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1.
Answer Key for Homework 5 (Due date: February 28, 2003)
Notice that you are working alone for this homework. If I were you, I would start doing it ASAP
even though you have lots of time. It will save you from last minute frustrations.
Make sure to write your name and section number on each page.
Staple everything together. Place it on the table with your section before class starts. There is a
penalty of 5 pt. if you do not staple them together.
Each question worth different with the total 100 pt. for the homework.
Exercise 5.1
(a) P(X=1,Y=1)=0.20
(b) P(X1 and Y1)=p(0,0)+p(0,1)+P(1,0)+p(1,1)=0.10+0.04+0.08+0.20=0.42
(c) P(X0 and Y0)=P(at least one hose in use at each pump)=p(1,1)+p(1,2)+p(2,1)+p(2,2)=0.70
(d) Marginal pmf of X is
x
0
1
2
otherwise
p(x)
0.16
0.34
0.50
0
Marginal pmf of Y is
y
0
1
p(y)
0.24
0.38
2
0.38
otherwise
0
(e) P(0,0)=0.10 P(X=0)P(Y=0)= 0.16(0.20). They are not independent.
(f) compute the marginal pmf of X-2Y
Let U=X-2Y. If X=0,1,2 and Y=0,1,2 then U=-4,-3,-2,-1,0,1,2
Marginal pmf of U is
u
-4
-3
-2
-1
0
1
p(u)
0.02
0.06
0.34
0.20
0.24
0.08
2
0.06
otherwise
0
U=-4 when X=0 and Y=2 only then P(U=-4)=P(X=0,Y=2)=0.02
U=-2 when( X=0 and Y=1) or (X=2 and Y=2) then
P(U=-2)=P(X=0,Y=1)+ P(X=2,Y=2)=0.04+0.30=0.34
(g) compute the marginal pmf of max(X,Y).
Let V=max(X,Y). If X=0,1,2 and Y=0,1,2 then V=0,1,2
Marginal pmf of V is
u
0
1
2
otherwise
p(u)
0.10
0.32
0.58
0
V=1 when ( X=1 and Y=1) or (X=0 and Y=1) or (X=1 and Y=0) then
P(V=1)=P(X=1,Y=1)+ P(X=0,Y=1)+P(X=1,Y=0)=0.20+0.04+0.08=0.32
(h) compute the Cov(X,Y)
Cov(X,Y)=E(XY)-E(X)E(Y)=[0+..+0+1(1)(0.2)+1(2)(0.06)+2(1)(0.14)+2(2)0.3] -(1.34)(1.14)
= 1.8-(1.34)(1.14)=0.2724
Where E(X)=0(0.16)+1(0.34)+2(0.50)=1.34 and E(Y)=0(0.24)+1(0.38)+2(0.38)=1.14
(i) compute E(X-2Y) and Var(X-2Y)
If you know E(X) and E(Y), E(X-2Y)= E(X)-2E(Y)=1.34-2(1.14)= -0.94
2
or E(U)=
 u p(u ) =-4(0.02)+….+2(0.06)=-0.94
u  2
If you know Var(X),Var(Y) and Cov(X,Y) then,
Var(X-2Y)= Var(X)+4Var(Y)-4Cov(X,Y)=[2.34-1.342]+4[1.9-1.142]-4(0.2724)=1.8564
2
or E(U2)=
u
2
 p(u ) =(-4)2(0.02)+….+22(0.06)=2.74 then Var(U)=2.74-(-0.94)2=1.8564
u  2
(j) compute the conditional pmf of Y given X=0
P(Y=y | X=0)=P(0,y)/0.16
= 0.10/0.16= 0.625
= 0.04/0.16= 0.25
= 0.02/0.16= 0.125
=0
when y=0
when y=1
when y=2
otherwise
2.
Exercise 5.2
(a) If X and Y are independent, the joint probabilities can be computed by P(x,y)=P(x)P(y) for all
x and y.
x 0
1
2
y
0
0.30
0.18
0.12
1
0.05
0.03
0.02
2
0.025 0.015 0.01
3
0.025 0.015 0.01
4
0.10
0.06
0.04
(b) P(X1,Y1)=p(0,0)+P(0,1)+p(1,0)+p(1,1)=0.56
(c) P(X+Y=0)=P(X=0,Y=0)=0.30
(d) P(X+Y1)= p(0,0)+P(0,1)+p(1,0)=0.53
3.
Exercise 5.9 .
(a)
 x3
2
K
(
x

y
)
dxdy

K
20 20
20  3  xy

380000 K

1
3
30 30
30
2
2
30
30
3

dy  K  19000  10 y 2 dy  K  19000 y  10 y 

 3

3
3  20


20 
20 
30
then K=3/380000
(b)
 x3
P( X  26, Y  26)    K ( x  y )dxdy   K   xy 2
 3
20 20
20 
26 26
26
2
2
26

dy  K  9576  6 y 2 dy


3

20 
20 
26
26
 9576
y3 
114912 K 114912
 K 
y  6  

 0.3024
3  20
3
380000
 3
30 x  2
(c) P( X
 Y  2)  1  
28 30
2
2
 K ( x  y )dydx  
22 20
 K (x
20 x  2
2
 y 2 )dydx =0.3593
30
 2
y3 
19000 
30

