Probability Theory Solution – Problem 73: Find the distribution of the total time spent at the root vertex

Question

On the 3-regular tree, consider the jump process starting from the root, where the transition rate along every edge is 1. Find the distribution of the total time spent at the root vertex.

Step-by-step solution

In a 3-regular tree, the root vertex has 3 adjacent child vertices, and every non-root vertex has 1 parent vertex and 2 child vertices. When the jump process is at the root vertex, there are 3 outgoing edges, each with transition rate 1, so the total transition rate at the root is 3. When the jump process is at a non-root vertex, there are also 3 outgoing edges (1 to the parent and 2 to children), each with transition rate 1, so the total transition rate is also 3. Let $T$ denote the total time the process spends at the root vertex. We analyze the composition of $T$ . Each time the process arrives at the root, it stays for a random duration before jumping to some child vertex. Let $\xi_k$ denote the holding time during the $k$ -th visit to the root ( $k \geq 1$ ). Then $T = \xi_1 + \xi_2 + \cdots$ For a continuous-time Markov chain, the holding time at any state follows an exponential distribution whose rate equals the total transition rate at that state. Since the total transition rate at the root is 3, each $\xi_k$ follows an exponential distribution with parameter 3, i.e., $\xi_k \sim \text{Exp}(3)$ , and all $\xi_k$ are mutually independent. Let $N$ denote the total number of times the process visits the root. We need to determine the distribution of $N$ . Define $p$ as the probability that the process, after leaving the root, eventually returns to the root. Since the 3-regular tree is infinite, the jump process is transient, so $p < 1$ . The distribution of $N$ is geometric: $P(N = k) = (1 - p) p^{k - 1}$ ( $k = 1, 2, \dots$ ), meaning the first failure to return to the root occurs after the $k$ -th departure, with the previous $k - 1$ departures all resulting in successful returns. $T$ is a sum of a random number of i.i.d. exponential random variables, which is a special case of a compound Poisson distribution. Compute the Laplace transform $L(s) = E[e^{-s T}]$ of $T$ using the law of total expectation: $L(s) = \sum_{k=1}^\infty P(N = k) E[e^{-s (\xi_1 + \cdots + \xi_k)}]$ Since the $\xi_k$ are i.i.d., $E[e^{-s (\xi_1 + \cdots + \xi_k)}] = [E[e^{-s \xi_1}]]^k$ . The Laplace transform of the exponential distribution is $E[e^{-s \xi_1}] = \frac{3}{3 + s}$ . Substituting: $L(s) = \sum_{k=1}^\infty (1 - p) p^{k - 1} \left( \frac{3}{3 + s} \right)^k = \frac{1 - p}{p} \sum_{k=1}^\infty \left( \frac{3 p}{3 + s} \right)^k$ Using the geometric series formula ( $|x| < 1$ implies $\sum_{k=1}^\infty x^k = \frac{x}{1 - x}$ ), we simplify: $L(s) = \frac{1 - p}{p} \cdot \frac{\frac{3 p}{3 + s}}{1 - \frac{3 p}{3 + s}} = \frac{3 (1 - p)}{3 + s - 3 p}$ Computation of $p$ : Starting from the root, the process first jumps to any one of the first-level child vertices. Let $q$ denote the probability of returning to the root from a first-level child vertex. By symmetry, $q$ is the same for all 3 first-level children, so $p = q$ . From a first-level child vertex, there are 3 possible transitions: 1 edge back to the root (rate 1) and 2 edges to second-level child vertices (rate 1 each). Let $q$ denote the probability of returning to the root from a second-level child vertex. Then: $q = \frac{1}{3} \cdot 1 + \frac{2}{3} \cdot q$ Solving gives $q = \frac{1}{2}$ , i.e., $p = \frac{1}{2}$ . Substituting into the Laplace transform: $L(s) = \frac{3 (1 - \frac{1}{2})}{3 + s - 3 \cdot \frac{1}{2}} = \frac{\frac{3}{2}}{\frac{3}{2} + s} = \frac{3}{3 + 2 s}$ This Laplace transform corresponds to an exponential distribution with parameter $\frac{3}{2}$ .

Final answer

The total time spent at the root vertex follows an exponential distribution with parameter $\frac{3}{2}$ , i.e., ${T \sim \text{Exp}\left( \frac{3}{2} \right)}$ .

