Homework 7: Stats 217

Due on August 5, 2022 at 11:59 pm. Each problem carries 10 points.

Note: The best 6 out of 7 homework scores will count towards the final grade. So, if you are
satisfied with the scores you obtained in the previous 6 homeworks, you may drop this one.

1. Shocks occur to a system according to a Poisson process of rate λ. Suppose that the sys-
tem survives each shock with probability α, independently of other shocks. What is the
probability that the system is surviving at time t?

2. Customers demanding service at a central processing facility arrive according to a Poisson

process of intensity λ = 8 per unit time. Independently, each customer is classified as high
priority with probability 0.2, or low priority with probability 0.8. What is the probability
that three high priority and five low priority customers arrive during the first unit of time?

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 Stats 217 Problem Set 7
Due: Friday, August 5th by 11:59pm on Gradescope.
Thanks to Sergio Charles for sharing solutions.
Question 1:
Proof Let N (t) be the number of times the system receives a shock by time t, then
{N (t)}t≥0 ∼ PP(λ) is a Poisson process. Furthermore, define the following Bernoulli ran-
dom variable:
(
1 if system survives the i-th shock with probability α
Yi = (1.1)
0 if system doesn’t survive the i-th shock with probability 1 − α
iid
so Yi ∼ Ber(α). We use Poisson thinning to determine the probability that the system survives
at time t. We can think of two cases: the system survives the i-th shock or it doesn’t survive
the i-th shock. Hence,
N (t)
X
N1 (t) = Yi (1.2)
i=1
is the number of shocks the system survives at time t. On the other hand,
N (t)
X
N0 (t) = (1 − Yi ) (1.3)
i=1
counts the number of shocks the system doesn’t survive at time t. Since we are determining
the probability that the system survives until time t, this means that it has survived all of the
N (t) shocks by time t, i.e. N1 (t) = N (t):
P(system is surviving at time t) = P(N1 (t) = N (t)) (1.4)
= P(N (t) − N1 (t) = 0) (1.5)
= P(N0 (t) = 0) (1.6)
where {N0 (t)}t≥0 ∼ PP((1 − α)λ) by Theorem 2.1 of Lecture 12. Thus,
P(N0 (t) = 0) = P(Poi((1 − α)λt) = 0) (1.7)
((1 − α)λt)0 e−((1−α)λt)
= (1.8)
0!
= e−(1−α)λt (1.9)
(α−1)λt
=e . (1.10)

Alternative proof : Note, we could have alternatively proved this as follows:

Proof We need to determine when P(N1 (t) = N (t)). However, N (t) is itself a random
variable, so we have to condition on it and marginalize it out to compute this probability as
follows:
∞
X
P(N1 (t) = N (t)) = P(N1 (t) = N (t)|N (t) = k)P(N (t) = k) (1.11)
k=0
∞
X
= P(N1 (t) = k)P(N (t) = k) (1.12)
k=0

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where
N (t)
X
N1 (t) = Yi ∼ Bin(N (t), α) (1.13)
i=1
is a sum of independent Bernoulli random variables and is, hence, distributed as a Binomial
random variable Bin(N (t), α). So
 
k k
P(N1 (t) = k) = α (1 − α)k−k = αk . (1.14)
k
Therefore,
∞
X
P(N1 (t) = N (t)) = αk P(N (t) = k). (1.15)
k=0

Since {N (t)}t≥0 ∼ PP(λ),

e−λt (λt)k
P(N (t) = k) = P(Poi(λt) = k) = (1.16)
k!
so
∞ −λt
X e (λt)k
P(N1 (t) = N (t)) = αk (1.17)
k!
k=0
∞ −λt
X e (αλt)k
(1.18)
k!
k=0
∞
−λt
X (αλt)k
=e (1.19)
k!
k=0
−λt αλt
=e e (1.20)
(α−1)λt
=e , (1.21)

which is the same result as before!

Question 2:
Proof Let N (t) be the number of customers who have arrived by time t, where {N (t)}t≥0 ∼
PP(λ) for λ = 8. Furthermore, we define the Bernoulli random variable:
(
1 if the i-th customer is high priority with probability p
Yi = (1.22)
0 if the i-th customer is low priority with probability 1 − p

where p = 0.2.
Using Poisson thinning, define
N (t)
X
N1 (t) := Yi , (1.23)
i=1
which is the number of high priority customers who have arrived by time t. Likewise, we define
N (t)
X
N0 (t) := (1 − Yi ), (1.24)
i=1

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which is the number of low priority customers who have arrived by time t.
Thus,

P(3 high priority and 5 low priority arrive during the first unit of time) = P(N0 (1) = 5, N1 (1) = 3)
(1.25)
By Theorem 2.1 of Lecture 12, recall that {N1 (t)}t≥0 ∼ PP(λp), {N1 (t)}t≥0 ∼ PP(λ(1 − p)),
and {N1 (t)}t≥0 ⊥⊥ {N0 (t)}t≥0 . Therefore, by independence of N0 (t) and N1 (t), we can write
the above equation as:

P(3 high priority and 5 low priority arrive during the first unit of time) (1.26)
= P(N0 (1) = 5, N1 (1) = 3) (1.27)
= P(N0 (1) = 5)P(N1 (1) = 3) (1.28)

where N0 (t) ∼ Poi(λ(1 − p)t) and N1 (t) ∼ Poi(λpt). Thus,

(λ(1 − p)t)k −λ(1−p)t

P(N0 (t) = k) = e (1.29)
k!
(λpt)k −λpt
P(N1 (t) = k) = e (1.30)
k!
where p = 0.2 and λ = 8. It follows that:

(8(1 − 0.2)(1))5 −8(1−0.2)(1) (8(0.2)(1))3 −8(0.2)(1)

P(N0 (1) = 5)P(N1 (1) = 3) = e × e (1.31)
5! 3!
(8(0.8))5 −8(0.8) (8(0.2))3 −8(0.2)
= e × e (1.32)
5! 3!
88 (0.8)5 (0.2)3 −8
= e (1.33)
5!3!
≈ 0.0205. (1.34)