Consider a jump process with state space . For , let be the transition probability matrix of the jump process at time . Prove that .

Stochastic Processes Solution – Problem 36: Prove that

Question

Consider a jump process with state space $\{1, \ldots, N\}$ . For $t \geq 0$ , let $\mathbf{P}(t)$ be the transition probability matrix of the jump process at time $t$ . Prove that $\det(\mathbf{P}(t)) > 0$ .

Step-by-step solution

Step 1. The transition probability matrix $P(t)$ of a continuous-time Markov chain satisfies: 1. $P(0) = I$ (identity matrix). 2. $P(t)$ is stochastic (row sums equal 1). 3. Chapman-Kolmogorov equation: $P(t+s) = P(t)P(s)$ .

Step 2. If the jump process has a conservative $Q$ -matrix (generator) satisfying $q_{ij} \ge 0$ ( $i \neq j$ ) and $q_{ii} = -\sum_{j \neq i} q_{ij}$ , then $P(t) = e^{tQ}$ for $t \ge 0$ .

Step 3. Since $P(t) = e^{tQ}$ , if $\lambda_1, \dots, \lambda_N$ are the eigenvalues of $Q$ , then the eigenvalues of $P(t)$ are $e^{t\lambda_1}, \dots, e^{t\lambda_N}$ . Each $e^{t\lambda_k} > 0$ since the exponential function is always positive. Thus $\det P(t) = e^{t\lambda_1} \cdots e^{t\lambda_N} = e^{t(\lambda_1 + \dots + \lambda_N)} > 0.$

Step 4. Even if $Q$ is not diagonalizable, $P(t) = e^{tQ}$ is always nonsingular because $\det e^{tQ} = e^{t \cdot \mathrm{tr}(Q)}.$ Since $\mathrm{tr}(Q) = \sum_{i=1}^N q_{ii}$ is a finite real number, $\det P(t) = e^{t \cdot \mathrm{tr}(Q)} > 0$ .

Step 5. By the matrix exponential determinant formula $\det e^A = e^{\mathrm{tr}(A)}$ , taking $A = tQ$ : $\det P(t) = e^{t \cdot \mathrm{tr}(Q)} > 0$ for all $t > 0$ .

Final answer

QED.

Marking scheme

The following is the grading rubric.

1. Checkpoints (max 7 pts total)

Score exactly one chain | take the maximum subtotal among chains; do not add points across chains.

Chain A: Matrix Exponential and Spectral Mapping (Algebraic Path - Recommended)

Establish the generator relation [3 pts] [additive]:
State that the transition matrix can be expressed as $P(t) = e^{tQ}$ (where $Q$ is the generator/ $Q$ -matrix);
Or write the Kolmogorov forward/backward equation $P'(t) = QP(t)$ or $P'(t) = P(t)Q$ , implying the exponential solution.
Use the determinant-trace/eigenvalue relation [3 pts] [additive]:
Cite the identity $\det(e^A) = e^{\mathrm{tr}(A)}$ (exponential form of Jacobi's formula);
Or use the spectral mapping theorem: if $\lambda_i$ are eigenvalues of $Q$ , then $e^{t\lambda_i}$ are eigenvalues of $P(t)$ , giving $\det P(t) = \prod e^{t\lambda_i} = e^{t\sum \lambda_i}$ .
*(Note: If the student assumes $Q$ is diagonalizable, no deduction as long as the conclusion is correct.)*
Positivity conclusion [1 pt] [additive]:
State that the exponential of a real number is always positive ( $e^{t \cdot \mathrm{tr}(Q)} > 0$ ).

Chain B: Differential Equation and Liouville's Formula (Analytic Path)

Establish the differential equation [3 pts] [additive]:
Write the ODE: $\frac{d}{dt} P(t) = QP(t)$ or $\frac{d}{dt} P(t) = P(t)Q$ .
Solve the determinant evolution [3 pts] [additive]:
Apply Liouville's formula: $\frac{d}{dt}(\det P(t)) = \mathrm{tr}(Q) \cdot \det P(t)$ ;
Or directly write the solution $\det P(t) = C e^{t \cdot \mathrm{tr}(Q)}$ .
Use the initial condition to conclude [1 pt] [additive]:
Use $P(0) = I \implies \det P(0) = 1$ to determine $C=1$ , and note the exponential is always positive.

Total (max 7)

2. Zero-credit items

Only restating problem conditions: Listing " $P(t)$ is a transition matrix" or "row sums are 1" without substantive derivation.
Irrelevant probabilistic properties: Only stating $p_{ij}(t) \ge 0$ or $\sum_j p_{ij}(t) = 1$ and claiming this implies positive determinant (logical error: nonneg matrices can have negative determinants).
Only discussing discrete time: Only discussing $P^n$ without addressing the continuous-time generator $Q$ .
False analogy: Claiming that since $P(0)=I$ and $P(t)$ is continuous, $\det P(t)$ is always positive (circular reasoning without proving $\det P(t) \neq 0$ ).

3. Deductions

Logical gap or ambiguity (max -1): Not stating $\det P(t)$ equals the product of eigenvalues, or not defining $Q$ .
Incorrect conclusion (cap at 3/7): Getting a wrong determinant expression but with correct earlier steps.
Conceptual confusion (flat -2): Confusing trace and determinant (e.g., writing $\det(e^A) = e^{\det(A)}$ ).

Stochastic Processes – Problem 36: Prove that