Marking scheme

The following is a marking scheme based on the official solution.

1. Checkpoints (max 7 pts)

Shared Prerequisites [max 4 pts]

*Regardless of the subsequent method used, the following steps are counted at most once.*

Holding rate at the root (1 pt)
Identify that the total transition rate at the root is 3, or equivalently that the single-visit holding time satisfies $\xi \sim \text{Exp}(3)$ .
*If not explicitly stated but the value $\frac{3}{3+s}$ or the mean $\frac{1}{3}$ is correctly used in subsequent calculations, this point may be awarded retroactively.*
Return probability equation (2 pts)
Establish a valid equation for the return probability $p$ (or $q$ ).
[additive] Correctly analyze that from a child vertex there is exactly one edge back to the root and two edges going deeper, earning 1 pt.
[additive] Write down the explicit equation (e.g., $p = \frac{1}{3} + \frac{2}{3}p^2$ or the form in the official solution), earning 1 pt.
Return probability value (1 pt)
Correctly solve to obtain $p = \frac{1}{2}$ .
*Note: If the student sets up an equation yielding $p=1$ (mistakenly concluding recurrence) and obtains $p=1$ , no credit is awarded for this checkpoint.*

Choose one of the following two chains for grading (take the higher score; do not mix chains)

Chain A: Transform Method / Random Sum Distribution [max 3 pts]

Distribution of the total number of visits (1 pt)
State that the total number of visits $N$ follows a geometric distribution with success/stopping probability $1-p$ .
*Acceptable forms include $P(N=k) = (1-p)p^{k-1}$ or an equivalent verbal description.*
Compound distribution computation (1 pt)
Set up the Laplace transform $L(s) = \sum P(N=k) [L_\xi(s)]^k$ and carry out the computation;
or cite the theorem that a geometric sum of i.i.d. exponential random variables is again exponential, and correctly state the parameter relationship.
Final result (1 pt)
Simplify to obtain $L(s) = \frac{3}{3+2s}$ or state that $T \sim \text{Exp}(1.5)$ .
*Writing only the distribution parameter $\lambda=1.5$ without the distribution name is also acceptable.*

Chain B: Mean Method + Distribution Argument [max 3 pts]

Exponential distribution argument (1 pt)
Must explicitly state or argue that a geometric sum of exponential random variables is again exponentially distributed (or use the memoryless property to justify this).
*Computing only the mean without explaining why the distribution is exponential earns no credit for this checkpoint, and the final result checkpoint cannot be awarded either.*
Total mean computation (1 pt)
Use $E[T] = E[N] \cdot E[\xi]$ or the recursive formula $E[T] = \frac{1}{3} + p E[T]$ to compute $E[T] = \frac{2}{3}$ .
Final result (1 pt)
From the mean, write the correct distribution $T \sim \text{Exp}(1.5)$ (i.e., parameter $\lambda = 3/2$ ).

Total verification: Shared prerequisites (4) + Chain A/B (3) = 7 pts

2. Zero-credit items

Merely copying the problem statement: Only listing the definition of a 3-regular tree or the given conditions without any probabilistic modeling.
Incorrect distributional assumption: Assuming without derivation that the total time follows a Poisson, normal, or deterministic distribution.
Recurrence error: Claiming $p=1$ (certain return), leading to infinite total time, without recognizing that this is a transient process.

3. Deductions

*Apply the most severe applicable deduction; the minimum score is 0.*

Arithmetic error: [-1]
When the approach is correct (e.g., the equation is set up properly) but an algebraic error leads to an incorrect value of $p$ or an incorrect integral.
Logic gap: [Cap at 5/7]
Using Chain B (mean method) and correctly computing the mean $2/3$ , but completely failing to mention why the resulting distribution is exponential.
Conceptual confusion: [-1]
Confusing the number of visits $N$ with the total time $T$ , e.g., directly reporting the expectation of $N$ as the final answer without multiplying by the single-visit holding time.
Serious error in the return probability equation: [Cap at 4/7]
Failing to establish an equation for $p$ , directly guessing the value of $p$ incorrectly, causing all subsequent work to be wrong.

Probability Theory – Problem 73: Find the distribution of the total time spent at the root vertex