suppose that customers arrive at facility according to poisson process having rate a 3 let nt bc

For the first question, we need to use the properties of Poisson process. The probability of hav

Suppose that customers arrive at facility according to...

Suppose that customers arrive at a facility according to a Poisson process with a rate λ = 3. Let Nt be the number of customers that have arrived up to time t and let Tn be the waiting time for the nth customer, n = 1, 2. Determine the following (conditional) probabilities and conditional expectations, where 0 < t1 < t2 < t3 < t4: (a) P(Nt = 5 | Nt > 1) (b) E[Nt | Nt > Nt-1] (c) E[Nt | T2 > 1] Consider a two-state Markov chain (Xn) with n > 0 on {0, 1} with transition matrix P: P = [a 1-a] [1-b b] Lemma: Let (Nt) be a Poisson process with parameter λ > 0, and let Xt = ENt; t ∈ R+. In other words, (Xt) is a two-state birth and death process. (a) Compute the mean return time E[T0 | X0 = 0] of Xt to state 0. (b) Compute the mean return time E[T1 | X0 = 1] of Xt to state 0. (c) Show that Xt is a two-state birth and death process and determine its generator matrix in terms of λ and α.

Transcript

00:01 Hello student, in the given question in part a we have to determine p of n of t which is equals to 5 given n of t is greater than 1.

00:10 So, we have to calculate the formula for conditional probability.

00:14 So, we have formula for conditional probability is probability of a given b which is equals to probability of a intersection b upon probability of b.

00:25 So, given that n t follows n of t follows poisson process poisson process with rate lambda is equals to 3 the number of customer that arrive up to date t.

00:43 So, n t follows the poisson distribution therefore n t follows the poisson distribution with parameter lambda of t.

00:59 So, using the information we can calculate the conditional probability as probability of n of t is which is equals to 5 given n of t is greater than 1 which is equals to probability of n of t equals to 5 intersection n of t is greater than 1 given upon probability of n of t is greater than 1.

01:20 Now, to find the probability of n of t which is equals to 5 intersection n of t is greater than 1 we use the fact that event n of t which is equals to 5 and n of t which is greater than 1 are mutually exclusive are mutually exclusive.

01:42 Therefore, the intersection of this event is 0.

01:45 So, probability of n of t is equals to 5 intersection n of t is greater than 1 which is equals to 0.

01:53 Now, to find to find probability of n of t is greater than 1.

02:01 So, we can use complementary probability here we can use complementary probability here as probability of n of t is greater than 1 which is equals to 1 minus probability of n of t is less than equals to 1 since n of t follows poisson distribution.

02:22 So, we can calculate the n of probability of n of t is less than equals to 1 which is equals to probability of n of t which is equals to 0 plus probability of n of t which is equals to 1.

02:36 So, which is equals to which is equals to e to the power minus lambda t multiply by 1 plus lambda of t.

02:49 Now, we can substitute this value into the formula for conditional probability as probability of n of t which is equals to 5 given n of t is greater than 1 which is equals to 0 upon 1 minus e to the power minus lambda into t multiply by 1 plus lambda of t...

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