lect03 - Lecture Notes 3 Two Random Variables • Joint,...

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Unformatted text preview: Lecture Notes 3 Two Random Variables • Joint, Marginal, and Conditional PMFs • Joint, Marginal, and Conditional CDFs, PDFs • One Discrete and one Continuous Random Variables • Signal Detection: MAP Rule • Functions of Two Random Variables EE 278: Two Random Variables 3 – 1 Joint, Marginal, and Conditional PMFs • Let X and Y be discrete random variables on the same probability space • They are completely specified by their joint pmf : p X,Y ( x, y ) = P { X = x, Y = y } , x ∈ X , y ∈ Y By axioms of probability, x ∈X y ∈Y p X,Y ( x, y ) = 1 • Example: Consider the pmf p ( x, y ) described by the following table x 1 2 . 5- 3 1 4 1 8 y- 1 1 8 1 4 2 1 8 1 8 EE 278: Two Random Variables 3 – 2 • To find p X ( x ) , the marginal pmf of X , we use the law of total probability p X ( x ) = y ∈Y p ( x, y ) , x ∈ X • The conditional pmf of X given Y = y is defined as p X | Y ( x | y ) = p X,Y ( x, y ) p Y ( y ) , p Y ( y ) = 0 , x ∈ X Check that if p Y ( y ) = 0 then p X | Y ( x | y ) is a pmf for X • Chain rule : p X,Y ( x, y ) = p X ( x ) p Y | X ( y | x ) = p Y ( y ) p X | Y ( x | y ) • X and Y are said to be independent if for every ( x, y ) ∈ X ×Y , p X,Y ( x, y ) = p X ( x ) p Y ( y ) , which is equivalent to p X | Y ( x | y ) = p X ( x ) , p Y ( y ) = 0 , x ∈ X EE 278: Two Random Variables 3 – 3 Bayes Rule for PMFs Given p X ( x ) and p Y | X ( y | x ) for every ( x, y ) ∈ X ×Y , we can find p X | Y ( x | y ) : p X | Y ( x | y ) = p X,Y ( x, y ) p Y ( y ) = p X ( x ) p Y | X ( y | x ) p Y ( y ) = p Y | X ( y | x ) ∑ x ∈X p X,Y ( x , y ) p X ( x ) = p Y | X ( y | x ) ∑ x ∈X p Y | X ( y | x ) p X ( x ) p X ( x ) The final formula is entirely in terms of the known quantities p X ( x ) and p Y | X ( y | x ) EE 278: Two Random Variables 3 – 4 Example: Binary Symmetric Channel Consider the following binary communication channel X ∈ { , 1 } Y ∈ { , 1 } Z ∈ { , 1 } The bit sent is X ∼ Bern( p ) , ≤ p ≤ 1 , the noise is Z ∼ Bern( ) , ≤ ≤ . 5 , the bit received is Y = ( X + Z ) mod 2 = X ⊕ Z , and X and Z are independent Find 1. p X | Y ( x | y ) 2. p Y ( y ) 3. P { X = Y } , the probability of error EE 278: Two Random Variables 3 – 5 1. To find p X | Y ( x | y ) we use Bayes rule p X | Y ( x | y ) = p Y | X ( y | x ) ∑ x ∈X p Y | X ( y | x ) p X ( x ) p X ( x ) We know p X ( x ) , but we need to find p Y | X ( y | x ) : p Y | X ( y | x ) = P { Y = y | X = x } = P { X ⊕ Z = y | X = x } = P { x ⊕ Z = y | X = x } = P { Z = y ⊕ x | X = x } = P { Z = y ⊕ x } since Z and X are independent = p Z ( y ⊕ x ) Therefore p Y | X (0 | 0) = p Z (0 ⊕ 0) = p Z (0) = 1- p Y | X (0 | 1) = p Z (0 ⊕ 1) = p Z (1) = p Y | X (1 | 0) = p Z (1 ⊕ 0) = p Z (1) = p Y | X (1 | 1) = p Z (1 ⊕ 1) = p Z (0) = 1- EE 278: Two Random Variables 3 – 6 Plugging into the Bayes rule equation, we obtain p X | Y (0 | 0) = p Y | X (0 | 0) p Y | X (0 | 0) p X (0) + p Y | X (0 | 1) p X (1) p X (0) = (1...
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This note was uploaded on 11/28/2009 for the course EE 278 taught by Professor Balajiprabhakar during the Fall '09 term at Stanford.

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lect03 - Lecture Notes 3 Two Random Variables • Joint,...

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