EL-630 Solutions to HW #11
2
2
1. X , Y ~ N ( X , Y , X , Y2 , XY = ) X ~ N ( X , X ), Y ~ N ( Y , Y2 )
(a) We need
+
Ecfw_Y | X = x = yf ( y | x)dy
Now
fY | X ( y | x ) =
Also
f XY ( x, y ) =
(1)
f XY ( x, y )
f X ( x)
1
e
2
2 X Y (1 )
(2)
( x X ) 2 2 (

EL-630 Solutions to HW #10
1. There are several ways to solve this problem. Let me show one of those:
XY =
Ecfw_( X X )(Y Y )
X Y
=
Cov ( X , Y )
X Y
.
Define Z = aX + Y where a is some constant.
Ecfw_Z = Z = Ecfw_aX + Y = Ecfw_aX + Ecfw_Y = a X +

EL-630 Solutions to HW #9
1. Note that X and Y both take values either from 0 to 1 or -1 to 0 simultaneously (Then both
of them are positive or negative at the same time. Hence Z = X + Y takes values from -1 to +1
only. When Z is negative, the line Z = X

EL-630 Solutions to HW #8
1. Here
f XY ( x, y ) = f X ( x ) fY ( y ) = e ( x + y )U ( x )U ( y )
(1)
(a) Define Z = X + Y takes values only from 0 to. Note that to both X and Y are positive
random variables hence
z
z
f Z ( z ) = 0 f X ( z y , y )dy = 0 e

EL-630 Solutions to HW #7
1. X ~ P()
Pcfw_ X = k = e
k
k!
,
k = 0,1, 2,
k +1
1 k
1
Ecfw_1/(1 + X ) =
e
= e
(k + 1)!
k ! k =0
k =0 1 + k
=
k
k! =
0
k =0
k +1
0!
+
1
1!
1
e
+
(k + 1)! = 1! +
k =0
2.
(k + 1)! =
k =0
2
2!
2
2!
1 + xy
,
f XY ( x, y )