EE 302 Division 1. Homework 10 Solutions. A summary of Poisson and Bernoulli processes: 0 Y1 Y2 Y3 Y4 Y5 T1 T2 T3 T4 T5 Bernoulli (p) Poisson (¸) 8 µ ¶ ( < t ¡ 1 k t¡k ¸ktk¡1e¡¸t p (1 ¡ p) t ¸ k (k¡1)! t ¸ 0 time of the k-th arrival pY (t)= k ¡ 1 fYk (t)= k : 0 t<0 0 ½ t<k ½ (1 ¡ p)t¡1pt¸ 1 ¸e¡¸t t ¸ 0 inter-arrival time pTk (t)= fTk (t)= 8 µ ¶0 t<1 0 t<0 ½ < ¿ n ¿¡n (¸¿)n ¡¸¿ number of arrivals p (1 ¡ p) 0 · n · ¿ n! e 0 · n PS (n)= n PN¿ (n)= within ¿ units of time ¿ : 0 n<0 0 otherwise Problem 1. Fred is giving out samples of canned dog food. He makes calls door to door, but he leaves a sample (one can) only on those calls for which the door is answered and a dog is in res- idence. On any call the probability of the door being answered is 3/4, and the probability that any household has a dog is 2/3. Assume that the events \Door answered" and \A dog lives here" are independent and also that the outcomes of all calls are independent. (a) Determine the probability that Fred gives away his ¯rst sample on his third call. Solution. This is a Bernoulli process: each call is a Bernoulli trial, and the calls are independent. The probability of success p is: p = P(fthe door is answered g\fa dog is in residenceg) = P(fthe door is answeredg)P(fa dog is in residenceg) 3 2 1 = ¢ = : 4 3 2 Y1, the time of the ¯rst success is a geometric random variable with parameter p, therefore we have: 3¡1 P(Y1 =3)=pY1 (3)=(1¡ 1=2) (1=2) = 1=8: (b) Given that he has given away exactly four samples on his ¯rst eight calls, determine the conditional probability that Fred will give away his ¯fth sample on his eleventh call. Solution. Due to the fresh-start property of the Bernoulli process, what will happen from the ninth call and thereafter will also be a Bernoulli process. So the answer to this problem 1 is equivalent to the probability that he will give out the ¯rst can on his third trial in the new process, which is exactly the same as Part (a), i.e. 1/8. (c) Determine the probability that he gives away his second sample on his ¯fth call. Solution. Recall that Yk, the arrival time of the k-th success in a Bernoulli process is a random variable with the following PMF: 8 µ ¶ < ¡ t 1 k ¡ t¡k ¸ ¡ p (1 p) ; if t k; pYk (t)=: k 1 0ift<k: Here we have k =2,t = 5, and hence the answer is: µ ¶µ ¶2 µ ¶3 4 1 1 P(fY2 =5g)=pY (5) = 1 ¡ 2 1 2 2 µ ¶5 1 1 =4 = : 2 8 (d) Given that he did not give away his second sample on his second call, determine the conditional probability that he will give away his second sample on his ¯fth call. Solution. This question can be formulated to: P(fY2 =5g\fY2 > 2g) P(fY2 =5g) P(fY2 =5gjfY2 > 2g)= = : P(fY2 > 2g) P(fY2 > 2g) Using the normalization axiom and the fact that Y2 cannot be less than 2, we have: P(fY2 > 2g)=1¡ P(fY2 · 2g)=1¡ P(fY2 =2g) Apply the PMF of Y2 to get: 1=8 1 P(fY2 =5gjfY2 > 2g)= µ ¶µ ¶2 = 1 1 6 1 ¡ 1 2 (e) We shall say that Fred \needs a new supply" immediately after the call on which he gives away his last can. If he starts out with two cans, determine the probability that he completes at least ¯ve calls before he needs a new supply. Solution. This problem is equivalent to determine the probability that the second success arrives on or after the ¯fth trial. It can be formulated as: P(fY2 ¸ 5g)=1¡ P(fY2 =2g) ¡ P(fY2 =3g) ¡ P(fY2 =4g) =1¡ pY2 (2) ¡ pY2 (3) ¡ pY2 (4) µ ¶µ ¶2 µ ¶µ ¶3 µ ¶µ ¶4 1 1 2 1 3 1 =1¡ ¡ ¡ 1 2 1 2 1 2 1 2 3 5 =1¡ ¡ ¡ = : 4 8 16 16 2 Problem 2. The PDF of the duration of the (independent) interarrival times between successive cars on the Trans-Australian Highway is given by ½ 1 ¡ t e 12 ;t¸ 0; fT (t)= 12 0;t<0; where these durations are measured in seconds. (a) An old wombat (on his way to the wombat meeting of Homework 9) requires 12 seconds to cross the highway, and he starts out immediately after a car goes by. What is the probability that he will survive? Solution. Notice that the arrival of cars is a Poisson process with arrival rate ¸ =1=12 cars per second (since the interarrival times are i.i.d. exponential random variables with parameter 