Assume that N(t) is a Poisson process with rate λ=2024. ComputeP(N(s) =1000,N(t) =2023),for any s≤t
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Assume that N(t) is a Poisson process with rate λ=2024. ComputeP(N(s) =1000,N(t) =2023),for any s≤t
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- Suppose Xn is the sample mean of a random sample of size n from a distribution that has a Gamma(2,1/2) (where a = 2, B = }) pdf f(x) = 4xe-2a, 0 < x < ∞, zero elsewhere. Use the Central Limit Theorem to deduce that the random variables Vn(Xn – 1) converges in distribution to N(0, 1/2).3. Jobs arrive in a computer queue in the manner of a Poisson process with intensity λ. The central processor handles them one by one in the order of their arrival, and each has an exponentially distributed runtime with parameter , the runtimes of different jobs being independent of each other and of the arrival process. Let X (t) be the number of jobs in the system (either running or waiting) at time t, where X (0) = 0. Explain why X is a Markov chain, and write down its generator. Show that a stationary distribution exists if and only if λ < μ, and find it in this case.Let X denote the length (in seconds) of the next smile of a ran- domly selected 8-week old baby. Suppose that X is uniformly distributed on the interval [0, 23]. (a) What is the probability that the next smile of a randomly selected 8-week old baby is between 15 and 25 seconds in length? (b) What is the moment generating function M(t) of X? No work is required for this part. 2.
- x(t) == ()-1 h(t)=pt .u(t) y(t) Z(t) is a sequence of Poisson impulses. X is the expected number of impulses per Find a unit time E[Ÿ(t)] b) Ryy (T).. c) Syy (w)According to the U.S. Customs and Border Protection Agency, the average airport wait time (time from arrival at the airport until the completion of security screening) at Chicago's O'Hare Inter- national airport is 32 minutes for a passenger arriving during the hours 4-5 PM. Assume the wait time is exponentially distributed so that p(t) = ke-kt for 0≤t<∞, where k is the reciprocal of the average wait time. Assume t is measured in minutes. 1. Sketch a labeled plot of the relevant pdf on the interval [0, 90] minutes (while remembering that it is defined for all t≥ 0.) 2. Set up integrals (you can write p(t) for the integrand) for the following probabilities: (a) Waiting longer than 60 minutes. (b) Waiting shorter than 30 minutes. (c) Waiting between 20 and 40 minutes. 3. Evaluate the probability that the waiting time is longer than 60 minutes (part 2(a) above) and convert to a percentage with two decimal places.The inhabitants of a city develop skin cancer at an approximate rate A. For those people who have developed skin cancer, some proportion p E (0, 1) will die from the disease. Assume a simple model such that n, the number of people who develop skin cancer, is distributed Poisson(A). Let X denote the people who die from skin cancer in this city. Then, assuming that every cancer patient is independent of the others, and that the proportion p is constant: n ~ Poisson(X) X\n - Binomial(n, p) Question 5: What is the distribution of n X? n – X|X ~ Binomial X(1-p) 1- p We do not have enough information to answer this question. n|X Poisson(Xp) On - X|X ~ Poisson(A(1 – p) 1 for n > x n|X ~ fn|x (n|x) = n! 1- Di-o i! 0.w.
- A9. If the total benefit function is B = 192M – 4M^2, the total damage function is D = A2, ⍺ = 0.5 and there is no discounting where: M is the level of emissions. A is the stock of the pollutant. ⍺ is the linear decay rate of A a) What are the steady-state levels of M and A? b) Assuming now that we have some discounting and r = 0.2, where r is the social consumption discount rate. What are the new steady state levels of M and A?Let i, denote the effective annual return achieved on an equity fund achieved between time (t-1) and time t. Annual log-returns on the fund, denoted by In(1+i), are assumed to form a series of independent and identically distributed Normal random variables with parameters u = 6% and σ = 14%. An investor has a liability of £10,000 payable at time 15. Calculate the amount of money that should be invested now so that the probability that the investor will be unable to meet the liability as it falls due is only 5%.Consider an infinite-server queueing system to which customers arrive according to a PP() (A > 0). Each customer's service time is exponentially distributed with rate u (u > 0). Since there are infinitely many servers, all customers begin their service immediately upon arrival (i.e., there is no queueing delay). Let X(t) be the number of customers in the system at time t. (a) Derive the limiting distribution (p) of {X(t) : t > 0}.
- The call centre decides to update its model to allow for the fact that calls arrive less regularly when the phonelines have only just opened. For the first hour, 0 ≤ t ≤ 60, the arrival of calls is now modelled as a time inhomogeneous Poisson process with rate function t for 0 t≤ 30 x(t) = 2 for 30The inhabitants of a city develop skin cancer at an approximate rate X. For those people who have developed skin cancer, some proportion p E (0, 1) will die from the disease. Assume a simple model such that n, the number of people who develop skin cancer, is distributed Poisson(A). Let X denote the people who die from skin cancer in this city. Then, assuming that every cancer patient is independent of the others, and that the proportion p is constant: ~ U Poisson(A) X\n - Binomial(n, p) Question 4: What is the marginal distribution of X? Hint: two primary ways to do this: 1. you can summate out n from the joint distribution directly (more straightforward, but tricky algebra) o Pay attention to the lower limit of n o Remember that i=0 2. use MGFS + iterated expectation: E[etx] = En|Exn[etx n|| and then recognize the distribution corresponding to the MGF (need to understand MGFS and iterated expectation). o Obtain the inner expectation by using the Binomial theorem. O X Binoтial(n, Ap)…Customers arrive at a facility according to a Poisson process N(t) of rate λ = 4.5 customers/hour. Each customer is admitted to the facility with probability p = 0.8. All customers, who are not admitted, leave and do not come back. Let X(t) be the number of customers admitted from time 0 to time t. Compute the following: (a) The mean value E[X(6.5)] = 23.4 (b) The probability P(X(6.5) = 23) = = (c) The conditional probability P(X(6.5) = 23 | N(6.5) = 30) = %. %. Note: For parts (b) and (c), enter your answers as percentages but do not type the percent sign.SEE MORE QUESTIONS