Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 1 / 28K (Applied Probability)

Table of contents
- a 28K
  - Solution a
- b 28K
  - Solution b
- c 28K
  - Solution c

a

Solution

Conditioned on the jump-chain states, the successive holding times are independent exponentials with rates

q_{i_{0}}, \dots, q_{i_{n}}

. Hence

q_{i_{n}} P (J_{n} \leq t < J_{n + 1} ∣ Y_{0} = i_{0}, \dots, Y_{n} = i_{n})

equals the integral over

s_{0}, \dots, s_{n - 1} \geq 0

with

\sum_{r < n} s_{r} \leq t

of

(\prod_{r = 0}^{n - 1} q_{i_{r}} e^{- q_{i_{r}} s_{r}}) q_{i_{n}} e^{- q_{i_{n}} (t - \sum_{r < n} s_{r})} .

This simplex integral is unchanged by reversing the

n + 1

time portions and the rates attached to them. It is therefore

q_{i_{0}} P (J_{n} \leq t < J_{n + 1} ∣ Y_{0} = i_{n}, \dots, Y_{n} = i_{0}),

as required. This is holding-time reversal along a fixed CTMC jump path.

b

Solution

The Markov property at deterministic times gives

P (Z_{n + 1} = j ∣ Z_{0}, \dots, Z_{n} = i) = P_{i} (X_{h} = j) .

Thus

(Z_{n})

is a discrete-time Markov chain with transition matrix

P (h) = (p_{ij} (h))_{i, j \in I};

when the minimal chain is nonexplosive,

P (h) = e^{h Q}

.

For an irreducible chain, state

i

is recurrent in continuous time exactly when

\int_{0}^{\infty} p_{ii} (t) d t = \infty,

and it is recurrent for the skeleton exactly when

\sum_{n \geq 0} p_{ii} (nh) = \infty

. If

t \in [nh, (n + 1) h]

, the event of staying at

i

supplies

e^{- q_{i} h} p_{ii} (nh) \leq p_{ii} (t) \leq e^{q_{i} h} p_{ii} ((n + 1) h) .

Integrating over each interval shows that the integral diverges exactly when the series does. Irreducibility then makes recurrence equivalent for the two chains. This is recurrence equivalence for a CTMC and its fixed-time skeleton.

c

Solution

On the finite state space,

P_{t} = e^{tQ}

and

E_{x} f (X_{t}) = (P_{t} f) (x) .

Therefore

lim_{t ↓ 0} \frac{E _{x} f ( X _{t} ) - f ( x )}{t} = (lim_{t ↓ 0} \frac{P _{t} - I}{t} f) (x) = Q f (x) .

The matrix semigroup satisfies

\frac{d}{d t} P_{t} f = P_{t} Q f

. Integrating from

0

to

t

gives the backward integral equation

E_{x} f (X_{t}) = f (x) + \int_{0}^{t} E_{x} [Q f (X_{s})] d s .

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