Let's go through this problem step-by-step to find $p_{11}(t) = P(X(t) = 1 | X(0) = 1)$.

Problem Breakdown

1. Process Definition: We are given a continuous-time process $X(t)$ that takes on the values 1 or 2.

2. Transition Probabilities: For a small time interval $h$, the transition probabilities are given by:

   \begin{cases} h & \text{if } j \neq k, \\ 1 - h & \text{if } j = k, \end{cases}$$ where $$j, k \in \{1, 2\}$$.

3. Matrix Formulation for $P(t)$: The probability matrix $P(t)$ is given as: $P(t) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \sum_{k=1}^{\infty} \frac{t^k}{k!} \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}^k.$

4. Goal: We need to find $p_{11}(t) = P(X(t) = 1 | X(0) = 1)$, which is the probability that $X(t) = 1$ given that $X(0) = 1$.

Solution Steps

To find $p_{11}(t)$, we will evaluate the probability matrix $P(t)$ and specifically focus on the $(1,1)$ entry of this matrix, as this entry represents $p_{11}(t)$.

Step 1: Analyze the Matrix Exponential Form

The expression given for $P(t)$ resembles the matrix exponential for a continuous-time Markov process. Specifically, we can rewrite $P(t)$ as: $P(t) = e^{tQ},$ where $Q = \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}$ is the generator matrix of the process.

Step 2: Compute $e^{tQ}$

To calculate $e^{tQ}$, we use the series expansion of the matrix exponential: $e^{tQ} = \sum_{k=0}^{\infty} \frac{(tQ)^k}{k!}.$

Since $Q$ is a 2x2 matrix, we can compute powers of $Q$:

• $Q^1 = \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}$.
• $Q^2 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$, since squaring $Q$ results in the zero matrix.

Thus, all higher powers $Q^k$ for $k \geq 2$ will be zero. This means the series truncates at $k=1$, and we get: $e^{tQ} = I + tQ = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + t \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}.$

Step 3: Calculate $P(t)$

Now, we substitute to find $P(t)$: $P(t) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + t \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 1 - t & t \\ t & 1 - t \end{pmatrix}.$

Step 4: Extract $p_{11}(t)$

The entry $p_{11}(t)$ is the $(1,1)$ element of $P(t)$: $p_{11}(t) = 1 - t.$

Final Answer

Therefore, the probability $p_{11}(t) = P(X(t) = 1 | X(0) = 1)$ is: $p_{11}(t) = 1 - t.$

Would you like further clarification on any part of this process, or perhaps more examples?

Here are five related questions that could extend this problem:

1. How would $p_{22}(t) = P(X(t) = 2 | X(0) = 2)$ be calculated using the same approach?
2. What is the long-term behavior of $p_{11}(t)$ as $t \to \infty$?
3. How does the generator matrix $Q$ relate to the rate of transitions between states?
4. Can this method be generalized to more than two states?
5. What changes would occur in the solution if the transition rates were different?

Tip: For continuous-time Markov chains, the generator matrix $Q$ determines the rate of change between states, and the matrix exponential $e^{tQ}$ is a powerful tool to calculate transition probabilities over time.