Probability and statistics
May 31, 2016
Lecturer: Prof. Dr. Mete SONER
Coordinator: Yilin WANG
Solution Series 12
Q1. If a random variable X has the χ2 distribution with m degrees of freedom, then the distribution of X 1/2 is called a chi (χ) distribution with m degrees of freedom. Determine the mean
of this distribution.
Solution:
Let X follows Chi-squared distribution χ2 (m) with m √
degrees of freedom which is also a
Gamma distribution with parameter (m/2, 1/2). Then X follows χ(m), has the expected
value:
Z
√
(1/2)m/2 ∞ √ m/2−1 −x/2
E( X) =
xx
e
dx
Γ(m/2) 0
Z
(1/2)m/2 ∞ (m+1)/2−1 −y
y
e (1/2)−(m+1)/2 dy
=
Γ(m/2) 0
√ Γ((m + 1)/2)
= 2
.
Γ(m/2)
Q2. Moore and McCabe (1999) describe an experiment conducted in Australia to study the
relationship between taste and the chemical composition of cheese. One chemical whose
concentration can affect taste is lactic acid. Cheese manufacturers who want to establish a
loyal customer purchases the cheese. The variation in concentrations of chemicals like lactic
acid can lead to variation in the taste of cheese. Suppose that we model the concentration
of lactic acid in several chunks of cheese as independent normal random variables with mean
µ and variance σ 2 = 0.09.
Suppose that we will sample 20 chunks of cheese. Let
T =
20
X
(Xi − µ)2 /20,
i=1
where Xi is the concentration of lactic acid in ith chunk. What number c satisfies P(T ≤
c) = 0.9?
Solution:
20
σ2 X
T =
20 i=1
1
Xi − µ
σ
2
.
Probability and statistics
May 31, 2016
To find the 0.9-quantile of T , remark that
P(T ≤ c) = P
20
20
T ≤ 2c .
σ2
σ
Where σ202 T ∼ χ2 (20). We find in the table of inverse c.d.f. of Chi-squared distribution that
the 0.9-quantile for χ2 (20) is 28.41. Hence
20
c = 28.41 ⇒ c = 28.41 ∗ 0.09/20 = 0.128.
σ2
Q3. When the motion of a microscopic particle in a liquid or a gas is observed, it is seen that
the motion is irregular because the particle collides frequently with other particles. The
probability model for this motion, which is called Brownian motion, is as follows: A coordinate system is chosen in the liquid or gas. Suppose that the particle is at the origin of
this coordinate system at time t = 0, and let (X, Y, Z) denote the coordinates of the particle
at any time t > 0. The random variables X, Y, and Z are i.i.d., and each of them has the
normal distribution with mean 0 and variance σ 2 t. Find the probability that at time t = 2
the particle will lie within a sphere whose center is at the origin and whose radius is 4σ.
Solution:
At time t = 2, X, Y, Z are i.i.d. normal distribution with mean 0 and variance 2σ 2 . Let
kx, y, zk denote the euclidean distance from (x, y, z) to 0,
P(kX, Y, Zk ≤ 4σ) = P((X 2 + Y 2 + Z 2 )/2σ 2 ≤ 8).
√
√
√
Since X 2 + Y 2 + Z 2 = 2σ 2 [(X/ 2σ)2 + (Y / 2σ)2 + (Z/ 2σ)2 ], we have
X2 + Y 2 + Z2
∼ χ2 (3).
2σ 2
The distribution of χ2 (3) is a Gamma distribution with parameter (3/2, 1/2) , with p.d.f.
(1/2)3/2 1/2 −x/2
f (x) =
x e
.
Γ(3/2)
We find that the probability of the particle to be in the ball of radius 4σ of the origine is the
probability of a χ2 (3) being less than 8:
Z
(1/2)3/2 8 1/2 −x/2
2
2
2
2
P((X + Y + Z )/2σ ≤ 8) =
x e
dx.
Γ(3/2) 0
Q4. Take x ∈ [0, 1]. We say that x is normal if for x in its binary form,
X
x=
xn 2−n xn ∈ {0, 1},
n∈N
we have that limn→∞
|{1≤k≤n:xk =1}|
n
= 12 .
2
Probability and statistics
May 31, 2016
(a) P
Prove that if we have a sequence (Un )n∈N i.i.d. Bernoulli with parameter 21 , then U =
−n
is an uniform random variable in [0, 1]
n∈N Un 2
(b) Prove that if U ∼ U (0, 1), P(U is normal) = 1.
Solution:
(a) First we P
have to prove that U is measurable. For this we just have to realize that
−n
U (m) := m
is measurable because it’s the finite sum of measurable function
n=1 Un 2
and we have that
Un → U.
Second we have to understand the measure that U produces in R. For this it’s enough to
show that the measure induced by U coincide with the uniform measure in the intervals
of the form
k k+1
,
,
2n 2n
for k ∈ N ∈ [0, 2n − 1]. This is because they generate the Borel σ-algebra.
Note that
P((∃n ∈ N)(∀m ≥ n)Xm = 1) = 0,
thanks to Borel-Cantelli Lemma. So we can work in
Ω̃ := Ω\{ω ∈ Ω : (∃n ∈ N)(∀m ≥ n)Xm = 1},
i.e. ourPprobability space is (Ω̃, Ã, P̃) where à = A |Ω̃ and P̃ := P |Ω̃ . Now we have that
i
n
if k = n−1
i=0 ki 2 ∈ {0, 1, .., 2 − 1}:
!
n −1
2\
k k+1
P U ∈ n, n
=P
{Ui+1 = kn−i }
2
2
i=0
=
n −1
2Y
i=0
−n
=2
P(Ui+1 = kn−i )
.
That is the probability of a uniform random variable to be in
k
2n
, k+1
.
n
2
(b) Take (Un )n∈N Bernoulli 12 i.i.d. Thanks to part (a) we have that U :=
uniform distributed and it’s normal iff
Pn
1
k=1 1{Uk =1}
→ .
n
2
Then:
Pn
1
k=1 1{Uk =1}
P(U is normal) = P
=
= 1.
n
2
Where in the last equality we have used the law of large numbers.
3
Pn
n=1
Un 2−n is