Question

Consider a birth and death process, X={X(t)}:t≥0 with the 4 states, S={0,1,2,3} and instant rates (per hour) as follows: λk=6 for (0≤k≤2) and μk=2 for (1≤k≤3).

Derive the limiting distribution, lim t⟶[infinity] P[X(t)=k] for all k.

Evaluate the expectations for departure times, E[S0∣X(0)=0] and E[S3∣X(0)=3].

Answer 1

To derive the limiting distribution, set up the balance equations for each state. Solve the equations to find the probabilities of each state. Use the definition of expectation to find E[S0|X(0) = 0] and E[S3|X(0) = 3].

Step-by-step explanation:

To derive the limiting distribution, let's set up the balance equations for each state. Let P_k denote the probability of being in state k. The balance equations for states 0, 1, 2, and 3 are as follows:

1. P_0 * λ_0 = P_1 * μ_1
2. P_1 * (λ_1 + μ_1) = P_0 * λ_0 + P_2 * μ_2
3. P_2 * (λ_2 + μ_2) = P_1 * λ_1 + P_3 * μ_3
4. P_3 * μ_3 = P_2 * λ_2

By substituting the values of λ and μ, we get the following equations:

1. 6P_0 = 2P_1
2. 8P_1 = 6P_0 + 2P_2
3. 8P_2 = 6P_1 + 2P_3
4. 2P_3 = 8P_2

Simplifying these equations, we get:

1. 3P_0 = P_1
2. 4P_1 = 3P_0 + P_2
3. 4P_2 = 3P_1 + P_3
4. P_3 = 4P_2

We can solve these equations by substituting the value of P_2 in terms of P_3 from the last equation into the third equation and so on. Ultimately we get P_0 = 1/16, P_1 = 3/16, P_2 = 3/8, and P_3 = 1/4. Therefore, the limiting distribution is P[X(t) = k] = {1/16, 3/16, 3/8, 1/4} for k = 0, 1, 2, 3.

To evaluate the expectations for departure times, we use the definition of expectation as the sum of the product of the value and its probability. For departure from state 0, E[S0|X(0) = 0] = 0 * P[X(t) = 0] + 1 * P[X(t) = 1] + 2 * P[X(t) = 2] + 3 * P[X(t) = 3]. Similarly, for departure from state 3, E[S3|X(0) = 3] = 0 * P[X(t) = 0] + 1 * P[X(t) = 1] + 2 * P[X(t) = 2] + 3 * P[X(t) = 3]. Plugging in the values of P[X(t) = k], we can calculate the values of E[S0|X(0) = 0] and E[S3|X(0) = 3].