03Summer-HE2-Soln - University of Illinois 1. Second Hour...

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University Second Hour Exam: Solutions ECE 313 of Illinois Page 1 of 1 Summer 2003 1. TRUE F X (b) is right continuous. FALSE It is possible that F X (a) = F X (b). TRUE This is the certain event. FALSE This need not hold. TRUE This is true, since X is a continuous random variable. TRUE This has to hold for all density functions. 2. P{ | X – 4 | > 3} = P{ X > 7} + P{ X < 1} = 1 – Φ 4 2 7 + Φ 4 2 1 = 1 – Φ (1.25) + Φ (–0.25) = 1 – Φ (1.25) + 1 – Φ (0.25) = 2 – 0.8944 – 0.5987 = 0.5069. P{ X < 3 | X > 2} = P{2 < X < 3}/P{ X > 2} = 2( Φ (0.25) – Φ (0)) = 2 Φ (0.25) – 1 = 1.1974– 1 = 0.1974 3.(a) The PDF of X has value 0.2 for –1 u 4. Using LOTUS we have E[ Y ] = E[| X –1|] = . ∫∫ = + = + 1 1 4 1 4 1 2 1 1 2 3 . 1 | ) 2 / ( 2 . 0 | ) 2 / ( 2 . 0 ) 1 ( 2 . 0 ) 1 ( 2 . 0 u u u u du u du u (b) Y takes on values in the range [0,3]. Thus, for v < 0 or v > 3, F Y (v) = 0. For any v, 0 v 2, F Y (v) = P[ Y v] = P[1 – v X v + 1] = 0.2(v + 1 – (1 – v)) = 0.4v, while for any v, 2 v 3, F Y (v) = P[ Y v] = P[– 1 X v + 1] = 0.2(v + 1 – (– 1)) = 0.2(v+2). Differentiating, we obtain f Y (v) =
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This note was uploaded on 09/08/2009 for the course GE 331 taught by Professor Negarkayavash during the Spring '09 term at University of Illinois at Urbana–Champaign.

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