1 ECE 340 Homework #8 Solutions
ECE 340: PROBABILISTIC METHODS IN ENGINEERING
SOLUTIONS TO HOMEWORK #8
5.1
a) A sample space may be defined as S={0,1,2}2, consisting of all the ordered pairs for which the first component is the total heads for Michael and the second component is the total heads for Carlos. The sample space S is shown in the table below.
# of heads of M #of heads of C
0 1 2
0 00 01 02
1 10 11 12
2 20 21 22
The range SXY = {(0,0), (1,0), (1,1), (2,0), (2,1), (2,2)} is shown in the second table below. Note that we can also put
S!" = { x, y :β x, y β 0,1,2 ! and x β₯ y}
# of heads of M #of heads of C
0 1 2
0 (0,0) (1,0) (2,0)
1 (1,0) (1,1) (2,1)
2 (2,0) (2,1) (2,2)
The mapping from S to SXY can be shown as the following: 00 (0,0) 01 (1,0) 10 (1,0) 02 (2,0) 20 (2,0) 11 (1,1)
2 ECE 340 Homework #8 Solutions
12 (2,1) 21 (2,1) 22 (2,2)
b)
π π,π = (0,0) = π 00 = π{ππππ} =12
!=
116
π π,π = (1,0) = π 10, 01 = π π»πππ,ππ»ππ,ππππ»,πππ»π =12
!Γ4 =
14
π π,π = (2,0) = π 02, 20 = π πππ»π»,π»π»ππ =12
!Γ2 =
18
π π,π = (1,1) = π 11 = π π»ππ»π,π»πππ»,ππ»ππ»,ππ»π»π =12
!Γ4 =
14
π π,π = (2,1) = π 12, 21 = π π»ππ»π»,ππ»π»π»,π»π»π»π,π»π»ππ» =12
!Γ4 =
14
π π,π = (2,2) = π 22 = π π»π»π»π» =12
!=
116
c) π π = π
= π π,π = 0,0 + π π,π = 1,1 +π π,π = 2,2
=116
+14+116
=38
d) If π βππππ = !! for Carlos , then
π π,π = (0,0) = π 00 = π{ππππ} = !!
!Γ !
!
!= !
!"
π π,π = (1,0) = π 10, 01 = π π»πππ,ππ»ππ,ππππ»,πππ»π
=14Γ34Γ
12
! Γ2 +
14
!Γ
12
!Γ2 =
18
π π,π = (2,0) = π 02, 20 = π πππ»π»,π»π»ππ =14
!Γ
12
!+
34
!Γ
12
!=1064
=532
π π,π = (1,1) = π 11 = π π»ππ»π,π»πππ»,ππ»ππ»,ππ»π»π =1264
π π,π = (2,1) = π 12, 21 = π π»ππ»π»,ππ»π»π»,π»π»π»π,π»π»ππ» =2464
π π,π = (2,2) = π 22 = π π»π»π»π» =964
π π = π = π π,π = 0,0 + π π,π = 1,1 + π π,π = 2,2 = !!
!"
3 ECE 340 Homework #8 Solutions
5.3 The input X to a communications channel is β-Ββ1β or β1β, with respective probabilities ΒΌ and ΒΎ. The output of the channel Y is equal to: the corresponding input X with probability 1 β p -Ββ pe; -ΒβX with probability p; 0 with probability pe.
a)
X 1-Ββp-Ββpe Y 1/4 -Ββ1 -Ββ1 p pe 0 p pe
3/4 1 1 1-Ββp-Ββpe
S = SXY = {(-Ββ1,-Ββ1), (-Ββ1,0), (-Ββ1,1), (1,-Ββ1), (1,0), (1,1)}
b)
P((X,Y) = (-Ββ1,-Ββ1)) = P{X = -Ββ1 & Y = -Ββ1} = P{Y = -Ββ1|X = -Ββ1} X P[X = -Ββ1] = (1-Ββp-Ββpe)(1/4) = 1/4(1-Ββp-Ββpe)
P((X,Y) = (-Ββ1,0)) = P{X = -Ββ1 & Y = 0} = P{Y = 0|X = -Ββ1} X P[X = -Ββ1] = (pe)(1/4) = 1/4pe
P((X,Y) = (-Ββ1,1)) = P{X = -Ββ1 & Y = 1} = P{Y = 1|X = -Ββ1} X P[X = -Ββ1] = (p)(1/4) = 1/4p
P((X,Y) = (1,-Ββ1)) = P{X = 1 & Y = -Ββ1} = P{Y = -Ββ1|X = 1} X P[X = 1] = (p)(3/4) = 3/4p
P((X,Y) = (1,0)) = P{X = 1 & Y = 0} = P{Y = 0|X = 1} X P[X = 1] = (pe)(3/4) = 3/4pe
P((X,Y) = (1,1)) = P{X = 1 & Y = 1} = P{Y = 1|X = 1} X P[X = 1] = (1-Ββp-Ββpe)(3/4) = 3/4(1-Ββp-Ββpe)
c)
P[X β Y] = 1 -Ββ P[X = Y] = 1 β [P(1,1) + P(-Ββ1,-Ββ1)] = 1 β [3/4(1-Ββp-Ββpe) + 1/4(1-Ββp-Ββpe)] = 1-Ββ(1-Ββp-Ββpe) = p+pe
P[Y = 0] = P((X,Y) = (-Ββ1,0)) + P((X,Y) = (1,0)) = P{X = -Ββ1 & Y = 0} + P{X = 1 & Y = 0}
= P{Y = 0|X = -Ββ1} X P[X = -Ββ1] + P{Y = 0|X = 1} X P[X = 1] = (pe)(1/4) + (pe)(3/4) = pe
4 ECE 340 Homework #8 Solutions
5.8 a)
d)
g)
h)
5 ECE 340 Homework #8 Solutions
5.9
a) and b) We show the joint pmf of π,π below by identifying the probability mass at the values π₯! , π¦! in the plane. We show the associated marginal pmfs of X and Y along the corresponding margins.
