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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 :∀ x, y ∈ 0,1,2 ! and x ≥ 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)


12 (2,1) 21 (2,1) 22 (2,2)

b)

𝑃 𝑋,𝑌 = (0,0) = 𝑃 00 = 𝑃{𝑇𝑇𝑇𝑇} =12

!=

116

𝑃 𝑋,𝑌 = (1,0) = 𝑃 10, 01 = 𝑃 𝐻𝑇𝑇𝑇,𝑇𝐻𝑇𝑇,𝑇𝑇𝑇𝐻,𝑇𝑇𝐻𝑇 =12

!×4 =

14

𝑃 𝑋,𝑌 = (2,0) = 𝑃 02, 20 = 𝑃 𝑇𝑇𝐻𝐻,𝐻𝐻𝑇𝑇 =12

!×2 =

18

𝑃 𝑋,𝑌 = (1,1) = 𝑃 11 = 𝑃 𝐻𝑇𝐻𝑇,𝐻𝑇𝑇𝐻,𝑇𝐻𝑇𝐻,𝑇𝐻𝐻𝑇 =12

!×4 =

14

𝑃 𝑋,𝑌 = (2,1) = 𝑃 12, 21 = 𝑃 𝐻𝑇𝐻𝐻,𝑇𝐻𝐻𝐻,𝐻𝐻𝐻𝑇,𝐻𝐻𝑇𝐻 =12

!×4 =

14

𝑃 𝑋,𝑌 = (2,2) = 𝑃 22 = 𝑃 𝐻𝐻𝐻𝐻 =12

!=

116

c) 𝑃 𝑋 = 𝑌

= 𝑃 𝑋,𝑌 = 0,0 + 𝑃 𝑋,𝑌 = 1,1 +𝑃 𝑋,𝑌 = 2,2

=116

+14+116

=38

d) If 𝑃 ℎ𝑒𝑎𝑑𝑠 = !! for Carlos , then

𝑃 𝑋,𝑌 = (0,0) = 𝑃 00 = 𝑃{𝑇𝑇𝑇𝑇} = !!

!× !

!

!= !

!"

𝑃 𝑋,𝑌 = (1,0) = 𝑃 10, 01 = 𝑃 𝐻𝑇𝑇𝑇,𝑇𝐻𝑇𝑇,𝑇𝑇𝑇𝐻,𝑇𝑇𝐻𝑇

=14×34×

12

! ×2 +

14

!×

12

!×2 =

18

𝑃 𝑋,𝑌 = (2,0) = 𝑃 02, 20 = 𝑃 𝑇𝑇𝐻𝐻,𝐻𝐻𝑇𝑇 =14

!×

12

!+

34

!×

12

!=1064

=532



𝑃 𝑋,𝑌 = (2,2) = 𝑃 22 = 𝑃 𝐻𝐻𝐻𝐻 =964

𝑃 𝑋 = 𝑌 = 𝑃 𝑋,𝑌 = 0,0 + 𝑃 𝑋,𝑌 = 1,1 + 𝑃 𝑋,𝑌 = 2,2 = !!

!"


5.3 The input X to a communications channel is “-‐1” or “1”, with respective probabilities ¼ and ¾. The output of the channel Y is equal to: the corresponding input X with probability 1 – p -‐ pe; -‐X with probability p; 0 with probability pe.

a)

X 1-‐p-‐pe Y 1/4 -‐1 -‐1 p pe 0 p pe

3/4 1 1 1-‐p-‐pe

S = SXY = {(-‐1,-‐1), (-‐1,0), (-‐1,1), (1,-‐1), (1,0), (1,1)}

b)

P((X,Y) = (-‐1,-‐1)) = P{X = -‐1 & Y = -‐1} = P{Y = -‐1|X = -‐1} X P[X = -‐1] = (1-‐p-‐pe)(1/4) = 1/4(1-‐p-‐pe)

P((X,Y) = (-‐1,0)) = P{X = -‐1 & Y = 0} = P{Y = 0|X = -‐1} X P[X = -‐1] = (pe)(1/4) = 1/4pe

