# See figure 37 figure 37a shows a a cs 3 2 xs 1 k 0 10

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Unformatted text preview: t < 0. FT (t) = c The complementary CDF, deﬁned by FT (t) = P {T > t} = 1 − FT (t), therefore satisﬁes: e−λt t ≥ 0 1 t < 0. c FT (t) = To ﬁnd the ﬁrst and second moments we use integration by parts: ∞ E [T ] = tλe−λt dt 0 = −te ∞ −λt ∞ + 0 1 1 = 0+ = . λ λ 0 e−λt dt 3.4. EXPONENTIAL DISTRIBUTION 79 ∞ E [T 2 ] = t2 λe−λt dt 0 ∞ = −t2 e−λt ∞ + 2te−λt dt 0 0 2 2 = 0 + 2 = 2. λ λ 2 1 1 1 Therefore, Var(T ) = E [T 2 ] − E [T ]2 = λ2 − λ2 = λ2 . The standard deviation of T is σT = λ . The standard deviation is equal to the mean. (Sound familiar? Recall that for p close to zero, the geometric distribution with parameter p has standard deviation nearly equal to the mean.) Example 3.4.1 Let T be an exponentially distributed random variable with parameter λ = ln 2. Find the simplest expression possible for P {T ≥ t} as a function of t for t ≥ 0, and ﬁnd P (T ≤ 1|T ≤ 2). Solution. c P {T ≥ t} = FT (t) = e−λt = e−(ln 2)t = 2−t , and P (T ≤ 1|T ≤ 2) = P {T ≤1,T ≤2} P {T ≤2} = P {T ≤1} P {T ≤2} = 1− 1 2 1− 1 4 2 = 3. Memoryless property of exponential distribution Suppose T is an exponentially distributed random variable with som...
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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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