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University of Illinois, Urbana–Champaign Fall 2007
ECE 313: Solutions to Homework 12
- The joint probability density function is illustrated in the following figure.
0
Y
X
1
1
f=1/
f=3/
(a) For 0 ≤ x ≤ 1, integrating the density function along the y-axis, we have
fX (x) =
1 −x
0
1
1 −x
Otherwise, fX (x) = 0.
(b) The probability is computed by integrating the joint density function over following in-
dicated region S 1 , S 2. Since the density is uniformly
1 2 in S 1 and
3 2 in S 2 , we just simply
multiply the area by the density.
0
Y
X
1
1
S
S
P (X + Y ≤ 3 /2) =
S 1 +
S 2 =
(c) In the same way, the area to integrate is the remaining part of the square cut by an
quarter arc.
0
Y
X
1
1
P (X
2
2 ≥ 1) =
π
(d) First, we need to find the conditional probability density function for 0 ≤ y ≤ 1. By
symmetry, the marginal pdf of fY (y) =
1 2
fX|Y =y(u|v) =
fX,Y (u, y)
fY (y)
1 2 1 2 +y^
0 ≤ u ≤ 1 − y 3 2 1 2 +y
1 − y < u ≤ 1
Then, the conditional expectation is computed by
E[X|Y = y] =
1 −y
0
u
1 + 2y
du +
1
1 −y
u
1 + 2y
du =
1 + 4y − 2 y 2
2 + 4y
for 0 ≤ y ≤ 1. Otherwise, E[X|Y = y] = 0.
- To obtain the CDF of Z = X 2 Y , we need to integrate the shaded area for any 0 < z < 1,
0
Y
X
1
1
x 2
z
z
FZ (z) =
z
0
1
0
2 xdydx +
1
√ z
∫ z x^2
0
2 xdydx = z − z ln z
Differentiate the CDF to get the pdf of Z,
fZ (z) = − ln z
for 0 < z < 1. Otherwise fZ (z) = 0.
- (a) E[Z] = 2E[X
2 ] − 2 E[Y
2 ] = 2(var(X) + (E[X])
2 − var(Y ) − (E[Y ])
2 ) = − 40
(b) cov(T, U ) = cov(2X + Y, 2 X − Y ) = E[(2X + Y )(2X − Y )] − E[2X + Y ]E[2X − Y ] =
4 E[X
2 ] − E[Y
2 ] − 4(E[X])
2
2 = 7
(c)
cov(X, Y ) = ρX,Y
var(X)var(Y ) = 0. 6
E[W ] = E[3X + Y + 2] = 3E[X] + E[Y ] + 2 = 9
var(W ) = var(3X + Y ) = 9var(X) + var(Y ) + 6cov(X, Y ) = 48. 6
(d) Linear combination of joint Gaussian r.v. is also a Gaussian r.v. So W is Gaussian with
mean 9 and variance 46.8. Pr(W > 0) = Pr(
W − 9 √
- 8
− 9 √
- 8
- (a)
var(X + Y ) = var(X) + var(Y ) + 2cov(X, Y ) = 36
var(X − Y ) = var(X) + var(Y ) − 2 cov(X, Y ) = 64
Solving for cov(X, Y ), we can get cov(X, Y ) = −7. With var(X) = 3var(Y ), further we
have var(X) = 37. 5 , var(Y ) = 12.5.
ρX,Y =
cov(X, Y ) √ var(X)var(Y )
The probability density function at X, Y space is scaled down by the absolute value
of the Jacobian determinant at the new space U, V.
fU,V (u, v) =
u (v+1)^2
u, v ∈ S 1
0 o/w
ii. We have x = u, y =
u v
. The Jacobian determinant is,
∣ ∣ ∣ ∣ ∣
∂u ∂x
∂u ∂y ∂v ∂x
∂v ∂y
y
− 1 −xy
− 2
= −xy
− 2 = −
v 2
u
The pdf becomes,
fU,V (u, v) =
u v^2
u, v ∈ S 2
0 o/w
iii. We have x = uv, y = u − uv. The Jacobian determinant is,
∣ ∣ ∣ ∣ ∣
∂u ∂x
∂u ∂y ∂v ∂x
∂v ∂y
y (x+y)^2
−x (x+y)^2
x + y
u
The pdf becomes,
fU,V (u, v) =
u u, v ∈ S 3
0 o/w
(b) The marginal pdf of U = X + Y from (i) can be found by integrating along the v-axis.
fU (u) =
0
u (v+1)^2 dv = u 0 ≤ u < 1
∫ 1 u− 1 u− 1
u (v+1)^2 dv = 2 − u 1 ≤ u ≤ 2
0 o/w
(c) In the same way, the marginal pdf of U from (iii) is
fU (u) =
1 0
udv = u 0 ≤ u < 1 ∫ (^1) /u
1 − 1 /u udv = 2 − u 1 ≤ u ≤ 2
0 o/w
We found the same answer as (b) regardless of the transformation of V.
