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Equal mean trajectories do not determine a master-equation state

Example

On states ${0, 1, 2}$ , compare $Q_{A} = 0$ with

$Q_{B} = (\begin{array}{ccc} 0 & 1 & 0 \\ 0 & - 2 & 0 \\ 0 & 1 & 0 \end{array}),$

both with initial vector $e_{1} = (0, 1, 0)^{T}$ . For $Q_{A}$ the law stays $e_{1}$ . For $Q_{B}$ it is

$p_{B} (t) = {(\frac{1 - e^{- 2 t}}{2}, e^{- 2 t}, \frac{1 - e^{- 2 t}}{2})}^{T} .$

Both mean trajectories equal $1$ , so both obey the same deterministic mean equation $m^{'} (t) = 0$ , $m (0) = 1$ . For every $t > 0$ the probability vectors differ. Thus an equation for the mean alone does not solve or determine the full master-equation state.

Facts & Assumptions

Given: The finite state set ${0, 1, 2}$ , initial vector $e_{1}$ , and generators $Q_{A}, Q_{B}$ displayed in the Example.

[F1]

Finite column generator: a finite column generator has nonnegative off-diagonal entries and every source column sums to zero.

[F2]

Finite master equation and uniformization formula: the full probability vector is the unique solution of $p^{'} = Q p$ from its initial vector.

Verification

technique · solve both finite master equations and compare their first moments

1.1givenF1F2algebra

$Q_{A} = 0$ is a finite column generator, and the constant vector $p_{A} (t) = e_{1}$ solves $p_{A}^{'} = Q_{A} p_{A}$ with the stated initial value. By [F2] it is the unique law, and its mean is $0 \cdot 0 + 1 \cdot 1 + 2 \cdot 0 = 1$ .

1.2givenF1F2algebra

The off-diagonal entries of $Q_{B}$ are nonnegative and its columns sum to zero, so [F1] applies. Writing $p_{B} = (p_{0}, p_{1}, p_{2})^{T}$ , its master equation is $p_{0}^{'} = p_{1}$ , $p_{1}^{'} = - 2 p_{1}$ , $p_{2}^{'} = p_{1}$ with initial values $(0, 1, 0)$ . Solving gives $p_{1} = e^{- 2 t}$ and $p_{0} = p_{2} = (1 - e^{- 2 t}) / 2$ ; direct differentiation verifies these formulas and their initial data, and [F2] makes this the unique law.

2.1step 1.1step 1.2algebra∎

The mean under $p_{B} (t)$ is $0 \cdot p_{0} + 1 \cdot p_{1} + 2 \cdot p_{2} = e^{- 2 t} + 1 - e^{- 2 t} = 1$ , the same as the mean under $p_{A} (t)$ , while $p_{B}$ has positive mass at $0$ for $t > 0$ and $p_{A}$ has none. Thus the same mean trajectory corresponds to two different full distributions.

Depends on

Used by

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Sources

Brian Munsky and Mustafa Khammash, The Finite State Projection Algorithm for the Solution of the Chemical Master Equation (standard reference, not scraped)