MAST30001 Stochastic Modelling
Tutorial Sheet 7
1. Let (Nt)t≥0 be a Poisson process with rate λ and let 0 < T1 < T2 < · · · be the times
of “arrivals” or jumps of (Nt)t≥0. Compute:
(a) P(N3 ≤ 2, N1 = 1),
The name of the game for all of these problems is to recast the event you want
to compute in terms of independent variables or events which are amenable
to the Poisson process description. Here, N1 and N3 − N1 are independent,
Poisson with respective means λ and 2λ, so that
P(N3 ≤ 2, N1 = 1) = P(N3 −N1 ≤ 1, N1 = 1) = P(N3 −N1 ≤ 1)P(N1 = 1)
=
(
e−2λ + 2λe−2λ
)
λe−λ.
(b) P(N3 ≤ 2, N1 ≤ 1),
Similar to (a),
P(N3 ≤ 2, N1 ≤ 1) = P(N3 ≤ 2, N1 = 1) + P(N3 ≤ 2, N1 = 0)
= [Ans. to (a)] + e−2λ
(
1 + 2λ+ (2λ)2/2
)
e−λ.
(c) P(N2 = 2, N1 = 2, N1/2 = 0),
Use that N2 −N1,N1 −N1/2, and N1/2) are independent and Poisson and that
P(N2 = 2, N1 = 2, N1/2 = 0) = P(N2 −N1 = 0, N1 −N1/2 = 2, N1/2 = 0)
to find
P(N2 = 2, N1 = 2, N1/2 = 0) = e−λ
(λ/2)2
2
e−λ/2e−λ/2.
(d) P(N7 −N3 = 2|N5 −N2 = 2),
Again, N7 −N5, N5 −N3, and N3 −N2 are independent and Poisson and
P(N7 −N3 = 2|N5 −N2 = 2) =
P(N7 −N3 = 2, N5 −N2 = 2)
P(N5 −N2 = 2)
.
The denominator of the fraction on the right hand side is easy. The numerator
equals
2∑
i=0
P(N7 −N5 = 2− i, N5 −N3 = i, N3 −N2 = 2− i)
=
2∑
i=0
(2λ)2−i
(2− i)!
e−2λ
(2λ)i
(i)!
e−2λ
(λ)2−i
(2− i)!
e−λ.
(e) the (joint) distribution function of (T1, T2),
The joint cdf P(T1 ≤ s, T2 ≤ t) is 0 if either t or s ≤ 0. Otherwise if t ≤ s this
is
P(T2 ≤ t) = P(Nt ≥ 2) = 1− e−λt[1 + λt].
Otherwise 0 ≤ s < t and we have
P(T1 ≤ s, T2 ≤ t) = P(Ns ≥ 1, Nt ≥ 2) = P(Nt ≥ 2)− P(Nt ≥ 2, Ns = 0)
= P(Nt ≥ 2)− P(Nt −Ns ≥ 2)P(Ns = 0)
= 1− e−λt[1 + λt]−
(
1− e−λ(t−s)[1 + λ(t− s)]
)
e−λs.
(f) the joint density of (T1, T2),
We differentiate the joint cdf with respect to s and t. The derivative with respect
to s is 0 for t < s, while for 0 < s < t it is equal to −λe−λt. Differentiating
with respect to t gives λ2e−λt on 0 < s < t <∞.
(g) the distribution of T1|{T2 = t2}.
We have seen in class that T1|{T2 = t} is uniform on [0, t]. Another way to see
this is that T1 must be less than T2, but the joint density above does not depend
on s, so the conditional distribution of T1 given {T2 = t} must not depend on
the value of s ∈ [0, t] (so it must be uniform). More explicitly, we can verify
this here by dividing the joint density above by the density of T2 (which, as the
sum of two independent exponential(λ) random variables is fT2(t) = λ
2te−λt at
t to give the conditional density λ2e−λt/(λ2te−λt) = 1/t.
2. Yeast microbes from the air outside of a culture float by according to a Poisson
process with rate 2 per minute. Each microbe that floats by joins the population
of the culture with probability p and with probability 1 − p the microbe doesn’t
join the culture, and this choice is made independent from the times of arrival and
choice to join of all other microbes.
(a) Find the probability that exactly four outside microbes float by in the first 3
minutes.
Let Nt be the number of microbes that float by up to time to t and let Mt be
the number that join the colony up to time t. Then Nt is Poisson with mean
2t and Mt is Poisson with mean 2pt, independent of the process Nt−Mt which
is Poisson with mean 2(1 − p)t. Also conditional on a Poisson process being
equal to k at time t, the distribution of the k points in the interval (0, t) are
the same as k i.i.d. variables that are uniform on (0, t). These facts and the
description of the process imply we have the following answers.....
e−664
4!
since Nt is a Poisson process.
(b) Find the probability that exactly four outside microbes join the culture in the
first 3 minutes.
e−6p(6p)4
4!
since Mt is a Poisson process.
(c) Given that 7 outside microbes have floated by the culture in first 3 minutes,
what is the probability that at least two of the seven join the culture?
1− 7p(1− p)6 − (1− p)7 since the description of the Mt process as a thinning
of Nt implies that Mt conditional on {Nt = 7} is Binomial.
(d) Given that 7 outside microbes have floated by the culture in first 3 minutes,
what is the probability that exactly 3 float by in the first 1 minute?(
7
3
)
(1/3)3(2/3)4 from the conditional Poisson process description. You can also
see this by writing it as
P(N1 = 3, N3 −N1 = 4)
P(N3 = 7)
,
and using the independence of the two events in the numerator etc.
(e) What is the probability that in the first 3 minutes, exactly four microbes join
the culture and 3 float by that don’t join the culture?
e−6p(6p)4
4!
· e
−6(1−p)(6(1−p))3
3!
(Mt and Nt −Mt are independent Poissons).
Assume now that a second strain of yeast microbes independently float by the culture
according to a Poisson process with rate 1, and each microbe joins the culture with
probability q, analogous to the previous process.
(f) What is the probability that exactly four yeast microbes (from either strain)
float by in the first 3 minutes?
Let Nt and Mt be respectively, the number of the first strain that float by, and
the number of this strain that float by and join the culture up to time t. Let
Kt and Lt be the analogous processes for the second strain. Then as before, all
of these processes are Poisson processes and with Nt having rate 2, Mt having
rate 2p, Kt having rate 1, Lt having rate q, and Nt,Mt are independent of
Kt, Lt. Because of independence, superposition of Poisson processes implies
that Nt +Kt and Mt +Lt are Poisson processes with rates 3 and 2p+ q. Using
these facts we find that
P(N3 +K3 = 4) =
e−994
4!
.
(g) What is the probability that exactly four yeast microbes (from either strain)
join the culture in the first 3 minutes?
P(L3 +M3 = 4) =
e−3(2p+q)(3(2p+ q))4
4!
.
3. Let U(1), . . . , U(n) be order statistics of independent variables, uniform on the interval
(0, 1). For 0 < x < y < 1 find:
(a) P(U(1) > x,U(n) < y),
The event {U(1) > x,U(n) < y} is the same as all the Ui’s are between x and y
which occurs with probability (y − x)n.
(b) P(U(1) < x,U(n) < y),
P(U(1) < x,U(n) < y) + P(U(1) > x,U(n) < y) = P(U(n) < y) = yn and then use
the previous answer to get yn − (y − x)n.
(c) P(U(k) < x,U(k+1) > y).
The event {U(k) < x,U(k+1) > y} is the same as k of the Ui’s are smaller than
x and the rest are larger than y, which occurs with probability(
n
k
)
xk(1− y)n−k.