Math255 Probability and Statistics Midterm 2 Solutions

İ.D. Bilkent University Fall 2021-2022
Math255 Probability and Statistics
Midterm 2 Solutions

Problem 1. [6 pts] Let (X, Y ) be uniformly distributed over the triangular region with corners at
(0, 0), (1, 0), and (0, 1) in the x − y plane, i.e.,
(
2, x ≥ 0, y ≥ 0, x + y ≤ 1,
fX,Y (x, y) =
0, otherwise.

(a) (3 pts) Compute the probability P 2X 2 > Y .





(b) (3 pts) Compute P Z ≤ 3/4 where Z is defined as Z = max(X, Y ).

(a) The probability in question is given by 2 (the density) times the area of the region defined by
the constraints x ≥ 0, y ≥ 0, x + y ≤ 1, and y ≤ 2x2 . In other words,
Z 1 Z min{1−x,2x2 }
2
P (2X > Y ) = 2 dy dx
0 0

Since (
2x2 , 0 ≤ x ≤ 12 ,
min{1 − x, 2x2 } =
1 − x, 21 ≤ x ≤ 1,
we have
Z 1/2 Z 2x2 Z 1 Z 1−x
2
P (2X > Y ) = 2 dy dx + 2 dy dx
0 0 1/2 0
Z 1/2 Z 1
2
= 4x dx + 2(1 − x) dx
0 1/2
1/2 1  3 
1 2 1 1

4 4 1 5
= x3 − (1 − x) 2
= + 1− = + =
3 0 1/2 3 2 2 6 4 12


(b) This time we seek the area of the region defined by the constraints x ≥ 0, y ≥ 0, x + y ≤ 1,
x ≤ 43 , and y ≤ 34 .
    Z 3 Z min(1−x, 3 )
3 3 3 4 4
P max(X, Y ) ≤ = P X ≤ ,Y ≤ = 2 dy dx
4 4 4 0 0
Z 1Z 3 Z 3 Z 1−x
4 4 4
= 2 dy dx + 2 dy dx
1
0 0 4
0
Z 1 Z 3
4 3 4
= 2· dx + 2(1 − x) dx
0 4 1
4
3
1 3 4 3 1 9 7
= · − (1 − x)2 = − + = .
4 2 1 8 16 16 8
4



Problem 2. [6 pts] Let (X, Y ) be independent random variables with
( (
1
1, 0 ≤ x ≤ 1, , −1 ≤ y ≤ 1,
fX (x) = and fY (y) = 2
0, otherwise, 0, otherwise.

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