MATHEMATICS 150.1Solutions for the 3rd Exam

1. The joint p.d.f. of X and Y is given by

fX,Y (x, y) = 3(x+ y)I(0,1)(x+ y)I(0,1)(x)I(0,1)(y)

(a) P [X + Y < 0.5]

(b) cov[X,Y ]

2. Suppose for the random variables X, Y , and Z, we are given that Var[X] = Var[Y ] = 1, Var[Z] = 2,cov[X,Y ]= −1, cov[Y,Z]= 1, and X and Z are independent. Find the covariance between X+2Y andY − 2Z.

3. Let X and Y follow the bivariate normal distribution with µX = 5, µY = 10, σX = 1, and σY = 5. Ifρ > 0, find ρ such that P [4 < Y < 16|X = 5] = 0.954.

4. Six cards are to be drawn from an ordinary deck of cards. Let X denote the number of low (1-5) redsuited cards drawn, Y be the number of high (6-10) black suited cards drawn, and Z be the numberof face cards drawn.

(a) Find the joint density function of X, Y , and Z.

(b) Determine P [X < Z|Y = 3] .

5. If the density function of X is given by fX(x) =x

3for x = 1, 2, and fY |X(y|x) is binomial with

parameters x and 0.5, find

(a) the joint density of X and Y ,

(b) cov[X,Y ]

6. Suppose that the joint moment generating function of X and Y is given by

mX,Y (t1, t2) =1

3(1− t2)2+

4et1

3(2− t2), for t2 < 1.

(a) Find Var[X] and Var[Y ]

(b) Find ρx,y

7. Let X1 and X2 be independent and both exponential distributed with parameter λ. If Y1 = X1 +X2

and Y2 =X1

X2.

(a) Find the joint p.d.f. of Y1 and Y2.

(b) Are Y1 and Y2 independent?

(c) Find P [Y2 ≥ 1|Y1 ≤ 2]

1. fX,Y (x, y) = 3(x+ y)I(0,1)(x+ y)I(0,1)(x)I(0,1)(y)

(a)

⇒ P

[X + Y <

1

2

]= P

[X <

1

2− Y

]=

∫ 12

0

∫ 12−y

0

3(x+ y)dxdy

=

∫ 12

0

3

(x2

2+ xy

)x= 12−y

x=0

dy

=

∫ 12

0

3

((12 − y

)22

+

(1

2− y)y

)dy =

∫ 12

0

3

(1

8− y

2+y2

2+y

2− y2

)dy

=

∫ 12

0

3

(1

8− y2

2

)dy = 3

(y

8− y3

6

) 12

0

= 3

(1

16− 1

48

)=

1

8

(b)

⇒ fX(x) =

∫ 1−x

0

3(x+ y)dy = 3

(xy +

y2

2

)y=1−x

y=0

= 3

(1

2− x2

2

)I(0,1)(x)

Similarly, ⇒ fY (y) =

∫ 1−y

0

3(x+ y)dx = 3

(1

2− y2

2

)I(0,1)(y)

⇒ µX =

∫ 1

0

x · 3(

1

2− x2

2

)dx =

3

2

(x2

2− x4

4

)1

0

=3

8= µY (Since X and Y are identically distributed)

⇒ E[XY ] =

∫ 1

0

∫ 1−y

0

xy · 3 (x+ y) dxdy =

∫ 1

0

∫ 1−y

0

(3x2y + 3xy2

)dxdy

=

∫ 1

0

3

[(1− y)3y

3+

(1− y)2y2

2

]dy =

∫ 1

0

(y − 3

2y2 +

1

2y4)dy

=

(y2

2− y3

2+y5

10

)1

0

=1

10

∴ cov[X,Y ] = E[XY ]− µXµY =1

10−(

3

8

)2

= − 13

320

2.

cov[X + 2Y, Y − 2Z] = E[(X + 2Y )(Y − 2Z)]− E[X + 2Y ]e[Y − 2Z]

= E[XY − 2XZ + 2Y 2 − 4Y Z]− (E[X] + 2E[Y ]) (E[Y ]− 2E[Z])

= E[XY ]− 2E[XZ] + 2E[Y 2]− 4E[Y Z]− µXµY + 2µXµZ − 2µY µY + 4µY µZ

= cov[X,Y ]− 2cov[X,Z] + 2V ar[Y ]− 4cov[Y, Z] = −1− 2(0) + 2(1)− 4(1) = −3

3.

fY |X(y|x) =fX , Y (x, y)

fX(x)=

1√

2πσY√

1− ρ2exp

{−1

2σ2Y (1− ρ2)

[y − µY −

ρσYσX

(x− µX)

]2}

=1

5√

2π√

1− ρ2exp

{−1

50 (1− ρ2)[y − 10]

2

}⇒ Let T = Y |X , then T ∼ N(10, 25(1− ρ2))

⇒ P [4 < Y < 16|X = 5] = P

[4− 10

5√

1− ρ2<

T − 10

5√

1− ρ2<

16− 10

5√

1− ρ2

]= 0.954

⇒ Φ

(6

5√

1− ρ2

)− Φ

(6

5√

1− ρ2

)= 0.954

⇒ 2Φ

(6

5√

1− ρ2

)= 1.954

⇒ Φ

(6

5√

1− ρ2

)= 0.977

⇒ 6

5√

1− ρ2= 2⇒

√1− ρ2 = 0.6

∴ ρ = 0.8

4. (a) fX,Y,Z(x, y, z) =

(10x

)(10y

)(12z

)(20

6− x− y − z

)(

526

)

