ࡱ> |~{[ +bjbj 0ΐΐ;!+( ( kkkkk8\$ &7;(ccc>>>$$$$$$$$'*b$k>>>>>$kkcc%###>kckc$#>$###cpٞ. z#$%0 &#*"*##^*k$>>#>>>>>$$#>>> &>>>>*>>>>>>>>>( 1: 9 Sums of exponential random variables and the Poisson distribution. We have seen how exponential random variables arise when we are modeling the time that it takes until something occurs. In particular, they are used to model the times between successive occurrences of some event, e.g. the arrival of a customer at a store. For this reason we are interested in sums of n exponential random variables since this will describe the time it takes for n occurrences of some event, e.g. the time it takes for the arrival of n customers at a store. If the exponential random variables are independent and identically distributed a nice formula for the distribution of the sum can be found and the resulting distribution is called the Erlang distribution Example 1. A bank is modeling the time T between customer arrivals. They assume T is an exponential random variable with mean 2 minutes. So the density function of T is (1) f(t) =  \eq \b\lc\{(\a( \f(e-t/2,2) for t > 0, , 0 for t < 0)) Suppose the times between arrivals are independent. The bank would like to know the following. a. The probability that the second customer arrives after 4 minutes from now. b. The probability that exactly one customer arrives within the next 4 minutes. Let T1 = time (in min) until the next customer arrives T2 = time between the arrival of the first and second customer S = T1 + T2 = time until the second customer arrives g(t) = probability density function of S For the first question we want to determine Pr{ 4 < S} =  \eq \i(4,(, )g(s) ds. Recall from the previous section that g(s) is the convolution of the density functions of T1 and T2 both of which are given f(t) in (1). Thus h(s) =  \eq \i(0,s, )f(t)g(s-t) dt =  \eq \i(0,s, ) \eq \f(e-t/2,2)  \eq \f(e-(s-t)/2,2) dt =  \eq \f(e-s/2,4)  \eq \i(0,s, )dt =  \eq \f(se-s/2,4) So Pr{ 4 < S} =  \eq \i(4,(, )g(s) ds =  \eq \i(4,(, ) \eq \f(se-s/2,4) ds = (-  \eq \f(se-s/2,2) - e-s/2) | \eq \o(4,() =  \eq \o(lim,s(()(-  \eq \f(se-s/2,2) - e-s/2) - (-  \eq \f(4e-4/2,2) - e-4/2) = 0 + 3e-2 ( 3(0.135) = 0.405 For the second question we want to know what is the probability that the second customer arrives after four minutes and the first customer arrives within 4 minutes, i.e. Pr{T1 ( 4 and S > 4}. Note that the event {T1( 4 and S > 4} is the same as the first customer arriving within 4 minutes and the second customer not arriving within 4 minutes, i.e. {T1( 4 and S > 4} = {T1( 4} {S ( 4}. Also, the event where the second customer arrives within 4 minutes is contained in the event where the first customer arrives within 4 minutes. So Pr{T1 ( 4 and S2 > 4} = Pr{T1( 4} Pr{S2 ( 4} =  \eq \i(0,4, ) \eq \f(e-s/2,2) ds -  \eq \i(0,4, ) \eq \f(se-s/2,4) ds = (- e-s/2) | \eq \o(0,4) - (-  \eq \f(se-s/2,2) - e-s/2) | \eq \o(0,4) =  \eq \f(se-s/2,2) | \eq \o(0,4) = 2e-2 ( 2(0.135) = 0.27 S is an example of an Erlang random variable. These have probability density functions of the form (2) f(t)= \eq \b\lc\{(\a\al(\f((ntn-1e-(t,(n-1)!) for t 0, , 0 for t < 0)) Here n is a positive integer and ( is a positive real number. These density functions are the density function of the sum of n independent exponential random variables with the same rates. Proposition 1. Let S = T1 + T2 + ... + Tn where T1, T2, ... , Tn are independent exponential random variables each with density function f(t) = le-lt for t 0 and f(t) = 0 for t < 0. Then S has density function h(t) of the form h(t) =  EQ \F(lntn-1e-lt,(n-1)!) for t 0 and h(t) = 0 for t < 0. Furthermore, the mean of S is n/(. Proof. This is proved by induction on n. It is clearly true for n = 1. We shall show it is