Lecture 19 - Exponential random variable and poisson process PDF

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Exponential random variable and poisson process...

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IEMS 315: Stochastic Models Ohad Perry

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Lecture 19: Summary - Exponential R.V. and Poisson Process Exponential Batteries A flashlight needs two batteries to be operational. Consider such a flashlight along with a set of n functional batteries, numbered 1 to n. Initially, batteries 1 and 2 are installed. Whenever a battery fails, it is immediately replaced by the lowest numbered functional battery that has not yet been put in use. Suppose that the lifetimes of the different batteries are independent exponential random variables each having rate µ. At a random time T a battery will fail and our stockpile of functional batteries will be empty. At that moment, exactly one of the batteries – which we call battery X – will not yet have failed. (a) What is P(X = n) ? The last battery is also the last functional battery with probability 1/2. The question is equivalent to asking which of two IID exponential r.v.’s is smaller since, when the last battery is inserted to the flashlight, the other active battery inside the flashlight has an exponential life time with rate µ, due to the memoryless property.

(b) What is P(X = 1) ? The last battery to be functional is battery 1 if it outlasted battery 2, then 3, 4, etc. Each time a battery is replaced, the remaining lifetime of battery 1 is exponential with rate µ, due to the memoryless property. Each time a battery is replaced, battery 1 has a probability 1/2 of outlasting the other battery in the flashlight, because the probability that a given battery lives longer than the other battery inside the flashlight is the probability that an exp(µ) random variable is the smaller of the two IID exp(µ) r.v.’s. This probability is µ/(µ + µ) = 1/2. Therefore, the answer is  n−1 1 , 2 since there will be n − 1 replacements.

(c) What is P(X = i), 1 < i < n? Use the same logic as in (b) above to get (1/2)n−i+1

i > 1.

(Battery i must “win” against the battery that was in the flashlight when it was inserted, and then against the remaining n − i batteries.)

(d) Find E[T ].

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IEMS 315, Lecture 19

T is the sum of n − 1 IID exponentials each having parameter 2µ: E[T ] =

n−1 . 2µ

(e) Find V ar(T ). For the same reason as (d) above, V ar(T ) =

n−1 4µ2

(e) What is the distribution of T ? The r.v. T , being a sum of n − 1 IID exponential r.v.’s with mean 1/(2µ), has the Gamma distribution with parameters n − 1 and 2µ. Since n − 1 is an integer, the Gamma distribution is called “Erlang”. (Gamma is a more general type of distribution than Erlang.)

Vacationing in Jamaica Seeking some rest after their hard studies, the students in IEMS 315 decided that, once the quarter is over, they will all buy new fishing rods and spend a weekend fishing in Negril, Jamaica. In addition to reminiscing about the good times they had during their M-W-F 9am stochastic classes, their goal is to catch grouper, snapper, tuna, sea bass, salmon, trout, Mahi-Mahi, swordfish, herring, mackerel, cod, halibut, tilapia, dolphins and blue whales. (Of course, any dolphin or blue whale caught will be set free immediately.) Suppose that fish are caught according to a Poisson process at a rate 3 per hour (where, for simplicity, “fish” also refers to dolphins and blue whales, although they are actually mammals). (a) How many students does it take to pull a blue whale out of the water? Many.

(b) What is the probability that exactly 4 fish are caught in a given 2-hour period? Let N (t) be the number of fish caught in the time interval [0, t). Then {N (t) : t ≥ 0} is a Poisson process with rate λ = 3 per hour, which means that P(N (2) = 4) =

e−(3·2)(3 · 2)4 e−6 64 ≈ 0.134. = 4! 4!

(c) What is the expected number of fish caught in a two-hour period?

IEMS 315, Lecture 19

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Since N (2) ∼ P oiss(6), E[N (2)] = 6.

(d) What is the variance of the number of fish caught in a two-hour period? A Poisson r.v. has variance that is equal to the mean, so that V ar(N (2)) = 6.

(e) What is the probability that at least two fish are caught in a two-hour period? P(N (2) ≥ 2) = 1 − P(N (2) = 0) − P(N (2) = 1) = 1 − e−6 − 6e−6 = 1 − 7e−6 . (f) What is the conditional probability that exactly 4 fish are caught in a given 2-hour period, given that 23 fish were caught in the previous two hours?

P(N (t + 2) − N (t) = 4|N (t) − N (t − 2) = 23) = P(N (t + 2) − N (t) = 4) = P(N (2) = 4) so the answer is just like in Part (b) above. The first equality follows from the independent increments property, and the second equality follows from the stationary increments property of the Poisson process.

