Homework 6RenewalMATH 56A: STOCHASTIC PROCESSESANSWERS TO HOMEWORKHomework 6RenewalThese problems are due Thursday, April 3. Answers will be posted the following week.First problem: Suppose that we have a renewal process with a uniform distributionf(t) =(1/10 if 0 < t ≤ 100 otherwise(i.e., we have a light bulb which lasts at most 10 days and can burn out at any time withequal probability during those 10 days)This means that F (t) = t/10 for 0 ≤ t ≤ 10.F (t) =t/10 if 0 < t ≤ 100 if t ≤ 01 if t > 10You walk into the warehouse after one year. The light bulbs are distributed according tothe equilibrium distribution.(1) What is the average age of a light bulb (in this equilibrium distribution)?t = 1year. The age of the light bulb is At. The density function for AtisfA(s) =1 − F (s)µ=1 − s/105=15−s50for 0 ≤ s ≤ 10. So, the average value of A isE(Et) =Z100s5−s250ds = 10 −1000150=103(2) If you pick a light bulb and it has been burning for 4 days, what is the probabilitythat it will burn for at least 4 more days.I decided this is not a good question because na¨ıve methods give the correct answerwhereas sophisticated methods give the wrong answer. The correct answer is 1/3.The na¨ıve explanation is that this is the conditional probability that a light bulbwill last for at least 8 days given that it has lasted for 4 days. This isP(T ≥ 8 | T ≥ 4) =P(T ≥ 8)P(T ≥ 4)=2/106/10=13But his calculation does not use the equilibrium distribution. A more sophisticatedinterpretation is that this is the conditional probability that C ≥ 8 given that C ≥ 4:P(C ≥ 8 | C ≥ 4) =P(C ≥ 8)P(C ≥ 4)=1But:P(C ≥ t) =Z10txf(x)µdx =Z10tx50dx = 1 −t2100So,P(C ≥ 8 | C ≥ 4) =P(C ≥ 8)P(C ≥ 4)=1 −641001 −16100=37However, this second calculation does not take into account that the condition is notjust that C ≥ 4 but also that you observed the light bulb when it was 4 days old.So, what I think is a very careful answer is:P(C ≥ 8 | 4 < A ≤ 4 + dt) =P(C ≥ 8 and 4 < A ≤ 4 + ds)P(4 < A ≤ 4 + ds)The probability that the age is 4 is the density function of AP(4 < A ≤ 4 + ds) = fA(4)ds =1 − F (4)µds =1 − 4/105ds =325dsP(C ≥ 8) =Z108fC(x)dxGiven the total life, the probability that you observe the light bulb when it is exactly4 days old is ds/C. So,P(C ≥ 8 | 4 < A ≤ 4 + dt) =Z108fC(x) dsdxx=Z108xf(x)µdsdxx=Z1081/105dsdx =125dsWhich givesP(C ≥ 8 | 4 < A ≤ 4 + dt) =ds/253ds/25=13Second problem: a) Calculate the probability distribution of Z = X + Y if X, Y areexponential variables with rate 2,3 resp.The density function of Z is the convolution fZ= fX∗ fY:fZ(z) =Zz0fX(x)fY(z − x)dx=Zz02e−2x3e−3(z−x)dx = 6Zz0e−3z+xdx = 6e−3z+x|z0= 6e−2z− 6e−3zb) Calculate the probability distribution of W = X − Y . [Hint: −Y is a random variablewith density function f(y) = 3e3yfor y < 0 and f(y) = 0 if y ≥ 0.Using the hint: fW= fX∗ f−Ywhich is given by:fW(w) =ZfX(x)f−Y(w − x)dxThe region of integration is given by x > 0 and y = w − x < 0 which means x > w. So,x > max(0, w). This makes the region of integration [w, ∞) if w ≥ 0 and [0, ∞) if w < 0.So, for w ≥ 0 we get:fW(w) =Z∞w2e−2x3e3(w−x)dx =6−5e3w−5x|∞w=65e−2wIf w < 0 thenfW(w) =Z∞02e−2x3e3(w−x)dx =6−5e3w−5x|0−∞=65e3wSo, the final answer is:fW(w) =(65e−2wif w ≥ 065e3wif w < 0Third problem: We have a water tower which gives water to a desert town. It has beenempty for a few day. :(Suppose that it showers from time to time. This event is a Poisson process with rateλ = 1/7 (once a week on average). When it showers it always dumps 5,000 gallons of waterinto the tank. The residents of this town use 1,000 gallons of water per day continuously (ata constant rate throughout the day when there is water).Convert this into an M/G/1 queueThe water in the tank is the queue. The residents together form the server which “serves”the inhabitants of the queue by drinking it. The water must be counted in units of 5,000gallons. Then each rainfall is one unit of water. The rainfalls are the people in line. It isgiven that these units of water enter the queue at the rate of λ = 1/7.and determine:(1) What is the meaning of Xn, Yn, Un, τ in this case?Xnis the number of times it rains during the time that the nth rainfall of wateris being used. This is the number of times it rains between 5n days and 5n + 5 daysafter the first rain.Ynis the number of units of water in the tower 5n days after the first rain.Unis 5 days for all n.τ is the number of the number of times it rains before the tower is empty. Anotherway to say it is that 5τ is the total length of time that there was water in the towerduring one period of time that there was water in the tower.(2) Compute µ.µ = E(Un) = 5 because it is always 5.(3) If there is a shower tomorrow, how long can the town expect to have water?This is the expected length of the first service periodE(S1) = E(τ)µ =µ1 − λµ=51 − 5/7=352= 1712The town can expect to have rain for 17 and a half
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