e2_07 - ECE 514 Fall 2007 Exam 2 Due 12:30pm at ECE front...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ECE 514 , Fall 2007 Exam 2: Due 12:30pm at ECE front office, October 25, 2007 Solutions (version: October 25, 2007, 21:11) 75 mins.; Total 50 pts. 1. (14 pts.) Consider three real random variables X , Y , and Z . a. Suppose that Z = X 2 + Y 2 and that X and Y have a joint density that is uniform on the region { [ x, y ] : x, y ∈ [- 1 , 1] } (i.e., the density function is constant in this region and is elsewhere). Find the CDF of Z , F Z ( z ) , for z ≤ 1 . b. Suppose X = Z 2 , Y = √ Z , and Z is uniformly distributed on [0 , 1] . Find the joint CDF of the pair ( X, Y ) , F X,Y ( x, y ) , x, y ∈ R . Are X and Y independent? Ans.: a. Because Z ≥ a.s., we deduce that if z < , then F Z ( z ) = 0 . So fix z ≥ . The value of the uniform density on the given region is 1 / 4 (because that’s the area of the region). Now, for z ≤ 1 , F Z ( z ) = P { X 2 + Y 2 ≤ z } , which is simply the 1 / 4 times the area of the circle { [ x, y ] : x 2 + y 2 ≤ z } , which is πz . So we have F Z ( z ) = πz/ 4 , ≤ z ≤ 1 . b. We have F X,Y ( x, y ) = P { X ≤ x, Y ≤ y } = P { Z 2 ≤ x, √ Z ≤ y } = P { Z ≤ √ x, Z ≤ y 2 } = P { Z ≤ min( √ x, y 2 ) } . If x ≤ or y ≤ , then F X,Y ( x, y ) = 0 . If x ≥ 1 and y ≥ 1 , then F X,Y ( x, y ) = 1 . Otherwise, F X,Y ( x, y ) = min( √ x, y 2 ) . This function cannot be written as the product of two functions, one....
View Full Document

{[ snackBarMessage ]}

Page1 / 3

e2_07 - ECE 514 Fall 2007 Exam 2 Due 12:30pm at ECE front...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online