pX(0) = P[X = 0] = pXY(0,0) = 1/16 pX(1) = P[X = 1] = pXY(1,0) + pXY(1,1) = ΒΌ + ΒΌ = 1/2
pX(2) = P[X = 2] = pXY(2,0) + pXY(2,1) + pXY(2,2) = 1/8 + ΒΌ + 1/16 = 7/16
pY(0) = P[Y = 0] = pXY(0,0) + pXY(1,0) + pXY(2,0) = 1/16 + ΒΌ + 1/8 = 7/16
pY(1) = P[Y = 1] = pXY(1,1) + pXY(2,1) = ΒΌ + ΒΌ = 1/2
pY(2) = P[Y = 2] = pXY(2,2) = 1/16
c)
π π,π = (0,0) = π 00 = π{ππππ} =14
!Γ
12
!=
164
π π,π = (1,0) = π 10, 01 = π π»πππ,ππ»ππ,ππππ»,πππ»π
=14Γ34Γ
12
! Γ2 +
14
!Γ
12
!Γ2 =
18
π π,π = (2,0) = π 02, 20 = π πππ»π»,π»π»ππ =14
!Γ
12
!+
34
!Γ
12
!=1064
=532
π π,π = (1,1) = π 11 = π π»ππ»π,π»πππ»,ππ»ππ»,ππ»π»π =1264
π π,π = (2,1) = π 12, 21 = π π»ππ»π»,ππ»π»π»,π»π»π»π,π»π»ππ» =2464
π π,π = (2,2) = π 22 = π π»π»π»π» =964
6 ECE 340 Homework #8 Solutions
pX(0) = P[X = 0] = pXY(0,0) = 1/64
pX(1) = P[X = 1] = pXY(1,0) + pXY(1,1) = 8/64 + 12/64 = 20/64
pX(2) = P[X = 2] = pXY(2,0) + pXY(2,1) + pXY(2,2) = 10/64 + 24/64 + 9/64 = 43/64
pY(0) = P[Y = 0] = pXY(0,0) + pXY(1,0) + pXY(2,0) = 1/64 + 8/64 + 10/64 = 19/64
pY(1) = P[Y = 1] = pXY(1,1) + pXY(2,1) = 12/64 + 24/64 = 36/64
pY(2) = P[Y = 2] = pXY(2,2) = 9/64
The sketches of pXY, pX and pY are similar to part (a) anb (b) above.
5.16 a) The joint cdf is:
πΉ!,! π₯!, π¦! = π{π β€ π₯!,π β€ π¦!}
Y 0 1/16 9/16 16/16 2 0 1/16 9/16 15/16 1 0 1/16 5/16 7/16
1 2 X 0 0 0 0
Joint cdf properties:
1. The joint cdf is a non-Ββdecreasing function of X and Y. a. In both the X and Y directions, the cdf is non-Ββdecreasing.
2. FX,Y(x1,-Βββ) = 0, FX,Y(-Βββ,y1) = 0, FX,Y(β,β) = 1. a. FX,Y(x1,-Βββ) = 0 for all columns, FX,Y(-Βββ,y1) = 0 for all rows, and FX,Y(β,β) = 1
3. The marginal cdfβs are the probabilities of the half planes defined by: FX(x1) = P[X β€ x1, Y < β] and FY(y1) = P[X < β, Y β€ y1].
a. lim!β! πΉX,Y(x1,n) = P[X β€ x1] = FX(x1) and b. lim!β! πΉX,Y(n,y1) = P[Y β€ y1] = FY(y1).
