 Let T0,m be the time at which the simple random walk reaches 0 for the
mth occasion. What is the generating function for T0,m.
Show that
P(Sn ≤ 0 for for all n ∈ {1, . . . , k} | T0,m = k) = 1
2
m.
1


### Problem 1: Time at Which Random Walk Reaches 0 for the mth Occasion

Let \( T_{0,m} \) be the time at which a simple symmetric random walk reaches 0 for the \(m\)-th time.

1. A **simple symmetric random walk** (SSRW) takes steps either \( +1 \) or \( -1 \) with equal probability (i.e., \( p = 1/2 \)).
2. The generating function for the time \( T_{0,m} \) is sought.

#### Generating Function for \( T_{0,m} \)

The generating function for \( T_{0,m} \) can be found by considering that the SSRW behaves like a recurrent process (i.e., it returns to the origin an infinite number of times). Let \( G(s) \) be the generating function for the return time to 0. 

For the 1st return to 0 (i.e., \( T_{0,1} \)), the generating function can be written as:

\[
G(s) = \sum_{n=1}^{\infty} P(T_0 = n) s^n
\]

Here, \( P(T_0 = n) \) is the probability that the random walk returns to 0 for the first time at time \( n \). This is a well-known result in random walk theory, and we can generalize this to the \( m \)-th return to the origin, \( T_{0,m} \).

For the \( m \)-th return, the generating function is the convolution of the generating function for the 1st return:

\[
G_m(s) = (G(s))^m
\]

Thus, the generating function for \( T_{0,m} \) is the \(m\)-fold convolution of \( G(s) \), which leads to:

\[
G_m(s) = \left( \sum_{n=1}^{\infty} P(T_0 = n) s^n \right)^m
\]

This gives the generating function for the \( m \)-th return time to 0.

---

### Problem 2: Probability \( P(S_n \leq 0 \ \forall n \in \{1, \ldots, k\} \ | \ T_{0,m} = k) \)

We want to show that:

\[
P(S_n \leq 0 \ \forall n \in \{1, \ldots, k\} \ | \ T_{0,m} = k) = \frac{1}{2^m}
\]

This is a **ballot theorem** type result. The idea is that after returning to 0 for the \(m\)-th time at time \(k\), the random walk must stay non-positive until time \(k\). 

This situation corresponds to counting paths where the random walk hits 0 exactly \(m\) times and stays non-positive throughout. The number of such paths corresponds to the number of **Dyck paths** (paths that never go above 0), and the probability of such paths is given by:

\[
P(S_n \leq 0 \ \forall n \leq k | T_{0,m} = k) = \frac{1}{2^m}
\]

This can be understood by the fact that after each return to 0, the walk has a 50% chance of staying below or equal to 0, so for \( m \) returns, the probability is \( \frac{1}{2^m} \).

---

### Summary

1. The generating function for the time at which a random walk reaches 0 for the \(m\)-th time, \( T_{0,m} \), is \( G_m(s) = \left( G(s) \right)^m \).
2. The probability that the walk stays non-positive from \( n = 1 \) to \( n = k \) given that it reaches 0 for the \(m\)-th time at time \( k \) is \( \frac{1}{2^m} \).

---

Do you want more details on the generating function derivation or further explanation of the probability result?

### Related Questions

1. What is the generating function for the first return to 0 in a simple random walk?
2. How is the convolution operation used in generating functions?
3. What is the significance of the ballot theorem in probability theory?
4. How does the number of returns to 0 affect the random walk’s overall behavior?
5. Can this approach be generalized to random walks on graphs?

### Tip:
Generating functions are powerful tools in probability theory as they encapsulate the entire distribution of a random variable and allow for solving complex recurrence relations.

Let T0,m be the time at which the simple random walk reaches 0 for the mth occasion. What is the generating function for T0,m? Show that P(Sn ≤ 0 for all n ∈ {1, . . . , k} | T0,m = k) = 1/2^m.

Explore the generating function for the time at which a simple random walk reaches 0 for the m-th occasion. Learn about the ballot theorem and how to prove that P(Sn ≤ 0 for all n ∈ {1, ..., k} | T0,m = k) = 1/2^m, including key formulas and probability theory concepts.

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