Final answer:
To derive the limiting distribution πk for all k such that 0 ≤ k ≤ N, we can use the properties of the exponential and Poisson distributions. By considering a continuous-time Markov chain with exponential holding times between states and knowing that the limiting distribution is πi = 1/λii for all states i, we can determine the value of the limiting distribution πk for the given problem.
Step-by-step explanation:
The limiting distribution πk can be derived by taking the limit as t approaches infinity of the probability of X(t) being equal to k. We can use the properties of the exponential and Poisson distributions to compute this limiting distribution.
First, let's consider a continuous-time Markov chain with exponential holding times between states.
The rate at which a transition occurs from state i to state j is denoted by λij.
When λij is constant for all states i and j, we have a homogeneous continuous-time Markov chain.
In this case, if the process is irreducible and has a unique stationary distribution, the limiting distribution is given by πi = 1/λii for all states i.
For the Poisson process, we have λij = 0 for all i ≠ j and λii = λ for all i.
Therefore, the limiting distribution is πi = 1/λ for all states i.
Now, let's apply this to the given problem. We have N = {2, 3, 5} and P(N) = 1/.
Since the probabilities must sum to 1, we can determine the value of P(0) as P(2) + P(3) + P(5) + P(0) = 1.
Using the property of the Poisson distribution mentioned earlier, we can write P(k) = 1/λ for all states k.
Therefore, P(0) = 1/λ. Substituting this into the equation P(2) + P(3) + P(5) + P(0) = 1, we can solve for the value of λ and then calculate the limiting distribution for all states k.