4. The joint cdf is continuous from both the βnorthβ and from the βeastβ. a. This joint cdf is continuous from both the βnorthβ and the βeastβ.
5. The probability of the rectangle {x1 < x β€ x2, y1 < y β€ y2} is given by:
7 ECE 340 Homework #8 Solutions
P[x1 < x β€ x2, y1 < y β€ y2] = FX,Y(x2,y2) -Ββ FX,Y(x2,y1) -Ββ FX,Y(x1,y2) -Ββ FX,Y(x1,y1)
b) Find the marginal cdf of X and of Y. FXY(x,y) = P{X β€ x, Y β€ y}
The marginal cdf of X is: FX(x) = FXY(x,β) = 0 for X < 0 1/16 for 0 β€ X < 1 9/16 for 1 β€ X < 2 16/16 for X β₯ 2
The marginal cdf of Y is: FY(y) = FXY(β,y) = 0 for Y < 0 7/16 for 0 β€ Y < 1 15/16 for 1 β€ Y < 2 16/16 for Y β₯ 2
5.26 a) 1 = π π₯ + π¦ ππ₯ππ¦ = π !!
!+ π₯π¦
!
!ππ¦ =!
!!!
!! π !
!+ π¦ ππ¦ = π !
!π¦ + !!
! !
!= π!
!
So, π = 1 b)
ππΏπ(π,π) = ππΏπ ππ, ππ π πππ πππ
!β
π
!β
For 0 < x < 1 and y > 1:
FXY(x,y) = !! !! (x,y) dxdy = !! !! (x + y) dxdy = [ !! !
! x2 + yx]10dy = !!+ π¦ !
! dy
= [!!y + !
!y2]x0 = [
!!(x) + !
!(x)2] -Ββ [!
!(0) + !
!(0)2] = !
!x + !
!x2 = !
!(x(x + 1))
For 0 < y < 1 and x > 1:
FXY(x,y) = !! !! (x,y) dxdy = !! !! (x + y) dydx = [ !! !
! y2 + yx]10dy = !!+ π₯ !
! dx
= [!!x + !
!x2]y0 = [
!!(y) + !
!(y)2] -Ββ [!
!(0) + !
!(0)2] = !
!y + !
!y2 = !
!(y(y + 1))
For x > 1 and y > 1:
FXY(x,y) = !! !! (x,y) dxdy = !! !! (x + y) dydx = [ !! !
! y2 + yx]10dx = !! !
! (1)2 + (1)x dx
= !! !
! + x dx = [!!(x) + !
!(x)2]10 = [
!!(1) + !
!(1)2] -Ββ [!
!(0) + !
!(0)2] = !
! + !
! = 1
If (x,y) is inside the unit interval:
FXY(x,y) = !! !! (x,y) dxdy = !! !! (z1 + z2)dz1dz2 = [ !!
!! z12+z2z1]y0dz2 = [!! (!
!(y)2+z2(y))
-Ββ (!!(0)2+z2(0))dz2
= !!!!y2 + z2y dz2 = [
!!y2z2 +
!!z22y]x0 = [(
!!y2(x) + !
!(x)2y) β (!
!y2(0) + !
!(0)2y)] = !
!y2x + !
!x2y
8 ECE 340 Homework #8 Solutions
c) πΉ! π₯ = lim!β! πΉ!" π₯, π¦ = πΉ!" π₯, 1 πππ 0 β€ π₯ β€ 1 β π! π₯ = !
!"πΉ! π₯ = π₯ + !
!, πππ 0 β€ π₯ β€ 1
Similarly π! π¦ = π¦ + !
!, πππ 0 β€ π¦ β€ 1
d) π π < π = π₯ + π¦ ππ₯ππ¦ = !!
!+ π₯π¦
!
!ππ¦ = !!!
!ππ¦!
!!!
!!
!! = !!
! !
!= !
!
π π < π! = π₯ + π¦ ππ¦ππ₯ = !!
!+ π₯π¦
!
!!
ππ₯ = !!
!+ π₯! ππ₯!
!!!
!!
!!! = !!
!"+ !!
! !
!= !
!"
5.45 π π₯, π¦ = π₯ + π¦ , 0 < π₯, π¦ < 1
π π₯ = π₯ +12, 0 < π₯ < 1
π π¦ = π¦ +12, 0 < π¦ < 1
π π₯, π¦ β π π₯ π(π¦) 0 < π₯, π¦ < 1 Thus, X and Y are not independent.
5.48 b) π 4π < 1,π < 0 = π π < !
!π π < 0 = !
!Γ !!Γ !!= !
!"
d) π πππ₯ π,π < !!= π π < !
! πππ π < !
!= π π < !
!π π < !
!= !
!Γ !!Γ(!
!Γ !!) = !
!
Top Related