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ECE313.Lecture34

# ECE313.Lecture34 - ECE 313 Probability with Engineering...

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Jointly Continuous Random Variables III Professor Dilip V. Sarwate Department of Electrical and Computer Engineering © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign. All Rights Reserved ECE 313 Probability with Engineering Applications

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ECE 313 - Lecture 34 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 2 of 41 Conditional pdf of X — Introduction For a continuous random variable X , the conditional pdf f X | A (u | A) describes the probabilistic behavior of X given that the event A has occurred P{u ≤ X ≤ u+ u | A} = P[{u ≤ X ≤ u+ u} A]/P(A) ≈ f X | A (u | A)• u In Lecture 29, we considered the case when A is specified in terms of X itself
ECE 313 - Lecture 34 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 3 of 41 X and Y are jointly continuous random variables, and suppose A = { Y α } Given that A occurred, ( X , Y ) must be in the orange shaded region shown Conditional pdf of X — given {Y ≤ α } v u α P{u ≤ X ≤ u+ u} = volume in green strip P[{u ≤ X ≤ u+ u} A] = volume in blue strip f X| A (u | A)• u vol. blue strip P(A) f X| A (u | A) = f X , Y (u,v)dv F Y ( α ) α

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ECE 313 - Lecture 34 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 4 of 41 The random point ( X , Y ) is uniformly distributed on region {(u,v): 0 < u < v < 1} f X , Y (u,v) = 2 for 0 < u < v < 1; green region Example D: conditional pdf given A u v 1 1 1/2 A = { Y ≤ 1/2}. P(A) = 1/4; orange region f X| A (u | A) = f X , Y (u,v)dv P(A) 1/2 f X| A (u | A) = 2•(1/2–u)/ (1/4) = 4(1–2u) for 0 < u < 1/2 and 0 otherwise u u
ECE 313 - Lecture 34 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 5 of 41 The random point ( X , Y ) is uniformly distributed on region {(u,v): 0 < u < v < 1} f X , Y (u,v) = 2 for 0 < u < v < 1; green region Example D: conditional pdf given A c u v 1 1 1/2 f X| A c (u | A c ) = f X , Y (u,v)dv P(A c ) 1/2 = 4/3 for 0 < u < 1/2 = 8/3(1–u) for 1/2 ≤ u < 1 = 0 otherwise A c = { Y > 1/2}. P(A c ) = 3/4; orange region

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ECE 313 - Lecture 34 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 6 of 41 u v 1 1 1/2 f X , Y (u,v) = 2 for 0 < u < v < 1; green region f X (u) = f X| A (u | A)P(A) + f X| A c (u | A c )P(A c ) Example D: unconditional pdf 1/2 A c = { Y > 1/2}. P(A c ) = 3/4; orange region 1 u u u 4 f X| A (u | A) 4/3 f X| A c (u | A c ) 2 f X (u)
ECE 313 - Lecture 34 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 7 of 41 Conditional pdf of X: given {Y = α } Suppose A = { Y = α } instead of { Y α } Now what is f X | A (u | A)? The conditioning event A has probability zero, so the usual definition will not work! Nonetheless, we have observed event A We wish to understand the probabilistic behavior of X under these circumstances We use a limiting argument to obtain the conditional pdf

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ECE 313 - Lecture 34 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 8 of 41 Conditional pdf of X: derivation
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ECE313.Lecture34 - ECE 313 Probability with Engineering...

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