EE562a_HW_3_Solutions

# EE562a_HW_3_Solutions - EE562A Homework 3 Solutions...

This preview shows pages 1–3. 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: EE562A Homework 3 Solutions Prepared by Hooman Shirani-Mehr Problem 1: Suppose X is a mean-zero random vector with covariance matrix K X = 1 4 3 4 20 22 3 22 36 Find a transformation A such that Y = AX has covariance matrix K Y = I . Solution H = h 11 h 21 h 22 h 31 h 32 h 33 ,H † = h 11 h 21 h 31 h 22 h 32 h 33 k 11 = 1 = h 2 11 ⇒ h 11 = 1 k 12 = 4 = h 11 h 21 ⇒ h 21 = 4 k 13 = 3 = h 11 h 31 ⇒ h 31 = 3 k 22 = 20 = h 2 21 + h 2 22 ⇒ h 22 = 2 k 23 = 22 = h 21 h 31 + h 22 h 32 ⇒ h 32 = 5 k 33 = 36 = h 2 31 + h 2 32 + h 2 33 ⇒ h 32 = √ 2 H = 1 4 2 3 5 √ 2 Problem 2: Let Z ( n ) be an i.i.d. Bernoulli sequence where P ( Z ( n ) = 1) = p, P ( Z ( n ) = − 1) = q = 1 − p. Let X ( n ) = n summationdisplay k =0 Z ( k ) where we take Z (0) = 0 = X (0) . Then X ( n ) is a discrete random walk. Find R X ( n,m ) for this random walk. Solution We can first assume n < m , then: R X ( n,m ) = E [ X ( n ) X ( m )] = E [ n summationdisplay k =0 Z ( k ) m summationdisplay k ′ =0 Z ( k ′ )] = n summationdisplay k =0 E [ Z ( k ) 2 ] + n summationdisplay k =0 m summationdisplay k ′ =0 ,k ′ negationslash = k E [ Z ( k ) Z ( k ′ )] = n. 1 + nm. (2 p − 1) 2 = n (1 + m (2 p − 1) 2 ) if m < n , we can simply switch m and n and the answer become m (1 + n (2 p − 1) 2 ) . Problem 3: Let X n , n ≥ 1 , denote a sequence of independent and identically distributed zero-mean unit-variance Gaussian random variables. Define for n ≥ 2 Y n = 1 2 ( X n + X n − 1 ) . a. Does this sequence converge in the mean square sense and, if so, to what limit? b. Does this sequence converge in distribution and, if so, to what distribution? Solution Since X n are i.i.d random variable, we immediately have: E [ Y n ] = 1 2 E [ X n + X n − 1 ] = 0 E [ Y 2 n ] = σ 2 Y n = 1 4 E [ X 2 n + X 2 n − 1 ] = 1 2 a) Since we don’t know what random variable Y n converges to(if exists). We can use the Cauchy’s criterion: If Y n converges to some random variable Y , the lim n,m →∞ E [ | Y n − Y m | 2 ] = 0 . And since Y n s are real in this problem, we have: lim n,m →∞ E [ | Y n − Y m | 2 ] = E [ Y 2 n ] + E [ Y 2 m ] − 2 E [ Y n Y m ] If we let m > n + 1 (we can always assume this), then E [ Y n Y m ] = E [ Y n ] E [ Y m ] = 0 . Then: lim n,m →∞ E [ | Y n − Y m | 2 ] = E [ Y 2 n ] + E [ Y 2 m ] = 1 negationslash = 0 So Y n doesn’t converge in mean square sense....
View Full Document

{[ snackBarMessage ]}

### Page1 / 6

EE562a_HW_3_Solutions - EE562A Homework 3 Solutions...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online