has independent increments if 0 t0 t1 tn then the

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Unformatted text preview: lity the lightbulb survives through the first k − 1 clicks and hλ is the probability it then fails during the k th click. Let T = Lh. Thus, T is the time until the lightbulb fails, whereas L is the number of clicks until the lightbulb fails. Note that P {T = (k + 1)h|T > kh} = λh for k ≥ 0. In words, the lifetime T is always a multiple of h, and the failure probability for each click is λh. Since the click duration is h, this corresponds to a failure rate of λ per unit time. The complementary CDF of T can be found as follows: P {T > c} = P {Lh > c} = P {L > c/h } = (1 − hλ) c/h . λ Recall from (2.5) that (1 − n )n → e−λ as n → ∞. Replacing n by 1/h implies that (1 − hλ)1/h → e−λ as h → 0, and therefore, (1 − hλ)c/h → e−λc as h → 0. Also, the difference between c/h and c/h is not important, because 1 ≥ (1 − hλ)c/h /(1 − hλ) c/h ≥ (1 − hλ) → 1 as h → 0. Therefore, P {T > c} = (1 − hλ) c/h → e−λc , as h ...
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This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Institute of Technology.

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