(g) What is the conditional expected number of fish that are caught in a given 2-hour period, given that 23 fish were caught in the previous two hours? By the independence of the disjoint time intervals, the conditional mean we are asked about is equal to the unconditional mean: E [N (t + 2) − N (t)|N (t) − N (t − 2) = 23] = E [N (t + 2) − N (t)] = E[N (2)] = 6. (h) What is the expected time until the fourth fish is caught? In a Poisson process, the times between successive events are IID exponential r.v.’s with rate λ. Here, the mean length of time between successive catches of fish is 1/λ = 1/3 hour = 20 minutes. Let Ti be the time interval between the (i − 1)st and ith fish. Then E [T1 + T2 + T3 + T4 ] = 4E [T1 ] = 4/3. Also, if we let Sn be the time at which the nth fish is caught, then Sn is Erlang with parameters n and λ and, in particular, E[Sn ] = n/λ. Then E[S4 ] = 4/3.

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IEMS 315, Lecture 19

(i) What is the variance of the time until the fourth fish is caught? The variance of the sum of independent r.v.’s is equal to the sum of the variances, so that V ar(T1 + T2 + T3 + T4 ) = 4V ar (T1 ) = 4/9. We can again use the fact that Sn is Erlang and has variance equal to n/λ2 : V ar (S4 ) = 4/9. (j) How likely would it be that more than 90 fish are caught in 27 hours? Here we use the normal approximation, as discussed in Lecture 15. We normalize by centering about the mean, which is E[N (27)] = 3 · 27 = 81, and dividing by the standard deviation, which is equal to 9 (since the variance of a Poisson r.v. is equal to its mean). Then   N (27) − 81 90 − 81 > ≈ P(N (0, 1) > 1) ≈ 0.16. P(N (27) > 90) = P 9 9

(k) How likely would it be that it would take less than 27 hours to catch 90 fish? Here we are asked to computed P(S90 < 27), where S90 is the time when the 90th fish is caught. Since S90 is a sum of 90 IID exponential r.v.’s, we can again use the normal approximation. The mean of each exponential is 1/3 and the variance is 1/9, so the mean of S90 is 30 and the variance is 10. Then   −3 S90 − 30 27 − 30 √ < √ ≈ P(N (0, 1) < − P(S90 < 27) = P ) ≈ 0.16. 3.162 10 10 Alternatively, we can use the inverse relation we studied in Lecture 16 to deduce the answer from the previous question: N (t) ≥ n if and only if Sn ≤ t.

Using this inverse relation does not give exactly the same outcome as in (j) above in terms of the normal approximation, but the normal approximations for (j) and (k) are close, and become closer to each other as n increases. (The normal approximation holds due to the CLT, so it is a limiting argument.)

(l) Suppose that exactly 4 fish are caught in a given two-hour period. What then is the probability that all four are caught in the first 30 minutes? Conditional on the number of events in the interval (a, b) being n, those n events are distributed as n IID uninform r.v.’s on [a, b]. Then all fish were caught in the first 1/2 hour if the last of them was caught in the first 1/2 hour. Letting UM := max{U1 , . . . , U4 }, where Ui ∼ U nif (0, 2), i = 1, . . . , 4, we have P(UM ≤ 1/2) = P(U1 ≤ 1/2, . . . , U4 ≤ 1/2) = (1/4)4 .

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(m) Suppose that exactly 4 fish are caught in a given two-hour period. What is the probability that the first fish is caught after 30 minutes? The first fish is caught after 1/2 hour if and only if all fish are caught after 1/2 hour, so the probability is (3/4)4 . In particular, letting Um denote the minimum of four IID U nif(0, 2), we have 4 4 Y Y 3/4 = (3/4)4 . P(Um > 1/2) = P(Ui > 1/2) = i=1

i=1

Suppose, in addition, that each fish caught is a grouper with probability 1/4, a snapper with probability 1/3 and a dolphin or a blue whale with 5/12, independently of all other fish caught. (Again, whales and dolphins are referred to as as “fish”.) (n) What is the probability that exactly 8 fish are caught in a given 2-hour period, with 3 of them being grouper and 5 being snapper? Here we use the splitting property of the Poisson process. Let NG (t) be the number of grouper caught in [0, t]; let NS (t) be the number of snapper caught in [0, t] and let NO (t) be the number of the sea mammals caught in [0, t]. Then NG (t), NS (t) and NO (t) are independent Poisson processes with respective rates λG := 3/4, λS := 1 and λO := 5/4. Then P(NG (2) = 3, NS (2) = 5, NO (2) = 0) = P(NG (2) = 3)P(NS (2) = 5)P(NO (2) = 0) e−1.5 (1.5)3 e−2 25 e−2.5 (2.5)0 = 0! 5! 3!

(o) What is the conditional probability that 8 grouper are caught in a 2-hour period, given that 3 snapper are caught in a subsequent 2-hour period? Here we have two disjoint intervals, but we also have two independent Poisson processes so that P(NG (2) = 8|NS (t + 2) − NS (t) = 3) = P(NG (2) = 8) =

e−1.5 (1.5)8 8!

(p) What is the conditional probability that 8 grouper are caught in a given 2-hour period, given that 14 snapper are caught in the same 2-hour period? Since the two Poisson processes are independent, the answer is the same as for the previous part....