true for n + 1 if it is true for n. Let Sn = T1 + T2 + ... + Tn and Sn+1 = T1 + T2 + ... + Tn+1. We are assuming the pdf of Sn is hn(t)=  EQ \F(lntn-1e-lt,(n-1)!). The pdf of Sn+1is hn+1(t) =  EQ \I(0,t, hn(r)f(t-r) dr) =  EQ \I(0,t, \F(lnrn-1e-lr,(n-1)!)le-l(t-r) dr) =  EQ \f(ln+1e-lt,(n-1)!) \I(0,t, rn-1 dr) =  EQ \f(ln+1tne-lt,n!) which proves the first result. The fact the mean of S is n/( follows from the fact that the mean of a sum is the sum of the means and the mean of an exponential random variable is 1/(. Poisson random variables. Let Sn = T1 + T2 + ... + Tn where T1, T2, ... , Tn are independent exponential random variables each with density function f(t) = le-lt for t 0 and f(t) = 0 for t < 0. To find the probability that S is between a and b one would have to integrate h(t) =  EQ \F(lntn-1e-lt,(n-1)!) between a and b. This integral can be done using integration by parts, but it takes some effort. It is usually easier to translate questions about S into questions about a related Poisson random variable. We let N(t) = max{n: Sn ( t} = the number of occurrences that have occurred by time t For each t this is a random variable. We want to calculate Pr{N(t) = n} for n = 0, 1, 2. ..... One has Theorem 2. Pr{N(t) = n} =  EQ \F(lntne-lt,n!) Proof. Note that N(t) = n if and only if Sn ( t and Sn+1 > t. So the event {N(t) = n} can be written as {N(t) = n} = {Sn ( t and Sn+1 > t} = {Sn ( t}  {Sn+1 ( t} where  is the difference between two sets, i.e A  B = {x: x ( A and x ( B} is the set of objects that are in A but not in B. If B is contained in A then Pr{A - B } = Pr{A} - Pr{B} since A is the union of B and A - B which are disjoint. The set {Sn+1 t} is contained in {Sn t}, because if Sn+1 t then Sn t. So (3) Pr{N(t) = n} = Pr{Sn ( t}  Pr{Sn+1 ( t} =  EQ \I(0,t, \F(lnsn-1e-ls,(n-1)!) ds) -  EQ \I(0,t, \F(ln+1sne-ls,n!) ds) In the second integral we integrate by parts letting u = sn/n! and dv = (n+1e-(s ds. 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A random variable M is a Poisson random variable if it only assumes values that are non-negative integers and there is a m > 0 such that Pr{M = n} =  EQ \F((ne-(,n!) for n=0, 1, 2, ..... Example 2. Suppose customers arrive at an bank at a rate of 30 per hour. Suppose the times between arrivals are exponentially distributed and independent. a. What is the probability that exactly 5 customers arrive in 10 minutes? b. What is the probability that at least 5 customers arrive in 10 minutes. c. What is the probability that the 5th customer arrives between 9 and 10 minutes. Solution. In this case ( = 1/2 if we measure time in minutes. Let N(t) be the number of customers that have arrived by time t. For part a we want Pr{N(10) = 5} =  EQ \F((1/2)5 105e-10/2,5!) =  EQ \F(55e-5,120) = (54e-5)/24 = 625e-5/24 0.175 For part b we want Pr{N(10) ( 5} = 1 - Pr{N(10) ( 4} = 1 - [ EQ \F(50e-5,0!) +  EQ \F(51e-5,1!) +  EQ \F(52e-5,2!) +  EQ \F(53e-5,3!) +  EQ \F(54e-5,4!)] = 1 - [1 + 5 +  EQ \F(25,2) +  EQ \F(125,6) +  EQ \F(625,24)] e-5 ( 0.56 For part c, note that if the 5th customer arrives between the 9th and 10th minute then the customers who have arrived by the 9th minute is 4 or less while the number of customers who have arrived th the 10th minute is at least 5. So the answer to part c is equal to Pr{N(9) ( 4 and N(10) ( 5} = Pr{N(9) ( 4 but not N(10) ( 4} Note that if N(10) ( 4 then N(9) ( 4. So Pr{N(9) ( 4 but not N(10) ( 4} = Pr{N(9) ( 4} - Pr{N(10) ( 4} = [ EQ \F(4.50e-4.5,0!) +  EQ \F(4.51e-4.5,1!) +  EQ \F(4.52e-4.5,2!) +  EQ \F(4.53e-4.5,3!) +  EQ \F(4.54e-4.5,4!)] - [ EQ \F(50e-5,0!) +  EQ \F(51e-5,1!) +  EQ \F(52e-5,2!) +  EQ \F(53e-5,3!) +  EQ \F(54e-5,4!)] ( 0.092 Problem 1. 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