MATH 255 - Probability and Statistics Midterm Exam II Solutions

Bilkent University Fall 2024


MATH 255 - Probability and Statistics

Midterm Exam II Solutions
24 November 2024


Problem 1. [20pt] The joint pdf of random variables X and Y is given by:

8uv u2 + v 2 ≤ 1, u ≥ 0, v ≥ 0
fX,Y (u, v) =
0 else

(a) Find the marginal pdf fX (x). Verify your answer by showing that fX (x) is a valid pdf.



4x(1 − x2 ) 0 ≤ x ≤ 1
fX (x) =
0 else


The marginal pdf of X can be written as:
√
∞ 1−x2
8x(1 − x2 )
Z Z
fX (x) = fX,Y (x, y)dy = 8xydy = = 4x(1 − x2 ), 0 ≤ x ≤ 1.
−∞ 0 2
Thus,

4x(1 − x2 ) 0≤x≤1
fX (x) =
0 else

The marginal fX (u) is a valid pdf if:
∞ 1 1
4x2 4x4
Z Z
fX (x)dx = 4x(1 − x2 )dx = − = 1.
−∞ 0 2 4 0

(b) Find fY |X (y|x).



8xy 2y
√

4x(1−x2 )
= 1−x2
0 < x < 1, 0 ≤ y ≤ 1 − x2
fY |X (y|x) =
0 else

The conditional distribution fY |X (y|x) is defined over the region where fX (x) > 0 as fol-
lows:
fX,Y (x, y)
fY |X (y|x) = 0<x<1
fX (x)

1

, Thus,

8xy 2y
√

4x(1−x2 )
= 1−x2
0 < x < 1, 0 ≤ y ≤ 1 − x2
fY |X (y|x) =
0 else

The conditional fY |X (y|x) is a valid pdf if:
√ √
∞ 1−x2 1−x2
y2
Z Z
2y 2
fY |X (y|x)dv = dy = × =1
−∞ 0 1 − x2 1 − x2 2 0

(c) Find P {X + Y ≤ 1}.


1
P {X + Y ≤ 1} =
3

Z 1 Z 1−x
P {X + Y ≤ 1} = P {Y ≤ 1 − X} = 8xydydx
0 0
√
since 1 − x ≤ 1 − x2 for 0 ≤ x ≤ 1. Thus,
Z 1  Z 1
(1 − x)2 x3
 
x 8 1
P {X + Y ≤ 1} = 8x dx = 8 − x2 + dx = =
0 2 0 2 2 24 3




2