Past exam of the options II course of the University of Oxford / 2006 / N / 2 / ii

Solution

Independent increments give independent Poisson variables

N [0, s]

and

N (s, t]

λ s

and

λ (t - s)

. Dividing their joint probability by

P (N [0, t] = n)

yields

P (N [0, s] = k ∣ N [0, t] = n) = (k n) (s / t)^{k} (1 - s / t)^{n - k}, 0 \leq k \leq n .

n, s / t

T_{i} > s

means that at most

i - 1

events have occurred by

s

. Consequently

P (T_{i} > s ∣ N [0, t] = n) = k = 0 \sum m i n (i - 1, n) (k n) (s / t)^{k} (1 - s / t)^{n - k} .

For

1 \leq i \leq n

, the conditional event time lies in

[0, t]

. Integrate its survival function and substitute

u = s / t

E (T_{i} ∣ N [0, t] = n) = t \sum_{k = 0}^{i - 1} (k n) \int_{0}^{1} u^{k} (1 - u)^{n - k} d u = t \sum_{k = 0}^{i - 1} \frac{1}{n + 1} = \frac{i t}{n + 1} .

The index range

i \leq n

is implicit in the source's last assertion. For

i > n

, the

i

th event is after

t

, and independent future increments instead give expectation

t + (i - n) / λ

; the displayed

i t / (n + 1)

formula does not apply.

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