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Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 356 / 1 / c / Solution 2026-09-28

The telegraph process has backward survival equations

\partial_{t} Q^{+} \partial_{t} Q^{-} = s \partial_{y} Q^{+} + λ (Q^{-} - Q^{+}), = - s \partial_{y} Q^{-} + λ (Q^{+} - Q^{-}) .

Initially

Q^{\pm} (y, 0) = 1

. Only a left-moving trajectory reaches the target, while reflection reverses a right-moving velocity at the outer wall, so the hyperbolic boundary conditions are

Q^{-} (0, t) = 0, Q^{+} (L, t) = Q^{-} (L, t) .

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Telegraph process Created 2026-09-28 Updated 2026-11-05

The Telegraph process moves at one of two opposite constant velocities and reverses direction at Poisson event times. Its position has finite propagation speed and persistent increments.

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