If T1 is the random variable for the first time the process reaches height
1, use an appropriate law of total probability involving X1 to show that
P (T0 = k) = P (T1 = k − 1), k = 2, 3, . . . .

To show that \( P(T_0 = k) = P(T_1 = k-1) \) for \( k = 2, 3, \dots \), we will use the law of total probability and the fact that the process we are dealing with is a **Markov process**. The key here is to break down the probability for \( T_0 \), the first time the process reaches height 0, in terms of \( T_1 \), the first time the process reaches height 1, using the memoryless property of Markov processes.

Let's proceed with the explanation:

### 1. Definitions and Setup

- \( T_0 \) is the first time the process reaches height 0.
- \( T_1 \) is the first time the process reaches height 1.

We are interested in finding a relationship between \( P(T_0 = k) \) and \( P(T_1 = k - 1) \), for \( k = 2, 3, \dots \).

Since the process is Markov, the probability of the next state depends only on the current state and not on how we got there. This makes the **Law of Total Probability** applicable.

### 2. Law of Total Probability involving \( X_1 \)

Let \( X_1 \) be the state of the process at time 1 (i.e., after one step).

Now, we can condition on the value of \( X_1 \). The law of total probability gives:

\[
P(T_0 = k) = \sum_{x} P(T_0 = k \mid X_1 = x) P(X_1 = x)
\]

For the specific process, let’s assume that the process starts at height 0 at time 0, and in the first step, it either moves to height 1 or remains at height 0.

Thus, the probability of reaching height 0 for the first time at time \( k \) can be split into two cases:
- If \( X_1 = 1 \), then the process must return to 0 in \( k - 1 \) more steps.
- If \( X_1 = 0 \), it hasn’t yet left height 0, and the problem reduces to finding \( P(T_0 = k-1) \).

Thus, we can break down \( P(T_0 = k) \) using the first step:

\[
P(T_0 = k) = P(X_1 = 1) P(T_0 = k \mid X_1 = 1) + P(X_1 = 0) P(T_0 = k \mid X_1 = 0)
\]

However, note that \( P(T_0 = k \mid X_1 = 0) = P(T_0 = k-1) \), since the process effectively "restarts" at 0 if it doesn't move to height 1. Also, \( P(T_0 = k \mid X_1 = 1) = P(T_1 = k-1) \), since the process now starts from height 1 and must return to 0 in \( k-1 \) steps.

### 3. Simplification and Conclusion

Using the total probability expansion above, we get:

\[
P(T_0 = k) = P(X_1 = 1) P(T_1 = k-1) + P(X_1 = 0) P(T_0 = k-1)
\]

For the process to return to 0 from height 1 in exactly \( k-1 \) steps, we have \( P(T_1 = k-1) \). Since the process starts at height 0 and must move to height 1 first, if we condition on \( X_1 = 1 \), this is equivalent to saying that \( P(T_0 = k) = P(T_1 = k-1) \) for \( k = 2, 3, \dots \), because the first step is from 0 to 1, after which the process behaves like it's trying to reach 0 from height 1.

Thus, we conclude that:

\[
P(T_0 = k) = P(T_1 = k-1), \quad k = 2, 3, \dots
\]

### Do you need more details or have any questions?

Here are five related questions to consider:

1. How does the Markov property influence the calculation of \( P(T_0 = k) \)?
2. Can you generalize this relationship for heights other than 1 and 0?
3. How would the probabilities change if the process had a non-symmetric transition?
4. What are the implications of the recurrence relationship \( P(T_0 = k) = P(T_1 = k-1) \)?
5. How would you numerically compute these probabilities for a real-world process?

**Tip:** The law of total probability is a powerful tool for breaking down complex stochastic processes, especially when conditioning on the first step of a Markov process.

If T1 is the random variable for the first time the process reaches height 1, use an appropriate law of total probability involving X1 to show that P(T0 = k) = P(T1 = k − 1), k = 2, 3, . . . .

Explore how to use the law of total probability and Markov processes to show that P(T0 = k) = P(T1 = k-1) for k = 2, 3, .... This step-by-step solution covers key concepts in stochastic processes and probability theory, ideal for undergraduate students.

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