M464 - Introduction To Probability II - Homework 8
Enrique Areyan
March 13, 2014
Chapter 5
2.4 Suppose that N points are uniformly distributed over the interval [0, N ). Determine the probability distribution for the
number of points in the interval [0, 1) as N → ∞.
Solution:
Let X1 , X2 , . . . , XN be N independent random variables with distribution Xi ∼ U nif orm([0, N )), for all i. Let,
(
1 if Xi ∈ [0, 1)
Yi =
0 otherwise
1
1−0
= , by the uniform cdf of uniform distribution on [0, N ).
N −0
N
N
P
1
1
Yi ∼ Binomial(N, ).
Thus, Yi ∼ Bernoulli(p = ). Now, let Z = number of points in the interval [0, 1). Then, Z =
N
N
i=1
µ
By fact given in class we know that Bin(m, ) =⇒ P ois(µ) as m → ∞. Applying this result we get:
m
Note that P r{Yi = 1} = P r{Xi ∈ [0, 1)} =
N
X
Z=
i=1
Yi ∼ Binomial(N,
1
) =⇒ Z ∼ P ois(1) as N → ∞
N
Hence, the probability distribution for the number of points in the interval [0, 1) is P ois(1) as N → ∞.
2.7 N bacteria are spread independently with uniform distribution on a microscope slide of area A. An arbitrary region
having area a is selected for observation. Determine the probability of k bacteria within the region of area a. Show that
as N → ∞ and a → 0 such that (a/A)N → c (0 < c < ∞), then p(k) → e−c ck /k!.
Solution: Fix an arbitrary region of area a. The proportion of bacteria in this area is
(
Yi =
a
. Let,
A
1 if bacteria i is in the region a, for i = 1, 2, . . . , N
0 otherwise
a
, i.e., the chance
A
a
of a bacteria being in the area a is equal to the proportion of bacteria in such area. Thus, Yi ∼ Bernoulli(p = ). Then,
A
Note that since bacteria is uniformly distributed over the entire region A, we have that P r{Yi = 1} =
Z = number of bacteria in region a =
N
X
Yi ∼ Binomial(N,
i=1
a
)
A
N
a k
a N −k
So, the probability of k bacteria within the region of area a is P r{Z = k} =
1−
, by Binomial pmf.
k
A
A
Now, let N → ∞ and a → 0 such that (a/A)N → c (0 < c < ∞). Then:
N
a k
a N −k
1−
k
A
A
=
a k N!
a N −k
1−
(N − k)!k! A
A
def. of binomial coefficient
=
N (N − 1)(N − 2) · · · (N − k + 1) a k a N −k
1−
k!
A
A
def. of factorial
=
N (N − 1)(N − 2) · · · (N − k + 1) c k c N −k
1−
k!
N
N
hypothesis
=
N (N − 1)(N − 2) · · · (N − k + 1) ck c N −k
1−
k
N
k!
N
pulling N k out
=
=
N N −1N −2
c N −k
N − k + 1 ck 1−
···
N N
N
N
k!
N
c N −k
1
2
k + 1 ck 1−
1 1−
1−
··· 1 −
N
N
N
k!
N
rearranging terms
simplifying
Now we can analyze each piece:
1
2
k+1
i
1 1−
1−
··· 1 −
→ 1 as N → ∞, because
→ 0 as N → ∞
N
N
N
N
c N
1−
c
c N −k
e−c
N
= = e−c as N → ∞, by limit definition of constant e and
→ 0 as N → ∞
1−
→
k
c
N
1
N
1−
N
ck
are constants and so remain unchanged by taking limits
k!
Thus,
p(k) =
1
2
k + 1 ck e−c ck
N
a k
a N −k
c N −k
=1 1−
1−
··· 1 −
→
1−
1−
A
A
N
N
N
k!
N
k!
k
as N → ∞ and a → 0 such that (a/A)N → c (0 < c < ∞).
Note that this is Poisson with parameter c.