P((X,Y) = (-‐1,1)) = P{X = -‐1 & Y = 1} = P{Y = 1|X = -‐1} X P[X = -‐1] = (p)(1/4) = 1/4p

P((X,Y) = (1,-‐1)) = P{X = 1 & Y = -‐1} = P{Y = -‐1|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-‐p-‐pe)(3/4) = 3/4(1-‐p-‐pe)

c)

P[X ≠ Y] = 1 -‐ P[X = Y] = 1 – [P(1,1) + P(-‐1,-‐1)] = 1 – [3/4(1-‐p-‐pe) + 1/4(1-‐p-‐pe)] = 1-‐(1-‐p-‐pe) = p+pe

P[Y = 0] = P((X,Y) = (-‐1,0)) + P((X,Y) = (1,0)) = P{X = -‐1 & Y = 0} + P{X = 1 & Y = 0}

= P{Y = 0|X = -‐1} X P[X = -‐1] + P{Y = 0|X = 1} X P[X = 1] = (pe)(1/4) + (pe)(3/4) = pe


5.8 a)

d)

g)

h)


5.9

a) and b) We show the joint pmf of 𝑋,𝑌 below by identifying the probability mass at the values 𝑥! , 𝑦! 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) = ¼ + ¼ = 1/2

pX(2) = P[X = 2] = pXY(2,0) + pXY(2,1) + pXY(2,2) = 1/8 + ¼ + 1/16 = 7/16

pY(0) = P[Y = 0] = pXY(0,0) + pXY(1,0) + pXY(2,0) = 1/16 + ¼ + 1/8 = 7/16

pY(1) = P[Y = 1] = pXY(1,1) + pXY(2,1) = ¼ + ¼ = 1/2

pY(2) = P[Y = 2] = pXY(2,2) = 1/16

c)

𝑃 𝑋,𝑌 = (0,0) = 𝑃 00 = 𝑃{𝑇𝑇𝑇𝑇} =14

!×

12

!=

164

𝑃 𝑋,𝑌 = (1,0) = 𝑃 10, 01 = 𝑃 𝐻𝑇𝑇𝑇,𝑇𝐻𝑇𝑇,𝑇𝑇𝑇𝐻,𝑇𝑇𝐻𝑇

=14×34×

12

! ×2 +

14

!×

12

!×2 =

18

𝑃 𝑋,𝑌 = (2,0) = 𝑃 02, 20 = 𝑃 𝑇𝑇𝐻𝐻,𝐻𝐻𝑇𝑇 =14

!×

12

!+

34

!×

12

!=1064

=532



𝑃 𝑋,𝑌 = (2,2) = 𝑃 22 = 𝑃 𝐻𝐻𝐻𝐻 =964


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:

𝐹!,! 𝑥!, 𝑦! = 𝑃{𝑋 ≤ 𝑥!,𝑌 ≤ 𝑦!}

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-‐decreasing function of X and Y. a. In both the X and Y directions, the cdf is non-‐decreasing.

2. FX,Y(x1,-‐∞) = 0, FX,Y(-‐∞,y1) = 0, FX,Y(∞,∞) = 1. a. FX,Y(x1,-‐∞) = 0 for all columns, FX,Y(-‐∞,y1) = 0 for all rows, and FX,Y(∞,∞) = 1

3. The marginal cdf’s are the probabilities of the half planes defined by: FX(x1) = P[X ≤ x1, Y < ∞] and FY(y1) = P[X < ∞, Y ≤ y1].

a. lim!→! 𝐹X,Y(x1,n) = P[X ≤ x1] = FX(x1) and b. lim!→! 𝐹X,Y(n,y1) = P[Y ≤ y1] = FY(y1).

4. The joint cdf is continuous from both the “north” and from the “east”. a. This joint cdf is continuous from both the “north” and the “east”.

