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[ ] [ ] [ ] [ ]
E X + Y + Z = E X + E Y + E Z = 0
a)
From Eq. 5.3 we have:
var X + Y + Z = var X + var Y + var Z + 2cov X Y , + 2cov X Z , +2cov Y Z ,
b)
From Eq. 5.3 we have:
var X + Y + Z = var X + var Y + var Z = 3
[ ] [ ]
1 1
n n
n i i
i i
E S E X E X nμ
= =
∑ ∑
( )
1 1 1
sum of diag.
sum of off-diag.
elements of element of K
covariance matrix K
var var cov ,
n n n
n k j k
k j k
S X X X
= = =
∑ ∑∑
2 2 2 2 1 2
2 2 2 2 2
1 2 2
n
n
X
n
C
σ ρσ ρ σ ρ σ
ρσ σ ρσ ρ σ
ρ σ σ
−
−
−
K
K
M O
M O
K
2 2
var 2 1
n
S = nσ + n − ρσ
Problem 1:[GAR]5.
Problem 2:[GAR]5.
[ ] [ ]
1 1
n n
n i i
i i
E S E X E X nμ
= =
∑ ∑
The same argument as problem 2
2 2 2 2 1 2
2 2 2 2 2
1 2 2
n
n
X
n
C
σ ρσ ρ σ ρ σ
ρσ σ ρσ ρ σ
ρ σ σ
−
−
−
K
K
M O
M O
K
1 1 1
2 2 2 2
1 0 1
var 2 2
j j n n
k
n
j k j
S n n
ρ
σ ρσ ρ σ ρσ
ρ
− − −
= = =
∑ ∑ ∑
1
2 2
n
n
n
ρ ρ
σ ρσ
ρ ρ ρ
−
a)
Z
M s
s s
α β
α β
b)
Z
a b
M s
α s β s
is partial fraction expansion, where b
α β
αβ
= , a
β α
αβ
2 2
Z
M s
s s
β α α β α β
α β α αβ β
1
t t
Z Z
f t M s e e t
α β
β α β α
αβ αβ
− − −
a)
Y = 0 ii)
Y = 0 iii)
Y = − X
b)
i) ( )
p y − = for y = −1,
⇒ Y = 1 or Y = − 1
( )
p y 0 = 1 for y = 0
⇒ Y = 0
( )
p y = for y = −1,
⇒ Y = 1 or Y = − 1
Problem 3: [GAR]5.
Problem 4: [GAR]5.
Problem 5: [GAR]4.
max y = 1 ⇒ Y = 1
( ) ( ) [ ]
2
2
2
ˆ
E Y Y E Y 1 E Y 2 E Y 1 0.
c)
1
0
x
x y
Y E Y x ydy
x x
2 2
2 1 1 1 1
2
0 0 0 0
x x x
E Y Y dx y x y dy dx y x y dy
x x x
2
2 1
0
x
x
E Y Y x x dx
x x
2 2
1 1
0 0
x x x
x
x x
dx dx
x x
1 1
0 0
x x x
dx dx
x x
ln ln3 0.
x
This is slightly better than the linear predictor.
a)
[ ]
0 otherwise
X
x
f x
[ ] ( )
2
10 10
4 4
X
x x
X = E x = xf x dx = dx = =
b)
Problem 7:
In homework set # 11 we saw that if Y = X + W and X & Ware independent
[ ]
w 1,
0 otherwise
W
f w
So:
[ ] [ ]
1,1 or 1, 1
0 otherwise
Y X
y x y x x
f y x
[ ]
XY X Y X
x x y x
f x y f x f y x
Now MMSE estimator is
XY
X Y
x x
Y
f x y
X E X Y y xf x y dx x dx
f y
∫ ∫
The one over which
XY
f x y is not zero is depicted.
To find
Y
f y we have
Y XY
x
f y = f x y dx
∫
So we have to fix y and for each y we have to find the range of x.
1 1
4 4
2
1 1
2
4 4
y y
Y XY
x x
y y
x x
y
y
y f y f x y dx dx x
y y
E X Y y x dx xdx x
y
y y y
= =
= =
∫ ∫
∫ ∫
1 1
1 1
2 2
1
2
1
y y
Y XY
x y y
y
y
y f y f x y dx dx
y y y
E X Y y x dx x y
y
= − −
−
∫ ∫
∫
10 10
1 1
2
10
2
1
Y XY
x y y
y
y
y f y f x y dx dx
y
E X Y y x dx x
y
y y y
= − −
−
∫ ∫
∫
Since X and W are independent the pdf of Y=X+W is the convolution of the pdf of X and the pdf of
W. So by finding the convolution of these two you could find the pdf of Y, too.
c)
we had
X = ay + b
[ ] [ ]
b = E x − aE y and
X
Y
a
σ
ρ
σ
X
C I
O
c)
since
[ ]
( )
[ ]
0 cov ,
i i j i j i j i j ij
E X = ⇒ X X = E X X − E X E X = E X X = r
So correlation Matrix is the same as covariance Matrix
X X
⇒ R = C = I
d)
2 2 2 2 2
1 2
1 1 1
... var
n n n
n i i i
i i i
E X X X E X E X X
= = =
∑ ∑ ∑
since
[ ]
i
E X =
2 2 2
1 2
1
n
n
i
E X X X n
=
∑
