MAST30001 Stochastic Modelling
Tutorial Sheet 10
1. The lifetime in hours of a saw blade in a lumber mill is a random variable with
density proportional to x(2− x) on 0 < x < 2. When the mill opened a new blade
was put in the saw. Every time a blade fails it is immediately replaced with a new
one. Since the mill is automated, it operates 24 hours a day, seven days a week.
(a) Model the number Nt of saw blades that have been replaced t hours after the
mill opened as a renewal process and determine the density, mean, and variance
of the random times between renewals.
(b) On average, about how many replacement blades for the saw will the mill go
through in the first 30 days of opening?
(c) Give an interval around your estimate from (b) that will have a 95% chance of
covering the true number of saws needed in the first 30 days.
(d) If you show up at the mill at the end of day 30, about what is the mean and
variance of the amount of time you’ll have to wait for the current blade in use
to be replaced?
2. A faculty course advisor is giving advice to students doing either course A or
course B. Course A is rather structured and so advice is simple. Course B has
more flexibility and so students need more time to select subjects. Moreover, in
some complicated situations advising time can be more than 15 minutes and for
these cases, the student is sent to talk to a faculty officer after 15 minutes. We
assume that the distributions of the time (in hours) that the advisor spends talking
to a given student have densities
(A) fA(x) = 32(1/4− x), x ∈ (0, 1/4),
(B) fB(x) = 32x, x ∈ (0, 1/4).
On each day, either only the course A or course B students can come to the course
advice session. We assume advisors work without breaks, there are always students
waiting for advice, and that the time that advice sessions last are i.i.d. Denote by
Nt the number of students an advisor has finished talking to t hours into the day.
Answer the following questions for both cases (A) and (B).
(a) About how many students can be advised in a typical 8 hour work day?
(b) Determine an interval that contains N8 with roughly 95% probability.
(c) What is the density of the limiting (as t→∞) residual lifetime TNt+1 − t?
(d) What is the joint density of the limiting (as t→∞) residual lifetime TNt+1− t
and age t− TNt?
3. (Thinned Renewal Process) Consider a renewal process with mean interarrival time
µ. Suppose that each event of this process is independently “counted” with proba-
bility p. Let Nt denote the number of counted events by time t; t > 0.
(a) Is (Nt)t≥0 a renewal process?
(b) What is limt→∞Nt/t?
4. (Delayed renewal process) A renewal process for which the time until the initial
renewal, say τ0, has a different distribution than the remaining interarrival times,
say τ1, τ2, . . ., is called a delayed (or a general) renewal process. If Nt is the number
of renewals of a delayed renewal process in [0, t], then show that limt→∞Nt/t→ 1/µ,
where µ = E[τ1].
5. The phenomenon in a renewal process (Nt)t≥0, that TNt+1 − TNt , the length of
a typical interval straddling a large value of t, is larger than the typical interval
length, τi, is sometimes called size biasing. Intervals aren’t selected at random to
straddle t, but longer intervals are more likely. Size biasing arises frequently in
statistical sampling. Consider a remote tourist destination that wants to estimate
the average length of a tourist’s visit. Two sampling schemes are up for discussion.
The first scheme uses the estimate of the average of the lengths of tourists’ visit from
a randomly chosen sample of tourists leaving the airport of the destination while
the second scheme uses the estimate of the average of the lengths of tourists’ visit
from a randomly chosen sample of tourists staying at hotels. These averages will
not be the same. Why? Which is bigger? Which scheme is the most appropriate to
estimate the average length of a tourist’s visit?