5. The probability of the rectangle {x1 < x ≤ x2, y1 < y ≤ y2} is given by:


P[x1 < x ≤ x2, y1 < y ≤ y2] = FX,Y(x2,y2) -‐ FX,Y(x2,y1) -‐ FX,Y(x1,y2) -‐ FX,Y(x1,y1)

b) Find the marginal cdf of X and of Y. FXY(x,y) = P{X ≤ x, Y ≤ y}

The marginal cdf of X is: FX(x) = FXY(x,∞) = 0 for X < 0 1/16 for 0 ≤ X < 1 9/16 for 1 ≤ X < 2 16/16 for X ≥ 2

The marginal cdf of Y is: FY(y) = FXY(∞,y) = 0 for Y < 0 7/16 for 0 ≤ Y < 1 15/16 for 1 ≤ Y < 2 16/16 for Y ≥ 2

5.26 a) 1 = 𝑘 𝑥 + 𝑦 𝑑𝑥𝑑𝑦 = 𝑘 !!

!+ 𝑥𝑦

!

!𝑑𝑦 =!

!!!

!! 𝑘 !

!+ 𝑦 𝑑𝑦 = 𝑘 !

!𝑦 + !!

! !

!= 𝑘!

!

So, 𝑘 = 1 b)

𝑭𝑿𝒀(𝒙,𝒚) = 𝒇𝑿𝒀 𝒛𝟏, 𝒛𝟐 𝒅𝒛𝟏𝒅𝒛𝟐𝒙

!∞

𝒚

!∞

For 0 < x < 1 and y > 1:

FXY(x,y) = !! !! (x,y) dxdy = !! !! (x + y) dxdy = [ !! !

! x2 + yx]10dy = !!+ 𝑦 !

! dy

= [!!y + !

!y2]x0 = [

!!(x) + !

!(x)2] -‐ [!

!(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 = !!+ 𝑥 !

! dx

= [!!x + !

!x2]y0 = [

!!(y) + !

!(y)2] -‐ [!

!(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] -‐ [!

!(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))

-‐ (!!(0)2+z2(0))dz2

= !!!!y2 + z2y dz2 = [

!!y2z2 +

!!z22y]x0 = [(

!!y2(x) + !

!(x)2y) – (!

!y2(0) + !

!(0)2y)] = !

!y2x + !

!x2y


c) 𝐹! 𝑥 = lim!→! 𝐹!" 𝑥, 𝑦 = 𝐹!" 𝑥, 1 𝑓𝑜𝑟 0 ≤ 𝑥 ≤ 1 ⇒ 𝑓! 𝑥 = !

!"𝐹! 𝑥 = 𝑥 + !

!, 𝑓𝑜𝑟 0 ≤ 𝑥 ≤ 1

Similarly 𝑓! 𝑦 = 𝑦 + !

!, 𝑓𝑜𝑟 0 ≤ 𝑦 ≤ 1

d) 𝑃 𝑋 < 𝑌 = 𝑥 + 𝑦 𝑑𝑥𝑑𝑦 = !!

!+ 𝑥𝑦

!

!𝑑𝑦 = !!!

!𝑑𝑦!

!!!

!!

!! = !!

! !

!= !

!

𝑃 𝑌 < 𝑋! = 𝑥 + 𝑦 𝑑𝑦𝑑𝑥 = !!

!+ 𝑥𝑦

!

!!

𝑑𝑥 = !!

!+ 𝑥! 𝑑𝑥!

!!!

!!

!!! = !!

!"+ !!

! !

!= !

!"

5.45 𝑓 𝑥, 𝑦 = 𝑥 + 𝑦 , 0 < 𝑥, 𝑦 < 1

𝑓 𝑥 = 𝑥 +12, 0 < 𝑥 < 1

𝑓 𝑦 = 𝑦 +12, 0 < 𝑦 < 1

𝑓 𝑥, 𝑦 ≠ 𝑓 𝑥 𝑓(𝑦) 0 < 𝑥, 𝑦 < 1 Thus, X and Y are not independent.

5.48 b) 𝑃 4𝑋 < 1,𝑌 < 0 = 𝑃 𝑋 < !

!𝑃 𝑌 < 0 = !

!× !!× !!= !

!"

d) 𝑃 𝑚𝑎𝑥 𝑋,𝑌 < !!= 𝑃 𝑋 < !

! 𝑎𝑛𝑑 𝑌 < !

!= 𝑃 𝑋 < !

!𝑃 𝑌 < !

!= !

!× !!×(!

!× !!) = !

!

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