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Unformatted text preview: EE 278 October 9, 2009 Statistical Signal Processing Handout #5 Homework #2 Solutions 1. The cdf of random variable X is given by F X ( x ) = braceleftBigg 1 3 + 2 3 ( x + 1) 2 1 x x < 1 a. Find the probabilities of the following events. A = { X > 1 3 } , B = { X  1 } , C = { X 1 3  < 1 } , D = { X < } . b. Does X have a pdf? Explain your answer. Solution (10 points) a. i. P( A ) = 1 P( A c ) = 1 P { X 1 3 } = 1 F X ( 1 3 ) = 0. ii. P( B ) = P { X  1 } + P { X +1 } = F X ( 1) + (1 F X (+1)) = 1 3 . iii. P( C ) = P { 2 3 < X < 4 3 } = F X ( 4 3 ) F X ( 2 3 ) = 1 11 27 = 16 27 . iv. P( D ) = F X (0) = 1. b. The random variable X is mixed. Therefore, it does not have a pdf. 2. Distance to nearest star. (Bonus) Let the random variable N be the number of stars in a region of space of volume V . Assume that N is a Poisson random variable with pmf p N ( n ) = e V ( V ) n n ! , n = 0 , 1 , 2 ,..., where is the density of stars in space. We choose an arbitrary point in space and define the random variable X to be the distance from the chosen point to the nearest star. Find the pdf of X in terms . Solution (10 points) The trick in this problem, as in many others, is to find a way to connect events regarding X with events regarding N . In our case, for x 0: F X ( x ) = P { X x } = 1 P { X > x } = 1 P { No stars within distance x } = 1 P { N = 0 in sphere of radius x centered at origin } = 1 e 4 3 x 3 . We obtain the pdf f X ( x ) by differentiating the cdf F X ( x ): f X ( x ) = 4 x 2 e 4 3 x 3 . 3. Additive Gaussian Noise channel. A communication channel has a realvalued input signal X and an output Y = X + 2 Z , where Z N (0 , . 09). Suppose that X = 1 is sent. Use the attached table of the Q ( ) function to find the probability of the event { Y > } . Solution (5 points) Since Z N (0 , . 09), Y = 1+2 Z N ( 1 , . 36). Now, we can express Y = 0 . 6 W 1, where W N (0 , 1). Thus P { Y > } = P { . 6 W 1 > } = P braceleftbigg W > 1 . 6 bracerightbigg Q (1 . 67) . 04457 . 4. Lognormal pdf. Let X N (0 , 2 ). Find the pdf of Y = e X . Solution (10 points) If Y = e X then P( Y y ) = 0 for y 0. For y > 0, P { Y y } = P { e X y } = P { X ln y } = F X (ln y ) F Y ( y ) = braceleftBigg y F X (ln y ) y > Obviously f Y ( y ) = 0 for y < 0. For y > 0, f Y ( y ) = d dy F Y ( y ) = F prime X (ln y ) d dy ln y = 1 y f X (ln y ) . Applying this result to X N (0 , 2 ), we obtain f Y ( y ) = y e (ln y ) 2 / 2 2 y 2 y > In this case Y is said to have a lognormal pdf. 5. Random phase signal. (Bonus) Let Y ( t ) = sin( t + ) be a sinusoidal signal with random phase U[ , ]. Find the pdf of the random variable Y ( t ) for fixed values of frequency and time t . Comment on the dependence of the pdf of Y ( t ) on time t ....
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 Fall '09
 BalajiPrabhakar
 Signal Processing

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