Bilkent University Spring 2022
MATH 255 - Probability and Statistics
Midterm Exam II Solutions
Problem 1) Suppose that X and Y have the joint PDF:
(
e−x if 0 ≤ y ≤ x
fX,Y (x, y) =
0 o.w.
Find the marginal PDF of X and the conditional PDF of Y given X.
(a) The marginal PDF of X is given by
Z ∞
fX (x) = fX,Y (x, y)dy
−∞
( Rx
−x
= 0 e dy x ≥ 0
0 o.w.
(
xe−x x ≥ 0
=
0 o.w.
(b) The conditional PDF of Y given X is undefined if x ≤ 0 . For x > 0, it can be written as
e−x
(
xe−x
x≥y≥0
fY |X (y|x) =
0 o.w.
(
1
x x≥y≥0
=
0 o.w.
1 09-25-2025 13:37:46 GMT -05:00
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, Problem 2) Suppose that the unit interval [0, 1] is randomly divided into three subintervals
by two points located as follows. The first one is uniformly distributed over the interval (0, 1).
Given that the first point is u ∈ (0, 1), the second point is also uniformly distributed but over
the interval (u, 1). Let X and Y be the length of the left and center subintervals as shown in
the following figure.
X Y
0 1
The The
first second
point point
Find their joint PDF and the expected value of X + Y .
(a) The support of fX,Y is the triangular region T = {(x, y) : x > 0, y > 0, and x + y < 1}. In
particular, fX,Y (x, y) = 0 if x ≤ 0 or if y ≥ 1. For 0 < x < 1, given X = x, the set of possible
values of Y is the interval (0, 1 − x). Since the second random point is uniformly distributed over
(x, 1), and Y is the difference between that point and the constant x, the conditional distribution of
Y is uniform over the interval (0, 1 − x). That is, if 0 < x < 1, then we have
(
1
fY |X (y|x) = 1−x 0 < y < 1 − x
0 o.w.
For 0 < x < 1, fX,Y (x, y) = fX (x)fX|Y (x|y) and fX,Y (x, y) = 0 otherwise. Therefore, we obtain
(
1
1−x (x, y) ∈ T
fX,Y (x, y) =
0 o.w.
(b) The expected value of X + Y can be written as
E[X + Y ] = E[X] + E[Y ]
(i)
= E[X] + E[E[Y |X]]
(ii) 1−X
= E[X] + E
2
1 1
= E[X] +
2 2
(iii) 3
= ,
4
where (i) follows from the law of iterated expectations, (ii) follows from the conditional PDF of Y
derived in part-(a), which turns out to be a uniform random variable over the interval (0, 1 − x),
and (iii) follows since X is a uniform random variable over the interval (0, 1).
2 09-25-2025 13:37:46 GMT -05:00
This study source was downloaded by 100000899606396 from CourseHero.com on
https://www.coursehero.com/file/177517434/MT2-solutions-Spring-2022pdf/
MATH 255 - Probability and Statistics
Midterm Exam II Solutions
Problem 1) Suppose that X and Y have the joint PDF:
(
e−x if 0 ≤ y ≤ x
fX,Y (x, y) =
0 o.w.
Find the marginal PDF of X and the conditional PDF of Y given X.
(a) The marginal PDF of X is given by
Z ∞
fX (x) = fX,Y (x, y)dy
−∞
( Rx
−x
= 0 e dy x ≥ 0
0 o.w.
(
xe−x x ≥ 0
=
0 o.w.
(b) The conditional PDF of Y given X is undefined if x ≤ 0 . For x > 0, it can be written as
e−x
(
xe−x
x≥y≥0
fY |X (y|x) =
0 o.w.
(
1
x x≥y≥0
=
0 o.w.
1 09-25-2025 13:37:46 GMT -05:00
This study source was downloaded by 100000899606396 from CourseHero.com on
https://www.coursehero.com/file/177517434/MT2-solutions-Spring-2022pdf/
, Problem 2) Suppose that the unit interval [0, 1] is randomly divided into three subintervals
by two points located as follows. The first one is uniformly distributed over the interval (0, 1).
Given that the first point is u ∈ (0, 1), the second point is also uniformly distributed but over
the interval (u, 1). Let X and Y be the length of the left and center subintervals as shown in
the following figure.
X Y
0 1
The The
first second
point point
Find their joint PDF and the expected value of X + Y .
(a) The support of fX,Y is the triangular region T = {(x, y) : x > 0, y > 0, and x + y < 1}. In
particular, fX,Y (x, y) = 0 if x ≤ 0 or if y ≥ 1. For 0 < x < 1, given X = x, the set of possible
values of Y is the interval (0, 1 − x). Since the second random point is uniformly distributed over
(x, 1), and Y is the difference between that point and the constant x, the conditional distribution of
Y is uniform over the interval (0, 1 − x). That is, if 0 < x < 1, then we have
(
1
fY |X (y|x) = 1−x 0 < y < 1 − x
0 o.w.
For 0 < x < 1, fX,Y (x, y) = fX (x)fX|Y (x|y) and fX,Y (x, y) = 0 otherwise. Therefore, we obtain
(
1
1−x (x, y) ∈ T
fX,Y (x, y) =
0 o.w.
(b) The expected value of X + Y can be written as
E[X + Y ] = E[X] + E[Y ]
(i)
= E[X] + E[E[Y |X]]
(ii) 1−X
= E[X] + E
2
1 1
= E[X] +
2 2
(iii) 3
= ,
4
where (i) follows from the law of iterated expectations, (ii) follows from the conditional PDF of Y
derived in part-(a), which turns out to be a uniform random variable over the interval (0, 1 − x),
and (iii) follows since X is a uniform random variable over the interval (0, 1).
2 09-25-2025 13:37:46 GMT -05:00
This study source was downloaded by 100000899606396 from CourseHero.com on
https://www.coursehero.com/file/177517434/MT2-solutions-Spring-2022pdf/
