Exponential Distribution Nur Aini Masruroh

Exponential distribution • Easy to work with mathematically • Often arises in practice as being the distribution of the amount of time until some specific event occurs – amount of time until an earthquake occurs, or until a war breaks out, or until a light bulb burns out • Has a unique property that makes it easy to analyze; It does not deteriorate with time – this means that if the lifetime of an item is exponentially distributed, then an item which has been in use for x (e.g. 10) hours, is as good as a new item in regards to the amount of time remaining until the item fails • It turns out that the exponential distribution is related to the stochastic process, the Poisson process.

Exponential distribution Recall definition: A continuous random variable X with parameter λ, λ>0, with probability density function Cumulative distribution function Mean = 1/λ, Variance = 1/λ2

Properties of exponential distribution Property 1: A random variable X is memoryless if P{X > s +t | X > t}= P{X > s}, s,t ≥ 0 From the definition of conditional distribution, we can rewrite

Example The amount of time a person spends in a bank is exponentially distributed with mean 10 minutes, λ=1/10. a) What is the probability that a customer will spend more than 15 minutes in the bank? b) What is the probability that a customer will spend more than 15 minutes in the bank given that he is still in the bank after 10 minutes?

Solution Let X represent the amount of time that the customer spends in the bank. Then P{X>15} = e-15λ = e-3/2≈ 0.22 Since the exponential distribution does not “remember” that the customer has already spent 10 minutes in the bank, P{X>15|X>10} = P{X>5} = e-5λ = e-1/2 = 0.604

Properties of exponential distribution Property 2 If X1, X2, …, Xn are independent identically distributed exponential random variables, each with mean (1/λ), then Y=X1+X2+…+Xn will have a Gamma distribution with parameters n, λ. The pdf is given by and E[Y] = (n/λ), V[Y] = (n/λ2)

Example The lifetime of an electronic tube is exponentially distributed with mean 1,000 hours. There are three of these tubes (original and 2 spares). When one burns out, it is replaced by a spare. What is the expected amount of time Y until the last spare burns out?

Solution The lifetimes X1, X2, X3 of the 3 tubes are independent exponential random variables each with parameter λ=1/1000 (since they do not interact with each other). The time Y until the last burn-out is given by Y=X1+X2+X3 From the property, Y must have a Gamma density with λ = 1/1000, n = 3 Hence, the expected time until the last bulb burns out E[Y] = (n/λ) = 3000

Properties of exponential distribution Property 3 If X1 and X2 are independent exponential random variables with means 1/λ1 and 1/λ2, then

Example Suppose a stereo system consists of two main parts, a radio and a speaker. If the lifetime of the radio is exponential with mean 1000 hours and the lifetime of the speaker is exponential with mean 500 hours, independent of the radio’s lifetime, then what is the probability that the system’s failure (when it occurs) will be caused by the radio failing?

Solution λ1=1/1000 λ2=1/500 The probability that the system’s failure will be caused by the radio failing is the same as the probability that the lifetime of the radio is less than the lifetime of the speakers. From the previous stated property of the Exponential distribution,

Summary of exponential distribution pdf, cdf, mean, variance Properties: Memoryless: P{X > s +t | X > t}= P{X > s}, s,t ≥ 0 Sum of n exponentials is Gamma distributed with mean E[Y] = (n/λ), V[Y] = (n/λ2) If X1 and X2 are independent exponential random variables with means 1/λ1 and 1/λ2, then