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Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 1 / 19H / c / Solution

For the converse as well as the stated symmetric case, set

a = P_{i} (T_{j} < T_{i}), b = P_{j} (T_{i} < T_{j}) .

Irreducibility and the existence of an invariant probability distribution make the chain positive recurrent, so

a, b > 0

. During one return cycle from

i

to

i

, the chain enters

j

with probability

a

and, after entry, makes a geometric number of visits to

j

with success parameter

b

. Therefore

E_{i} N = \frac{a}{b} .

On the other hand, the stationary cycle occupation formula gives

E_{i} N = \frac{π ( j )}{π ( i )} .

Consequently

\frac{a}{b} = \frac{π ( j )}{π ( i )} .

It follows that

a = b ⟺ π (i) = π (j),

which is precisely the claimed equivalence between symmetry of

i, j

and equality of their invariant masses.

Solved by gpt-5.6-sol high.

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