MATH-341: FALL SEMESTER 2017
HOMEWORK 12
CORRESPONDING QUIZ ON 12/6
ANSWER KEY
The following questions cover R Section 6.4:
1. Suppose that the pdf of a random variable X is:
f (x) = 3x2 1{0<x<1} .
Let Y = 1 − X 2 . Find the pdf of Y .
First, we have:
FY (y) = Pr(Y ≤ y)
=
Pr(1 − X 2 ≤ y)
=
Pr(1 − y ≤ X 2 )
1 − Pr(1 − y > X 2 )
p
p
= 1 − Pr( 1 − y > X > − 1 − y)
p
= 1 − Pr(X < 1 − y)
=
where the last line follows from the support of X: 0 < X < 1. From this, we have:
fY (y)
d
FY (y)
dy
p
d
=
[1 − Pr(X < 1 − y)]
dy
"Z √
#
1−y
d
2
3x dx
= −
dy 0
#
" √
1−y
d
= −
x3
dy
0
=
d
[(1 − y)3/2 ]
dy
3p
1−y
2
= −
=
Also, since Y = 1 − X 2 , this tells us that the support of Y must be 0 < Y < 1. Therefore,
3p
1 − y · 1{0<y<1}
fY (y) =
2
2. Suppose that a random variable X can have each of the seven values -3,-2,-1,0,1,2,3 with equal
probability. Determine the pmf of Y = X 2 − X.
First we note that:
0 for x = 0 or x = 1
2 for x = −1 or x = 2
y=
6 for x = −2 or x = 3
12 for x = −3
1
Thus,
Pr(Y = 0)
=
Pr(Y = 2)
=
Pr(Y = 6)
=
Pr(Y = 12)
=
2
7
2
7
2
7
1
7
3. Suppose that the pdf of X is:
f (x) =
1
x · 1{0<x<2}
2
Determine the pdf of Y = 4 − X 3 .
FY (y) = Pr(Y ≤ y)
=
Pr(4 − X 3 ≤ y)
Pr(4 − y ≤ X 3 )
1
= Pr (4 − y) 3 ≤ X
Z 2
1
=
x dx
1
(4−y) 3 2
=
=
x2
4
2
1
(4−y) 3
2
=
⇒ fY (y)
=
=
=
1−
(4 − y) 3
4
d
FY (y)
dy
"
#
2
d
(4 − y) 3
1−
dy
4
1
1
(4 − y)− 3
6
Also, the support of Y is −4 < Y < 4, so we have:
fY (y)
=
1
1
(4 − y)− 3 1{−4<y<4}
6
4. Suppose that the pdf of X is:
f (x) = e−x 1{x>0}
2
Determine the pdf of Y = X 1/2 .
FY (y)
=
=
Pr(Y ≤ y)
1
Pr X 2 ≤ y
Pr(X ≤ y 2 )
Z y2
=
e−x dx
=
0
y2
=
−e
−x
0
2
1 − e−y
i
2
d h
1 − e−y
=
dy
=
⇒ fY (y)
=
2ye−y
2
Coming from the support of X: X > 0, we thus know that the support of Y is Y > 0, so:
fY (y)
=
2
2ye−y 1{y>0}
5. Let X have the uniform distribution on the interval [a, b], and let c be some constant such
that c > 0. Let Y = cX + d where d is also a constant. Show that Y also has a uniform
distribution, specifically on the interval [ca + d, cb + d].
If X is uniform over the interval [a, b], then:
fX (x)
=
1
1{a≤x≤b}
b−a
Now, let Y = cX + d. This function of X is monotonic, and note that:
Y
= cX + d
Y − d = cX
Y −d
= X
c
y−d
⇒ h−1 (y) =
c
3
so we can use:
fY (y)
=
=
=
=
=
=
fX [h−1 (x)]
dh−1
dy
1
dh−1
1{a≤h−1 (x)≤b}
b−a
dy
d y−d
1
×
1
y−d
b − a {a≤ c ≤b} dy c
1
1
1{ca+d≤y≤cb+d} ×
b−a
c
1
1{ca+d≤y≤cb+d}
cb − ca
1
1{ca+d≤y≤cb+d}
(cb + d) − (ca + d)
where this is the pdf of a uniform random variable between ca + d and cb + d as required.
6. Suppose that you would like to simulate a variable X with the following pdf:
f (x) =
1
(2 − x)1{0<x<2}
2
Show how you can use the Probability Integral Transformation (as shown in class, and a
specific one shown in Example 6.5 on page 306), to transform a uniform random variable into
one with this pdf.
First, we find the cdf of X:
Z
FX (t)
t
=
−∞
t
Z
1
(2 − x)1{0<x<2} dx
2
1
(2 − x) dx
2
=
0
= x−
x2
4
t
0
t2
= t−
4
Now, using the Probability Integral Transform, we have Y = FX (x), where we showed that
this Y has a Uniform[0,1] distribution.
−1
Therefore, if Y = FX (x), then we note that X = FX
(y). So, we take the cdf of X and solve
4
for x:
y
−4y
4 − 4y
4 − 4y
p
± 4 − 4y
p
− 4 − 4y
p
2−2 1−y
x2
4
= x2 − 4x
= x−
= x2 − 4x + 4
=
(x − 2)2
=
(x − 2)
= x − 2 (since 0 < x < 2, so x − 2 must be negative)
= x
So, if we can simulate values of the variable Y which come from a Uniform distribution over
the interval
[0, 1], then we would just take these values, and plug them into the expression
√
2 − 2 1 − y, which will then give us values of x that will have the distribution of the variable
X given by the pdf fX above.
7. Read and understand each question on the In-class activity from 11/27/17
See class notes.
8. Read and understand each question on the In-class activity from 12/1/17
See class notes.
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