⇒ fY (y) =

(10y

)(42

6− y

)(

526

)

Thus, fX,Z|Y (x, z|y) =fX,Y,Z(x, y, z)

fY (y)=

(10x

)(12z

)(20

6− x− y − z

)(

426− y

)

Moreover, fX,Z|Y (x, z|3) =fX,Y,Z(x, 3, z)

fY (3)=

(10x

)(12z

)(20

3− x− z

)(

423

)So P [X < Z|Y ] = fX,Z|Y (0, 1|3) + fX,Z|Y (0, 2|3) + fX,Z|Y (0, 3|3) + fX,Z|Y (1, 2|3)

=

(100

)(121

)(202

)+

(100

)(122

)(201

)+

(100

)(123

)(200

)+

(101

)(122

)(200

)(

423

)=

16

41

5. fX(x) =x

3for x = 1, 2, and fY |X(y|x) =

(xy

)(1

2

)x(a)

fX,Y (x, y) = fY |X(y|x · fX(x) =

(xy

)(1

2

)x· x

3

=x!

(x− y)!y!

(1

2

)x (x3

), for x = 1, 2, and y = 0, . . . , x

(b)

⇒ fY (y) =∑x

fX,Y (x, y) =1!

(1− y)!y!

(1

2

)(1

3

)I{0,1}(y) +

2!

(2− y)!y!

(1

2

)2(2

3

)I{0,1,2}(y)

⇒ E[Y ] =∑y

yfY (y) = 1

(1

6+

1

3

)+ 2

(1

6

)=

5

6

⇒ E[X] =∑x

xfX(x) =

(1

3

)+ 2

(2

3

)=

5

3

While E[XY ] =∑x

∑y

xyfX,Y (x, y)

= (1 · 1)1!

1!0!

(1

2

)(1

3

)+ (2 · 1)

2!

1!1!

(1

2

)2(1

3

)+ (2 · 2)

2!

0!2!

(1

2

)2(1

3

)=

3

2

∴ cov[X,Y ] =3

2−(

5

6

)(5

3

)=

1

9

6. mX,Y (t1, t2) =1

3(1− t2)2+

4et1

3(2− t2)

(a) V ar[X] = E[X2]− (E[X])2

E[X] =∂m

∂t1

∣∣∣∣t1=t2=0

=4et1

3(2− t2)

∣∣∣∣t1=t2=0

=4

6

E[X2] =∂2m

∂t21

∣∣∣∣t1=t2=0

=4et1

3(2− t2)

∣∣∣∣t1=t2=0

=4

6

∴ V ar[X] =4

6−(

4

6

)2

=2

9

V ar[Y ] = E[Y 2]− (E[Y ])2

E[Y ] =∂m

∂t2

∣∣∣∣t1=t2=0

=2

3(1− t2)3+

4et1

3(2− t2)2

∣∣∣∣t1=t2=0

=2

3+

1

3= 1

E[Y 2] =∂2m

∂t21

∣∣∣∣t1=t2=0

=2

(1− t2)4+

8et1

3(2− t2)3

∣∣∣∣t1=t2=0

= 2 +1

3=

7

3

∴ V ar[Y ] =7

3− (1)

2=

4

3

(b)

E[XY ] =∂2m

∂t1t2

∣∣∣∣t1=t2=0

=4et1

3(2− t2)2

∣∣∣∣t1=t2=0

=1

3

⇒ cov[X,Y ] =1

3−(

2

3

)(1) = −1

3

∴ ρX,Y =cov[X,Y ]

σXσY=− 1

3√43

√2

9= −1

2

√3

2

7. X1, X2 ∼ exp(λ)⇒ fX1

= λe−λx1 and fX2= λe−λx2

(a) Y1 = X1 +X2 and Y2 =X1

X2

⇒ J(x1, x2) =

∣∣∣∣∣∣1 1(1

x2

) (−x1x2

) ∣∣∣∣∣∣ = −x1x2− 1

x2= − (x1 + x2)

(x2)2

⇒ X1 =Y1Y2

1 + Y2, X2 =

Y11 + Y2

fX1,X2(x1, x2) = λ exp[−λx1] · λ exp[−λx2]

= λ2 exp[−λ(x1 + x2)]

fY1,Y2(y1, y2) = λ2 exp

[−λ(y1 + y1y2

1 + y2

)]· (x2)

2

x1 + x2

= λ2 exp(−λy1) ·

y21(1 + y2)2

y1

=λ2 exp(−λy1)y1

(1 + y2)2

=λ2y1e

−λy1

Γ(2)· 1

(1 + y2)2

(b) ∴ y1 ∼ Γ(2, λ), and fY2(y2) =1

(1 + y2)2.

fY1,Y2(y1, y2) = fY1(y1) · fY2(y2), hence Y1 and Y2 are independent.

(c)

P [Y2 ≥ 1|Y1 ≤ 2] = P [Y2 ≥ 1] ( since Y1 and Y2 are independent )

=

∫ +∞

1

1

(1 + y2)2 dy2 =

−1

1 + y2

∣∣∣∣+∞1

=1

2