Page 332 - A First Course In Stochastic Models

RUIN PROBABILITIES 327

The ruin probability as waiting-time probability
A less obvious context in which the ruin probability appears is the M/G/1 queue.
Customers arrive at a single server station according to a Poisson process with rate
λ. The service or work requirements of the successive customers are independent
random variables having a common probability distribution function B(x) with
finite mean µ. The server works at a rate of σ > 0. It is assumed that σ > λµ.
For n = 1, 2, . . . define the random variable D n by
D n = the delay in queue of the nth customer (excluding service time).

Assuming that service is in order of arrival, lim n→∞ P {D n ≤ x} exists for all x.
Moreover, letting

W q (x) = lim P {D n ≤ x},
n→∞
it holds that

W q (x) = 1 − Q(σx), x ≥ 0. (8.4.2)
A proof of these statements goes as follows. Let τ n denote the time between the
arrival of the nth and (n + 1)th customers for n = 1, 2, . . . with the convention
that the 0th customer arrives at epoch 0. Then

D n + X n /σ − τ n if D n + X n /σ − τ n ≥ 0,
D n+1 =
0 if D n + X n /σ − τ n < 0.
Hence, letting U n = X n /σ − τ n for n ≥ 1, we have

D n+1 = max(0, D n + U n ).
Substituting this equation in itself, it follows that

D n+1 = max{0, U n + max(0, D n−1 + U n−1 )}
= max(0, U n , U n + U n−1 + D n−1 ), n ≥ 1.
By a repeated application of this equation and by D 1 = 0, we find

max(0, U n , U n + U n−1 + D n−1 )

= max(0, U n , U n + U n−1 , . . . , U n + U n−1 + · · · + U 1 ), n ≥ 1.
Since the random variables U 1 , U 2 , . . . are independent and identically distributed,
(U n , . . . , U 1 ) has the same joint distribution as (U 1 , . . . , U n ). Thus

D n+1 = max(0, U 1 , U 1 + U 2 , . . . , U 1 + · · · + U n ), n ≥ 1.