{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# hw2b - ECE514 Random Process Fall 2010 HW2 Due Contact...

This preview shows pages 1–3. Sign up to view the full content.

ECE514 Random Process Fall 2010 HW2 - Due: September 30, 2010 Contact [email protected] 1. A classical approach to generate an RV X with arbitrary CDF F X ( · ) is to apply the inverse CDF to a uniformly distributed RV, i.e., X = F - 1 X ( U ) , U [0 , 1] . This approach was outlined in class, appears in Chapter 1, and is related to previous homework and quiz questions. Using this transformation F - 1 : U [0 , 1] X R , we can leverage the progress that has been made in development of random number generators for uniformly distributed outputs. In this question, we will show how to construct a function that maps a single uniformly distributed RV to K dependent RV’s that satisfy an arbitrary CDF. To show this result, we will proceed in several steps. First, we show that K independent uniform RV’s can be mapped to an arbitrary K -RV CDF, i.e., ( X 1 , . . . , X K ) = g ( U 1 , . . . , U K ) , where U k U [0 , 1] , k ∈ { 1 , . . . , K } , ( U k ) K k =1 are independent, ( X 1 , . . . , X K ) F X 1 ,...,X K ( x 1 , . . . , x K ) for arbitrary CDF F ( · ) , and g : [0 , 1] K R K . Construct ( x 1 , . . . , x K ) iteratively as below. 1. Let g 1 ( · ) = F - 1 X 1 ( · ) be the inverse finction of F X 1 ( · ) , the marginal CDF of the K -dimensional CDF F X 1 ,...,X K ( x 1 , . . . , x K ) , i.e., F X 1 ( c ) = lim c 2 →∞ ,...,c K →∞ F X 1 ,...,X K ( c, c 2 , . . . , c K ) = F X 1 ,...,X K ( c, + , . . . , + ) . 2. Compute U 1 U [0 , 1] and x 1 = g 1 ( U 1 ) . 3. Initialize k = 2 . 4. Define g k ( · ) = F - 1 X k | X 1 ,...,X k - 1 ( ·| X 1 = x 1 , . . . , X k - 1 = x k - 1 ) as the inverse finction of the marginal CDF of X k conditioned on all previous values ( X 1 = x 1 , . . . , X k - 1 = x k - 1 ) that have been generated, F X k | X 1 ,...,X k - 1 ( X k c k | X 1 = x 1 , . . . , X k - 1 = x k - 1 ) = c u k = -∞ f X 1 ,...,X k ( X 1 = x 1 , . . . , X k - 1 = x k - 1 , X k = u k ) du k u k = -∞ f X 1 ,...,X k ( X 1 = x 1 , . . . , X k - 1 = x k - 1 , X k = u k ) du k (1) 5. Compute U k U [0 , 1] and x k = g k ( U k ) . 6. If k = K then halt; else go to Step 4. We now prove that this construction generates ( X 1 , . . . , X k ) with the correct CDF. Our proof is via induction. (a) Prove the basis of the induction, that is, prove that Step 2 generates X 1 whose marginal distribution follows that of the K -dimensional CDF. Hint: this part is straightforward from previous material. Answer : In the review material on probability (Chapter 1), we saw that in order to generate a random variable (RV) with cumulative distribution function (CDF) F X ( · ) , it suffices to first generate a uniformly distributed RV U [0 , 1] and map it to X via the inverse CDF. More formally, F X ( · ) : R [0 , 1] , F - 1 X ( · ) : U [0 , 1] R , X F X ( · ) , U U [0 , 1] , u = F X ( x ) , x = F - 1 X ( u ) , and F - 1 X ( F X ( x )) = x . Therefore, generating U [0 , 1] using a random number generator (RNG) and passing the result through F - 1 X ( · ) generates X with the requisite distribution. (b) Prove the inductive step. That is, for 1 < k K we assume that ( X 1 , . . . , X k - 1 ) were generated such that their CDF follows the ( k - 1) -dimensional marginal CDF of the K -dimensional CDF. Based 1

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

View Full Document
ECE514 Random Process Fall 2010 HW2 - Due: September 30, 2010 Contact [email protected] on this inductive assumption, we aim to prove that ( X 1 , . . . , X k ) follows the k -dimensional marginal of the K -dimensional CDF.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern