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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Ask a homework question - tutors are online