part38

# part38 - EE 5375/7375 Random Processes Homework#4 solutions...

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EE 5375/7375 Random Processes September 30, 2003 Homework #4 solutions Problem 1. textbook problem 6.4 A discrete-time random process is deﬁned by X n = s n for n 0 where s is selected at random from the interval (0,1). (a) Sketch a sample path of the process. (b) Find the PDF of X n . (c) Find the joint PDF of X n and X n +1 . (d) Find the mean and autocovariance functions of X n . (a) The sample path will decrease geometrically by X 0 = 1, X 1 = s , X 2 = x 2 , and so on, as shown in Fig. 1. (b) The PDF of X n at a speciﬁc n is P ( X n x ) = P ( s n x ) = P ( s x 1 /n ) = x 1 /n The last step follows because s is uniformly distributed in (0,1). (c) By deﬁnition, the joint PDF is P ( X n x, X n +1 y ) = P ( s n x, s n +1 y ) = P ( s x 1 n , x y 1 n +1 ) = P ( s min( x 1 n , y 1 n +1 )) = min( x 1 n , y 1 n +1 ) We use the minimum of x 1 n and y 1 n +1 because the joint event that s is less than both of them means that s must be less than the smaller of them. Again, the last step follows because s is uniformly distributed. (d) The mean of X n depends on knowledge of the pdf of s which is f s ( s ) = 1 over the interval (0,1): E ( X n ) = E ( s n ) = Z 1 0 s n f s ( s ) ds = Z 1 0 s n ds = 1 n + 1 The autocovariance function for lag k is deﬁned as C X ( n, n + k ) = E ( X n X n + k ) - E ( X n ) E ( X n + k ) = E ( s n s n + k ) - 1 n + 1 1 n + k + 1 = E ( s 2 n + k ) - 1 ( n + 1)( n + k + 1) = Z 1 0 s 2 n + k ds - 1 ( n + 1)( n + k + 1) = 1 2 n + k + 1 - 1 ( n + 1)( n + k + 1) Problem 2. textbook problem 6.11 Find an expression for E ( | X t 2 - X t 1 | 2 ) in terms of autocorrelation function. The absolute value does not matter because it is squared. E ( | X t 2 - X t 1 | 2 ) = E (( X t 2 - X t 1 ) 2 ) = E ( X 2 t 2 - 2 X t 1 X t 2 + X 2 t 1 ) = E ( X 2 t 2 ) - 2 E ( X t 1 X t 2 ) + E ( X 2 t 1 ) = R X ( t 2 , t 2 ) - 2 R X ( t 1 , t 2 ) + R X ( t 1 , t 1 ) 1

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where R X ( t, s ) is the autocorrelation function. Problem 3. textbook problem 6.20 Let Y n be the process that results when individual 1’s in a Bernoulli process are erased with probability α . Find the PMF of S 0 n , the counting process for Y n . Does Y n have independent and stationary increments? Let I n be a Bernoulli process where a 1 occurs with probability p and a 0 occurs with probability 1 - p . Whenever a 1 occurs, it is not erased with probability (1 - α ) in the derived process Y n . The probability of a 1 in Y n is therefore (1 - α ) p . The counting process S 0 n gives the number of ones in the ﬁrst n Bernoulli trials where each trial has probability (1 - α ) p of success. Therefore S 0 n is binomial with parameters ( n, (1 - α ) p ) and has PMF P ( S 0 n = x ) = ± n x ² ((1 - α ) p ) x (1 - (1 - α ) p ) n - x Y n has independent and stationary increments because it is a Bernoulii process (sequence of independent Bernoulli random variables). Problem 4. textbook problem 6.24
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part38 - EE 5375/7375 Random Processes Homework#4 solutions...

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