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Fall 2010 HW10 - University of Illinois Fall 2010 ECE 313...

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University of Illinois Fall 2010 ECE 313: Problem Set 10: Solutions Joint distributions and independence of random variables 1. [A joint pmf] (a) The marginal pmfs are the column and row sums shown in the table below. u=0 u=1 u=2 u=3 Row sum = p Y ( v ) v=4 0 0.1 0.1 0.2 0.4 v=5 0.2 0 0 0 0.2 v=6 0 0.2 0.1 0.1 0.4 Column sum p X ( u ) 0.2 0.3 0.2 0.3 (b) The possible values of Z are 5 through 9; ( p Z (5) , p Z (6) , p Z (7) , p Z (8) , p Z (9)) = (0 . 3 , 0 . 1 , 0 . 4 , 0 . 1 , 0 . 1) . (c) No. For example, p X,Y (0 , 4) = 0 6 = 0 . 08 = p X (0) p Y (4) . (d) Normalizing the column for u = 3 yields that p Y | X ( v | 3) is equal to 2 3 for v = 4 , 1 3 for v = 6 , and zero for other values of v. Therefore, E [ Y | X = 3] = 4 · 2 3 + 6 · 1 3 = 14 3 = 4 . 666 . . . . 2. [A joint distribution] (a) Clearly f X ( u ) = 0 for u < 0 . For u 0 , f X ( u ) = Z 0 ve - (1+ u ) v dv = 1 (1 + u ) 2 Z 0 (1 + u ) 2 ve - (1+ u ) v dv = 1 (1 + u ) 2 , where we used the fact that the gamma density with parameters r = 2 and λ = 1 + u integrates to one. Clearly f Y ( v ) = 0 for v < 0 . For v > 0 , f V ( v ) = Z 0 ve - (1+ u ) v du = ve - v Z 0 e - uv du = ve - v 1 v = e - v . That is, V has the exponential distribution with parameter one.
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