Gong, @G. E&CE 411, Spring 2006, Handout 1
1
Joint Gaussian Random Variables
1 The Case of Two Random Variables
The random variables X and Y are said to be Gaussian if their joint density function is given by 1 1 f (x, y) = exp{- S} (1) 2X Y 2(1 - 2 ) where S= and = x - X X
2
(x - X )(y - Y ) - 2 + X Y Cov(X, Y ) . X Y
y - Y Y
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Example 2.1 : Let X and Y be two jointly Gaussian random variables with means mX and 2 2 mY , variances X and Y , respectively, and correlation coefficient X,Y . (a) Show that the conditional probability density function fX|Y (x|y) is a Gaussian distribux 2 tion with mean X + Y (y - Y ) and variance X (1 - 2 ). X,Y (b) What happens if = 0? What if = 1? Solution. (a) According to the definition, we have fX,Y (x, y) 2Y = f X|Y (x|y) = exp[-A] fY (y) 2X Y 1 - 2 X,Y where A = = = Thus f X|Y (x|y) = 2X 1 1 - 2 X,Y exp - 1 X x - X + (y - Y ) 2 ) 2 Y 2(1 - X,Y X
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(x - X )2 (y - Y )2 (x - X )(y - Y ) (y - Y )2 + - 2 - 2 2 2 2(1 - 2 )X Y 2(1 - 2 )X 2(1 - 2 )Y 2Y X,Y X,Y X,Y 1 2 2(1 - 2 )X X,Y (x - X )2 +
2 (y - Y )2 X 2 (x - X )(y - Y )X X,Y - 2 2 Y Y 2
1 X x - X + (y - Y ) 2 ) 2 Y 2(1 - X,Y X
@G. Gong, E&CE 411, Spring 2006, Handout 1
2 which is a Gaussian PDF with mean mX + (y - mY )X /Y and variance (1 - 2 )X . X,Y
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(b) = If 0 then f X|Y (x|y) = fX (x) which implies that Y does not provide any information about X or X, Y are independent. If = 1 then the variance of f X|Y (x|y) is zero which means that X|Y is deterministic. This is to be expected since = 1 implies a linear relation X = AY + b so that knowledge of Y provides all the information about X.
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A General Case
Let X1 , X2 , , Xn ) be n random variables with means 1 , 2 , , n , respectively. We define the random vector X = (X1 , X2 , , Xn ), the vector of the means m = (1 , 2 , , n ), and the n n covariance matrix C = (Cij ) where Cij = Cov(Xi , Xj ) = E[Xi - i )(Xj - j ) ], then we shall say that the random variables {Xi } are jointly Gaussian if their joint probability density function is given b...