1
Mid-term Examination
ST305 Statistical Inference
1. [2 marks] An experiment consists of ipping two coins. Write down the
sample space for the outcome, which represents the results of the both coins.
1. Solution.
The sample space
S = {(h, h), (h, t), (t, h), (t, t)},
where t = tail and h = head.
2. [6 marks] Suppose the sample space is S = {(a, a), (b, b), (a, b), (b, a)}, and
dene the events, A = {(a, b), (b, a)}, B = {(a, a)} and C = {(a, a), (b, b), (a, b)}.
Find A ∩ B , A ∪ B , and C c .
2.
Solution.
A∩B =∅
A ∪ B = {(a, b), (b, a), (a, a)}
C c = {(b, a)}.
3. [7 marks] Consider two independent events A and B . Suppose P (A) = 0.1
and P (B) = 0.2. Calculate P (Ac B c ).
3.
Solution.
A and B is independent =⇒ Ac and B c are also independent
P (Ac B c ) = P (Ac )P (B c ) = (1 − P (A))(1 − P (B)) = (1 − 0.1)(1 − 0.2) =
0.72.
2
4. [6 marks] Suppose
P {X = i} =
n
i
0.4i 0.610−i ,
i = 0, 1, . . . , n.
Calculate P (X ≤ 1) and P (X ≥ 1).
4.
Solution.
P (X ≤ 1) = P (X = 0) + P (X = 1) = 0.40 0.610 + 100.40.69 =
P (X ≥ 1) = 1 − P (X = 0) = 1 − 0.40 0.610 =
5. [10 marks] Suppose the random variable X has the following probability
density function
1
f (x) = e−x + e−2x , x ≥ 0.
2
Calculate P (X > 5) and E[X].
5.
Solution.
∞
Z
5
1
1
1
( e−x + e−2x )dx = e−5 + e−2∗5 =
2
2
2
Z
∞
P (X > 5) =
1
x( e−x + e−2x )dx
2
Z0 ∞
Z ∞
1 −x
=
x e dx +
e−2x )dx
2
0
Z0 ∞
Z ∞
1 −x
1
=
(−x) de +
(− )de−2x
2
2
0
0
1 1
= +
2 4
E[X] =
6. [5 marks] Let X be a Poisson random variable with mean 5 and variance
10. Suppose Y = 2X + 10. Calculate E[Y ] and V ar(Y ).
3
6.
Solution.
E[Y ] = E[2X + 10] = 2E[X] + 10 = 2 ∗ 5 + 10 = 20
V ar(Y ) = V ar(2X + 10) = 4V ar(X) = 4 ∗ 10 = 40
7. [6 marks] Let Z be a standard normal random variable. Calculate
(a) P (X ≤ 2), and
(b) P (X ≤ −2).
7. Solution.
(a)
(b) P (X ≤ −2) = 1 − P (X ≤ 2) =
8. [12 marks] Suppose X is a random variable with the following probability
density function:
f (x) =
(x−5)2
1
√ e− 2×100 ,
10 2π
−∞ < x < ∞
Calculate P (5 ≤ X ≤ 15).
8.
.
Solution.
X ∼ N (5, 100)
(X − 5)/10 ∼ N (0, 1)
P (−1 ≤ X ≤ 1) = P (5 − 5)/10 ≤ X ≤ (15 − 5)/10) = P (0 ≤ X ≤ 1) =
P (X ≤ 1) − P (X ≤ 0) =
9. (a) [10 marks] Let X be a random variable with a probability mass function:
−2 n
fX (n) =
e 2
,
n!
n = 0, 1, . . . .
Show that the moment generating function of X is MX (t) = e2(e −1) .
Provide the full details of derivation.
t
4
(b) [9 marks] Suppose the moment generating function of a random variable
t
Y is MY (t) = e2(e −1)+2t . Find the probability density/mass function of Y .
9.
Solution.
MX (t) = E[etX ]
∞
X
=
etn fX (n)
=
0
∞
X
0
∞
X
−2 n
tn e 2
e
n!
t
t
e−2+2e e−2e (2et )n
=
n!
0
P
t
∞ −2et
(2et )n
e−2+2e
0 e
=
n!
−2+2et
=e
.
(b) MY (t) = e2(e −1) e2t = MX (t)e2t
=⇒ Y = X + 2.
Thus,
t
fY (n) = P (Y = n)
= P (X + 2 = n)
= P (X = n − 2)
e−2 2x
, n = 2, 3, . . .
=
(n − 2)!
10. [8 marks] Joe is 80% certain that his missing remote controlled is in
one of the two drawers of his TV cabinet, being 30% certain it is in the left
drawer and 50% certain it is in the right drawer. If a search of the left drawer
does not nd the remote controller, what is the conditional probability that
it is in the right drawer?
5
10.
Solution.
L: the event that the it is in the left drawer
R: the event that it is in the right drawer
The desired probability:
P (R|Lc ) =
P (R)
50%
5
P (RLc )
=
=
=
.
P (Lc )
1 − P (L)
1 − 30%
7
11. Suppose the moment generating function of a random variable X is
MX (t) =
0.1
1 − 0.1t
4
if t < 10.
Calculate E[X] and V ar(X).
11.Solution.
E[X] =
E[X 2 ] =
d
MX (t)
dt
d2
MX (t)
dt2
=
t=0
=
t=0
4 ∗ 0.1
(1 − 0.1t)5
= 0.4
t=0
4 ∗ 5 ∗ 0.1 ∗ 0.1
(1 − 0.1t)6
= 0.2.
t=0
V ar(X) = E[X 2 ] − (E[X])2 = 0.2 − 0.42 = 0.04.
11.
Solution.
2
E[X] = m′ (t)|t=0 = 2tet |t=0 = 0
2
E[X 2 ] = m′′ (t)|t=0 = 4t2 e−t |t=0 = 0
12. [9 marks] Recall that family of pdfs/pmfs is called an
family if it has the form:
f (x; θ) = h(x)c(θ) exp
k
X
i=1
!
wi (θ)ti (x) IA (x).
exponential
6
Consider the distribution with its probability density function given by:
g(x; θ) = θ−1 exp(1 − (x/θ)), 0 < θ < x < ∞.
Show whether this distribution belong to the exponential family for not.
Please provide the full details of derivation.
12.
We can write
Solution.
x f (x; θ) = θ−1 exp 1 −
I[θ,∞) (x)
θ
= h(x)c(θ) exp(w(θ)t(x))I[θ,∞) (x),
where
h(x) = e1 ,
c(θ) = θ−1 ,
w(θ) = θ−1 ,
The indicator function I[θ,∞) (x):
• is a function of both θ and x
=⇒ Thus, this is NOT an exponential family.
t(x) = −x.