- Since X, Y, Z are iid exponential r.v. of λ = 1, the joint pdf is
fX,Y,Z (x, y, z) =
e
−(x+y+z) 0 ≤ x, y, z
0 o/w
And X =
U +V −W 2
, Y =
W +U −V 2
, Z =
V +W −U 2
. Using Jacobian transform of three r.v.s, we
have the scalar (^) ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∂u ∂x
∂u ∂y
∂u ∂z ∂v ∂x
∂v ∂y
∂v ∂z ∂w ∂x
∂w ∂y
∂w ∂z
Therefore, the new pdf at U, V, W space is
fU,V,W (u, v, w) =
1 2
e −(u+v+w)/ 2 |u − v| ≤ w ≤ u + v, 0 ≤ u, v, w
0 o/w
- We know the best mean square estimator for X
3 given Y is g(Y ) = E[X
3 |Y ]. To find it, we
first find the conditional pdf of X given Y = y, which is fX|Y (x|y) =
e −(x/y)
y for x and y both
positive. Then, we have E[X 3 |Y ] =
∞ 0
x 3 e
−(x/y)
y
dx = 6y 3 for y is positive.
- (a)
f (a, b, c) = E[(X − a − bY − cZ)
2 ] =
E[X
2 ] − bE[XY ] − cE[XZ] − aE[X] + b
2 E[Y
2 ] − bE[XY ] + cbE[Y Z] + abE[Y ]
− cE[XZ] + cbE[ZY ] + c
2 E[Z
2 ] + baE[Z] − aE[X] + bcE[Y ] + acE[Z] + a
2
To find the minimum, we take partial differential on all three variables and make them
equal zero. Then we have three equations with three unknown.
∂f
∂a
= 0 E[X] = a + bE[Y ] + cE[Z]
∂f
∂b
= 0 E[XZ] = aE[Z] + bE[Y Z] + cE[Z
2 ]
∂f
∂c
= 0 E[XY ] = aE[Y ] + bE[Y
2 ] + cE[Y Z]
Solving for a ∗ , b ∗ , c ∗ , we have
a
∗ = E[X] − b
∗ E[Y ] − c
∗ E[Z]
b
∗
cov(Y, Z)cov(X, Z) − cov(X, Y )var(Z)
(cov(Y, Z))^2 − varY varZ
c
∗
cov(Y, Z)cov(X, Y ) − cov(X, Z)var(Y )
(cov(Y, Z)) 2 − varY varZ
Alternative solution: The orthogonal principle says, if the mean square error for some
a
∗ , b
∗ , c
∗ from r.v. X to the linear r.v. vector space a + bY + cZ spanned by r.v. 1, Y, Z is
minimized, the error vector from X to a ∗
- b ∗ Y + c ∗ Z must be orthogonal to the space,
thus all basis of the space. Therefore, X − a
∗ − b
∗ Y − c
∗ Z is orthogonal to 1, Y, Z. In
this sense, the following equations are hold.
E[X − a
∗ − b
∗ Y − c
∗ Z] = 0 (1)
E[(X − a
∗ − b
∗ Y − c
∗ Z)Y ] = 0 (2)
E[(X − a
∗ − b
∗ Y − c
∗ Z)Z] = 0 (3)
Immediately, using (1), we get
a
∗ = E[X] − b
∗ E[Y ] − c
∗ E[Z] (4)
Using (1) and cov(X, Y ) = E[XY ] − E[X]E[Y ], (2) and (3) becomes
cov(X − a
∗ − b
∗ Y − c
∗ Z, Y ) = cov(X − b ∗ Y − c
∗ Z, Y ) = 0 (5)
cov(X − a
∗ − b
∗ Y − c
∗ Z, Z) = cov(X − b ∗ Y − c
∗ Z, Z) = 0 (6)
Further, (5) (6) can be simplified by the linearity of covariance.
cov(X, Y ) − b
∗ var(Y ) − c
∗ cov(Z, Y ) = 0 (7)
cov(X, Z) − b
∗ cov(Y, Z) − c
∗ var(Z) = 0 (8)
