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hw2-sol - EE 278 Statistical Signal Processing October 9...

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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 0 0 x < - 1 a. Find the probabilities of the following events. A = { X > 1 3 } , B = {| X | ≥ 1 } , C = {| X - 1 3 | < 1 } , D = { X < 0 } . 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 - ρV ( ρV ) n n ! , n = 0 , 1 , 2 , . . . , where ρ 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 ρ . 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 - ρ 4 3 πx 3 . We obtain the pdf f X ( x ) by differentiating the cdf F X ( x ): f X ( x ) = 4 πρx 2 e - ρ 4 3 πx 3 . 3. Additive Gaussian Noise channel. A communication channel has a real-valued input signal X and an output Y = X + 2 Z , where Z ∼ N (0 , 0 . 09). Suppose that X = - 1 is sent. Use the attached table of the Q ( · ) function to find the probability of the event { Y > 0 } .
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Solution (5 points) Since Z ∼ N (0 , 0 . 09), Y = - 1+2 Z ∼ N ( - 1 , 0 . 36). Now, we can express Y = 0 . 6 W - 1, where W ∼ N (0 , 1). Thus P { Y > 0 } = P { 0 . 6 W - 1 > 0 } = P braceleftbigg W > 1 0 . 6 bracerightbigg Q (1 . 67) 0 . 04457 . 4. Lognormal pdf. Let X ∼ N (0 , σ 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 0 y 0 F X (ln y ) y > 0 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 , σ 2 ), we obtain f Y ( y ) = 0 y 0 e - (ln y ) 2 / 2 σ 2 y 2 π σ y > 0 In this case Y is said to have a lognormal pdf. 5. Random phase signal. (Bonus) Let Y ( t ) = sin( ωt + Θ) be a sinusoidal signal with random phase Θ U[ - π, π ] . Find the pdf of the random variable Y ( t ) for fixed values of frequency ω and time t . Comment on the dependence of the pdf of Y ( t ) on time t . Solution (15 points) The general formula for calculating the pdf of a differentiable function of a continuous random variable is given in the lecture notes: f Y ( y ) = summationdisplay θ k : f Θ ( θ k )= y f Θ ( θ ) | dy/dθ | θ = θ k . We can apply that general formula to this special case. Here f Θ ( θ ) = 1 2 π , - π θ π .
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