ECE109 Disc6.pdf

# ECE109 Disc6.pdf - UC San Diego J Connelly ECE 109...

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UC San Diego J. Connelly ECE 109 Discussion 6 Notes Problem 6.1 A random variable X has probability density function f X ( u )= braceleftbigg u/ 2 if 0 u 2 0 else X is then used to create a new random variable Z , where Z = braceleftbigg X +1 if X 1 X - 1 if X > 1 . (a) Find the CDF of X (b) Find the probability Z [1 / 2 , 3 / 2] . (c) Find the probability X > 1 given Z 1 . (d) Find the expected values of X and Z . Solutions (a) F X ( u )= P ( X u )= integraldisplay u -∞ f X ( v ) dv = 0 u < 0 integraltext u 0 v/ 2 dv 0 u 2 1 u > 2 = 0 u < 0 u 2 / 4 0 u 2 1 u > 2 (b) Z is also a continuous random variable; however, Z is deterministic, once we know X . P (1 / 2 Z 3 / 2)= P ( { 1 / 2 Z 3 / 2 } ∩ { X 1 } )+ P ( { 1 / 2 Z 3 / 2 } ∩ { X > 1 } ) = P ( { 1 / 2 X +1 3 / 2 } ∩ { X 1 } )+ P ( { 1 / 2 X - 1 3 / 2 } ∩ { X > 1 } ) = P ( {- 1 / 2 X 1 / 2 } ∩ { X 1 } )+ P ( { 3 / 2 X 5 / 2 } ∩ { X > 1 } ) = P (0 X 1 / 2)+ P (3 / 2 X 2) =( F X (1 / 2) - F X (0))+( F X (2) - F X (3 / 2)) =(1 / 2) 2 / 4 - 0+1 - (3 / 2) 2 / 4= 1 2 (c) P ( X > 1 | Z 1)= P ( X > 1 Z 1) P ( Z 1) = P ( X > 1 X - 1 1) P ( Z 1) = P ( X > 1 X 2) P ( Z 1) =0 where the last equality comes from the fact P ( X 2)=0 .

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• Spring '08
• KennethZeger
• Variance, Probability theory, probability density function

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