  K 10 x 2 
(d) f(x)=  K ( x  y )dy  K  x y 
x 2  0.05

3  20
3  380000


20
30
2
2
where 20x30
30
 2
x3 
19000 
30

  K 10 y 2 
(e) f(y)=  K ( x  y )dx  K  y x 
y 2  0.05

3  20
3  380000


20
30
2
2
where 20x30 then f(x,y)f(x)f(y) . X and Y are not independent.
(f) First compute Cov(X,Y)=E(XY)-E(X)E(Y)=641.4474-25.329(25.329)=-0.1082
30
 30 x 4
x2 
 30

2

where E(X)=  xf ( x)dx   x
x  0.05 dx  
 0.05 
380000
380000
4
2  20



20
30
= 25.329
30
 30 x 5
x3 
 30

E ( X )   x f ( x)dx   x 
x 2  0.05 dx  
 0.05 
3  20
 380000

 380000 5
20
30
2
2
2
= 649.825
30
 30 y 4
y2 
 30

and E(Y)=  yf ( y )dy   y
y 2  0.05 dy  
 0.05 
380000
2  20

 380000 4
20 
30
= 25.329
30
 30 y 5
y3 
 30

2
E (Y )   y f ( y )dy   y 
y  0.05 dy  
 0.05 
3  20
 380000

 380000 5
20
30
2
2
2
= 649.825
 x4
x2 3
E(XY)=   xy( K ( x  y )) dxdy   K 
y
y
 4
2
20 20
20 
30 30
30
2
2

dy

20 
30
30

y2
y4 

 250  =81250000K=641.4474
  K 162500 y  250 y dy  K 162500
2
4  20

20
30

3
then Corr(X,Y)=

Cov( x, y )
Var ( X ) Var (Y )

 0.1082
649.825  25.329 2 649.825  25.329 2
= -0.0131
(g) compute the conditional pdf of X given Y=y
f(x|y)=
f ( x, y )

f ( y)
K (x2  y 2 )
3( x 2  y 2 )

where 20y30
19000  (30 y 2  19000)

2
K 10 y 

3 

4.
Exercise 5.41 (a)
_
_
x
possible pairs
(1,1)
(1,2),(2,1)
(1,3),(3,1),(2,2)
(1,4),(4,1),(2,3),(3,2)
(3,3),(2,4),(4,2)
(3,4),(4,3)
(4,4)
1
1.5
2
2.5
3
3.5
4
P( x )
0.4(0.4)=0.16
2(0.4)(0.3)=0.24
2(0.4)(0.2)+0.3(0.3)=0.25
2(0.4)(0.1)+2(0.3)(0.2)=0.20
0.2(0.2)+2(0.3)(0.1)=0.10
2(0.2)(0.1)=0.04
0.1(0.1)=0.01
_
(b) P( x 2.5)=0.16+0.24+0.25+0.20=0.85
(c)
R
possible pairs
0
(1,1),(2,2),(3,3),(4,4)
1
(1,2),(2,1),(2,3),(3,2),(3,4),(4,3)
2
(1,3),(3,1),(2,4),(4,2)
3
(1,4),(4,1)
5.
P(R)
0.4(0.4)+0.3(0.3)+0.2(0.2)+0.1(0.1)=0.30
2(0.4)(0.3)+2(0.3)(0.2)+2(0.2)(0.1)=0.40
2(0.4)(0.2)+2(0.3)(0.1)=0.22
2(0.4)(0.1)=0.08
Exercise 5.59 (a)(c)
(a) E(X1+X2+X3)=60+60+60=180

and
Var(X1+X2+X3)=15+15+15=45
then
200  180 
 =P(Z2.98)=0.9986
45 

 150  180
200  180 
Z
 =P(-4.47Z2.98)
P(150X1+X2+X3200)= P
45
45 

P(X1+X2+X3200)= P Z

=0.9986-0=0.9986
(c) E(X1-0.5X2-0.5X3)=60-0.5(60)-0.5(60)=0 and Var(X1+X2+X3)=15+(0.5)215+(0.5)215=22.5
  10  0
then P(-10 X1-0.5X2-0.5X35)= P

=0.8531-0.0174=0.8357
22.5
Z
50 
 =P(-2.11Z1.05)
22.5 