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NTUEE Probability and Statistics - Part 3. Homework Solution
Problem 1.
Suppose you roll two four-faced dice, with faces labeled 1, 2, 3, 4, and each equally
likely to appear on top. Let X denote the smaller of the two numbers that appear on
top. (If both dice show the same number, then X is equal to that common number.)
Please find the probability mass function (PMF) of X.
Sol.
X = 4 corresponds to: (4, 4).
X = 3 corresponds to: (3, 3), (3, 4), (4, 3).
X = 2 corresponds to: (2, 2), (2, 3), (2, 4), (3, 2), (4, 2).
X = 1 corresponds to: (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (3, 1), (4, 1).
So pX (4) = 1/16, pX (3) = 3/16, pX (2) = 5/16, pX (1) = 7/16, and pX = 0 otherwise.
Problem 2.
According to the PMF in Problem 1, find P[X < 4|X > 1].
Sol.
P[X < 4|X > 1] =
P[1 < X < 4]
8/16
8
=
= .
P[X > 1]
9/16
9
Problem 3.
According to the PMF in Problem 1, find E(2X ).
Sol.
E(2X ) = 21 ·
5
3
1
14 + 20 + 24 + 16
74
37
7
+ 22 ·
+ 23 ·
+ 24 ·
=
=
= .
16
16
16
16
16
16
8
Problem 4.
Given that E[X + 4] = 10 and E[(X + 4)2 ] = 116, what is the standard deviation of
X?
Sol. E[X + 4] = E[X] + 4 = 10 ⇒ E[X] = 6,
2
E[X 2 p
+ 8X + 16] =pE[X 2 ] + 8E[X] + 16
√ = 116 ⇒ E[X ] = 52,
σX = Var(X) = E[X 2 ] − E 2 [X] = 16 = 4.
Problem 5.
Let X be a continuous random variable with PDF f (x) = 4xe−2x for x > 0 and
1
f (x) = 0 otherwise.R Please find the mean and the variance of X.
∞
Sol. The identity 0 tn e−t dt = n! follows from integration by parts. By the substitution t = 2x,
Z ∞
Z
1 ∞ 2 −t
2!
2 −2x
4x e dx =
E(X) =
= 1.
t e dt =
2 0
2
0
Similarly,
2
Z
∞
4x e
E(X ) =
0
2
3 −2x
1
dx =
4
∞
Z
t3 e−t dt =
0
3!
= 3/2.
4
2
So Var(X) = E(X ) − E (X) = 3/2 − 1 = 1/2.
Problem 6.
Let Y = X 4 , where X follows an exponential distribution with mean equal to
Find the probability density function of Y .
Sol.
Since X is exponential distributed,
λx
λe
x ≥ 0,
fX (x) =
0
otherwise.
R t1/4
1
.
λ
1/4
λeλx dx = 1 − e−λt . We have
0
y < 0,
0
FY (y) =
=
1/4
4
−λy
P[X ≤ y]
1−e
y ≥ 0.
For t ≥ 0, P[X 4 ≤ t] =
0
Taking derivative with respect to y,
λ −3/4 −λy1/4
y
e
y ≥ 0,
4
fY (y) =
0
y < 0.
Problem 7.
Let U be a uniform random variable on [0, 1], and let V = U1 . (a) Find the density
function of V . (b) What is the mean of V ?
Sol.
First we find the CDF,
Z 1
FV (v) = P(V ≤ v) = P (1/U ≤ v) = P (U ≥ 1/v) =
dv = 1 − 1/v
1/v
2
0
for v ≥ 1. Thus, fV (v) = Fv (v) = 1/v 2 for v ≥ 1 and fV (v) = 0 otherwise. We have
Z ∞
Z ∞
v
E(V ) =
vfV (v)dv =
dv = ln v|∞
1 = ∞,
2
v
−∞
1
so the mean does not exist.
Problem 8.
Given a random variable X with CDF FX and let g be the CDF of X, i.e. g(x) =
FX (x). Suppose that g is strictly increasing. We can define another random variable
Z by Z = g(X). Please find the CDF of Z.
Sol.
For 0 ≤ z ≤ 1, we have
FZ (z) = P[Z ≤ z] = P[FX (X) ≤ z]
= P[X ≤ FX−1 (z)]
= FX (FX−1 (z))
=z
since FX is strictly increasing.
Thus,

 0 if z < 0,
z if 0 ≤ z ≤ 1,
FZ (z) =

1 if z > 1.
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