1/12). Therefore, the number N(t) of arrivals during a time interval of duration t, is a Poisson random variable with parameter t=12. Note that, due to the fresh-start property, no matter when the wombat starts crossing the highway, what will happen afterwards is still a Poisson process with the same rate. Therefore, this wombat will survive if and only if there is no cars during the 12 seconds it takes for him to cross the highway: With the given PDF, we have: 1 0 ¡12¢ 1 (12 ¢ 12 ) ¡1 P(wombat survives) = P(0 cars in 12 seconds) = e 12 = e : 0! Equivalently, he will survive if it takes more than 12 seconds for the ¯rst car to arrive: Z 1 ¡ 12 ¡1 P(wombat survives) = P(T>12) = fT (t)dt = e 12 = e : 12 (b) Another old wombat, slower but tougher, requires 24 seconds to cross the road, but it takes two cars to kill him. (A single car won't even slow him down.) If he starts out at an arbitrary time, determine the probability that he survives. Solution. This wombat will survive if and only if there are fewer than two cars during the 24 seconds it takes to cross. The probability of this is: 1 0 1 1 ¡24¢ 1 (24 ¢ 12 ) ¡24¢ 1 (24 ¢ 12 ) ¡2 P(N(24) · 1) = P(N(24) = 0) + P(N(24) = 1) = e 12 + e 12 =3e : 0! 1! Equivalently, the wombat will survive if and only if Y2, the time of the second arrival, is greater than 24 seconds: P(fwombat survivesg)=P(fY2 > 24g) Z 1 = fY2 (t)dt 24 Z 1 = ¸2te¡¸tdt 24 =3e¡2: 3 (c) If both these wombats start out at the same time, immediately after a car goes by, what is the probability that exactly one of them survives? (Hint. Consider random variables N1 = the number of cars in the ¯rst 12 seconds and N2 = the number of cars in the second 12 seconds.) Solution. Observe the following: fexactly one survivesg = fonly the 1st survivesg[fonly the second survivesg = fN1 = 0 and N2 > 1g[fN1 = 1 and N2 =0g Due to the memorylessness of Poisson process, N1 and N2 are independent, and they both Poisson random variables with parameter ¸ ¢ 12 = 1. Therefore we have: P(fexactly one survivesg)=P(fN1 = 0 and N2 > 1g)+P(fN1 = 1 and N2 =0g) = P(fN1 =0g)P(fN2 > 1g)+P(fN1 =1g)P(fN2 =0g) = pN1 (0)(1 ¡ pN2 (0) ¡ pN2 (1)) + pN1 (1)pN2 (0) = e¡1(1 ¡ e¡1 ¡ e¡1)+e¡1e¡1 = e¡1(1 ¡ e¡1): Problem 3. Eight light bulbs are turned on at t = 0. The lifetime of any particular bulb is independent of the lifetimes of all other bulbs and is described by the exponential PDF with parameter ¸. Let Y be the time until the third failure. (a) Find E[Y ]. Solution. The failure of a bulb can be thought of as the ¯rst arrival of a Poisson process with parameter ¸. Since the eight bulbs are independent, we have eight independent Poisson process running in parallel at the very beginning. The ¯rst burnout can be viewed as the ¯rst arrival of the merged process, which is also a Poisson process, with parameter 8¸.SoT1, the time of the ¯rst burnout is an exponential random variable with parameter 8¸. After that, due to the memorylessness of exponential distribution, the remaining lifetimes of the other 7 bulbs are still independent exponential random variables. Therefore we have 7 inde- pendent Poisson process running. The remaining time T2 of the next burnout is an exponential random variable with parameter 7¸. By the same argument, the time between the second and the third burnout is an exponential random variable with parameter 6¸. Now we can represent Y as: Y = T1 + T2 + T3: Because linearity of expectation, we have: 1 1 1 E[Y ]=E[T1]+E[T2]+E[T3]= + + : 8¸ 7¸ 6¸ (b) Find var(Y ). Solution. Due to the memorylessness of Poisson process, T1, T2 and T3 are independent. Therefore: var(Y )=var(T1)+var(T2)+var(T3) µ ¶2 µ ¶2 µ ¶2 1 1 1 = + + 8¸ 7¸ 6¸ 1 1 1 = + + : 64¸2 49¸2 36¸2 4 (c) Find the transform associated with Y . Solution. Because of independence, we have: MY (s)=MT1 (s)MT2 (s)MT3 (s) 8¸ 7¸ 6¸ = ¢ ¢ : 8¸ ¡ s 7¸ ¡ s 6¸ ¡ s Problem 4. Arrivals of certain events at points in time are known to constitute a Poisson process, but it is not known which of two possible values of ¸, the average arrival rate, describes the process.