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.