MTH5121 Probability Models
Solutions to Exercise Sheet 9
1. Suppose that N (t) is a Poisson Process of rate 0.4.
a) Find the probability that there are no arrivals in the first 10 units of time.
b) Find the probability that there are is exactly one arrival between times 5 and 10,
and exactly 2 between times 10 and 15.
c) Let A be the event that there are no arrivals between times 0 and 10, and let B be
the event there are exactly 2 arrivals between time 5 and 15.
i. Do you think A and B are independent? A guess is fine for this part.
ii. Suppose both events occur. What can you say about the number of arrivals
between times 10 and 15?
iii. Use part ii. to find P(A ∩ B).
Solution:
a) The number of arrivals in an interval (s, t) is Poisson distributed with mean λ(t − s).
Hence the number of arrivals in the first 10 units of time is Poisson distributed with
mean 0.4 × (10 − 0) = 4. Thus the probability that there are no arrivals in the first
10 units is the probability a Poisson mean 4 random variable is zero, which is e−4 .
b) Similarly the number of arrivals between times 5 and 10 is Poisson distributed with
mean 0.4 × (10 − 5) = 2. Thus the probability that there is exactly one arrival
between times 5 and 10 is 1 × e−2 = 2e−2 .
Again similarly the number of arrivals between times 10 and 15 is Poisson distributed
with mean 0.4×(15−10) = 2. Thus the probability that there are exactly two arrivals
between times 10 and 15 is 22 e−2 /2! = 2e−2 .
Finally by property 5 of the definition of the Poisson Process the number of arrivals
between time 5 and 10, and the number of arrivals between times 10 and 15 are
independent. Hence the probability that there is exactly one arrival between time 5
and 10, and exactly two between times 10 and 15 is 2e−2 × 2e−2 = 4e−4 .
c)
i. They are not independent since the intervals overlap (this is not a proof but is
a good guide).
ii. Since there are no arrivals between times 0 and 10, the two arrivals between
times 5 and 15 must both be after time 10. Therefore there must be two arrivals
between times 10 and 15.
iii. By ii. if A and B both occur is the same as saying that there are zero arrivals
between times 0 and 10 and two arrivals between times 10 and 15. These
two intervals are disjoint so the numbers of arrivals in these two intervals are
independent: i.e.,
P(A ∩ B) = P(zero arrivals in (0, 10)) × P(2 arrivals in (10, 15))
We calculate these two probabilities exactly as above.
The number of arrivals in (0, 10) is Po((10 − 0) × 0.4) = Po(4). Hence
P(zero arrivals in (0, 10)) = e−4 .
The number of arrivals in (10, 15) is Po((15 − 10) × 0.4) = Po(2). Hence,
P(2 arrivals in (10, 15)) = 22 e−2 /2! = 2e−2 .
Combining these gives
P(A ∩ B) = e−4 × 2e−2 = 2e−6 .
2. Suppose X1 , X2 , X3 , . . . are independent Bernoulli random variables each with probability
of success 2/3 and that N is an independent Poisson random variable with mean 12. Let
P
X= N
i=1 Xi .
Find the probability generating function GX of X and thus deduce that the distribution
of X is Po(8).
Solution: Star by recalling Theorems 22 from the notes. The pgf of each Xi is GXi (t) = q+pt,
and the pgf of N is GN (t) = e12(t−1) . Thus, by Theorem 22, the pgf GX of X is given by
GX (t) = GN (GXi (t)) = GN (q + pt) = e12(q+pt−1) = e12(pt−p) = e12p(t−1) .
This is the pgf of a Poisson random variable with mean 12p so X ∼ Po(12p).
3. Suppose that cars arrive at a T-junction according to a Poisson Process of rate 2. Each
car independently randomly chooses to turn left with probability 2/3 and otherwise turns
right.
Find the distribution of the number of cars that turn left between time 3 and time 9.
Hint: You can use the previous question: let Xi be the random variable that is 1 if the
ith car after time 3 turns left and 0 otherwise.
Solution: Let N be the (random variable) number that arrive at the T-Junction. Thus
N is Poisson distributed with mean 2 × (9 − 3) = 12. Let Xi be a random variable which
is 1 if the ith car arriving turns left and zero otherwise. The number of cars that turn
P
left is then N
i=1 Xi . Hence the number of cars turning left is Poisson distributed with
mean 12 × 2/3 = 8.
Comment: This is an important property of Poisson Processes: if we have a Poisson
process but remove a fixed proportion at random then the remaining arrivals also form a
Poisson Process.