The binormal joint density function is

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Problem 4.27: The binormal joint density function is
1
1
fx,y (x, y)
exp
2(1 – p2)
2n0102/1- p2
[(r – m1)²
of
(y – m2)? 2p(x – m1)(y – m2)]|
exp{-(z – m)C-'(z – m)'}
(27)"det(C)
where z = [r y], m = [m1 m2] and
of
C =
c-( )
ρσισ
1) With
c-()
4.
C =
9
we obtain of = 4, o = 9 and pơ102 = -4. Thus
2) The transformation Z = 2X +Y, W = X – 2Y is written in matrix notation as
()-(: )G)-()
2
1
= A
W
The ditribution fz,w(z, w) is binormal with mean m' = mA', and covariance matrix C' =
ACA'. Hence
C' = ({
9 2
2
4 -4
2
1
1 -2
9
-2
2 56
The off-diagonal elements of C' are equal to po zow = COV(Z,W). Thus COV (Z, W) = 2.
3) Z will be Gaussian with variance o? = 9 and mean
%3D
mz = [ m1
m2
= 4
21
Transcribed Image Text:Problem 4.27: The binormal joint density function is 1 1 fx,y (x, y) exp 2(1 – p2) 2n0102/1- p2 [(r – m1)² of (y – m2)? 2p(x – m1)(y – m2)]| exp{-(z – m)C-'(z – m)'} (27)"det(C) where z = [r y], m = [m1 m2] and of C = c-( ) ρσισ 1) With c-() 4. C = 9 we obtain of = 4, o = 9 and pơ102 = -4. Thus 2) The transformation Z = 2X +Y, W = X – 2Y is written in matrix notation as ()-(: )G)-() 2 1 = A W The ditribution fz,w(z, w) is binormal with mean m' = mA', and covariance matrix C' = ACA'. Hence C' = ({ 9 2 2 4 -4 2 1 1 -2 9 -2 2 56 The off-diagonal elements of C' are equal to po zow = COV(Z,W). Thus COV (Z, W) = 2. 3) Z will be Gaussian with variance o? = 9 and mean %3D mz = [ m1 m2 = 4 21
4.27 Random variables X and Y are jointly Gaussian with
m = [1 2]
-4
4
C =
9.
1. Find the correlation coefficient between X and Y.
2. If Z = 2X + Y and W = X – 2Y, find COV(Z, W).
3. Find the PDF of Z.
Transcribed Image Text:4.27 Random variables X and Y are jointly Gaussian with m = [1 2] -4 4 C = 9. 1. Find the correlation coefficient between X and Y. 2. If Z = 2X + Y and W = X – 2Y, find COV(Z, W). 3. Find the PDF of Z.
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