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K. Two Random Variables.
1. Regression (Summary).
2. Covariance (  xy and
s xy )
3. The Correlation Coefficient (  xy and rxy )
4. Functions of Two Random Variables.
5. Sums of Random Variables, Independence.
In the following problems (I) check for independence, (ii) Compute
Covx, y  , Corr x, y  and, (iii) Compute Ex  y  and


Var x  y  : D&C pg. 221 3, 4, 7, 14. In problem 3, find the following: Px  y  4 , P x  y  4 x  0
Downing and Clark (formerly pg. 348 now posted at end of 251hwkadd) Old Computational Problem 1: For the sample data below b)
Compute Cov x, y and Corr x, y . c) Compute the mean of x  y and Var x  y .








x 34 26 9 30 47 10 34 34 45 10 47 32 47 8 45
y 6 57 89 60 95 42 31 28 90 25 45 23 52 95 48
Text 5.8 (Compute
Var x  y  ), 5.9, 5.11, 5.16, 5.17 [5.8, 5.9, 5.10, 5.13] (5.8, 5.10), K1-K4 .
The Downing and Clark problems are in this document.
----------------------------------------------------------------------------------------------------------------------------Downing and Clark, pg. 219, Computational Problem 3: Using the joint probability table below, (a)
check for independence, (b) Compute Covx, y  , Corr x, y  , (c) Compute Ex  y  and Var x  y  . (d)

y

Find Px  y  4 and P x  y  4 x  0 .
0
2
4
6
 2  .1
0 .1 .2


x
4  0 .1 0 .1
5  .1
0 .1 .2


Solution: a) Check for independence: Look at the upper left hand probability below. Its value is .1 and it
represents Px  2   y  0 . If x and y were independent, we would have Px  2   y  0
 Px  2  P y  0  .4.2  .08 . Since this is not true, x and y are not independent.
b) Compute Covx, y  and Corr x, y  .
y
2
x
4
5
P y 
yP y 
y P y 
2
0 2
 .1 0

 0 .1
 .1 0

.2 .1
0
.1
.2
6
.2

.1
.2

.5
0 0.2 0.8 3.0
0
.4 3.2
18
Px  xPx  x 2 Px 
.4  0.8
1.6
.2
0.8
3.2
.4
2.0
10 .0
1.0
2.0
14 .8
4.0
21 .6
 Px  1 ,   Ex   xPx  2.0 , E x    x
 E  y    yP y   4.0 and E y    y P y   21 .6
To summarize -
y
4
.1
2
x
2
2
2
Px   14 .8 ,
 P y   1 ,
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E xy  

.1 20  0 22  .1 24  .2 26
xyPxy     040  .142
 044
 .146
  .150
 052
 .154
 .256
 0  0.8  2.4
 0

  0  0.8
 0  2.4  8.0
0
 2  6.0
 0
 
 xy  Covxy  Exy   x  y  8  24  0 ,  x2  E x 2   x2  14.8  2.02  10 .8 and
 
 y2  E y 2   y2  21.6  4.02  5.6 . So that  xy 
 xy
0

 0.
 x y
10 .8 5.6
c) Compute Ex  y  and Var x  y  .
Ex  y   Ex  E y    x   y  2.0  4.0  6.0
and Var x  y    x2   y2  2 xy  Var x   Var  y   2Covx, y   10 .8  5.6  20  16 .4


d) Find Px  y  4 , P x  y  4 x  0
y
6 . If we compute the sum of x  y and
.2

.1 0 .1
0 .1 .2

y
place it in our joint probability table in place of the probabilities, we get
0
2
4
6 .
2  2
0 2 4


x
4  4
6 8 10 
5  5
7 9 11


The first five of these, -2, 0, 2, 4, and 4 are less than or equal to 4. If we read their probabilities off the joint
probability table, we get Px  y  4  .1  0  .1  .2  0  .4 .
P A  B 
From the multiplication Rule we know P A B  
, so
P B 
Px  y  4  x  0
. The sums for which x is zero or lower are those in the top
P x y  4x 0 
P x  0 
row, and we have already found that Px  0  Px  2  .4 . But the only sums for which both
Remember our original probabilities:

0

2
.1

x
4  0
5  .1

2
0
4
.1

x  y  4 and x  0 are exactly the same sums in the top row, so that Px  y  4  x  0  .4 too.


Thus P x  y  4 x  0 
Px  y  4  x  0 .4
  1.
P x  0 
.4
251solnK1 3/28/06 (Open this document in 'Page Layout' view!)
Downing and Clark, pg. 219, Computational Problem 4: Using the joint probability table below, (a)
check for independence, (b) Compute Covx, y  , Corr x, y  , (c) Compute Ex  y  and Var x  y  .
y
2
1
x
0
1
2
3
.15

.05
0

.15
0

5
7
.05 0 

0 .20 
.10 0 

0 .20 
.10 0 
Solution: a) Check for independence: The big giveaway in any of these is the presence of a zero. Look at
the lower right hand corner. The probability there is 0  Px  2   y  7 . If x and y were
independent, we would have to have either Px  2  0 or P y  3  0 because Px  2   y  7 
would equal Px  2  P y  7 . Since this is not true, x and y are not independent.
b) Compute Covx, y  and Corr x, y  .
y
x
2
1
0
1
2
P y 
yP y 
y 2 P y 
3
5
.15 .05

.05 0
 0 .10

.15 0
 0 .10

.35 .25
1.05 1.25
3.15 6.25
xPx 
 0.40
 .025
0
0.35
0.20
 0.10
x 2 Px 
0.80
0.25
0
0.35
0.40
1.80
 xPx  0.10 , E x    x Px  1.80 ,  P y   1 ,
 E  y    yP y   5.9 and E y    y P y   41 .8 .
To summarize -
y
7
Px 

0
.20

.20 
.25

0
.10

.20 
.35

0 
.10
.40 1.00
3.60
5.9
32 .4 41 .8
 Px   1 , 
x
 E x  
2
E xy  

2
2
2
 0
 .15 23  .05 25  0 29   0.9  0.5
 .05 13


 0 15  .20  19   0.15
 0  1.8

xyPxy     003
 .10 05
 009    0
0
 0  0.1

 

 015
 .20 19  0.45
 0  1.8
  .1513
  023
 .10 25
 029   0  1.0
 0
 
 xy  Covxy  Exy  x  y  0.1   0.105.90  0.49 ,  x2  E x 2   x2  1.8  0.10 2  1.79 and
 
 y2  E y 2   y2  41.8  5.92  6.99 . So that  xy 
c) Compute Ex  y  and Var x  y  .
 xy
0.49

 .1385 .
 x y
1.79 6.99
Ex  y   Ex  E y   x   y  0.10  5.90  5.80 and
Var x  y    x2   y2  2 xy  Var x   Var  y   2Covx, y   1.79  6.99  20.49   9.76
251solnK1 3/28/06 (Open this document in 'Page Layout' view!)
Downing and Clark, pg. 219, Computational Problem 7: Using the joint probability table below, (a)
check for independence, (b) Compute Covx, y  , Corr x, y  , (c) Compute Ex  y  and Var x  y  .
y






1
2
x
3
4
1
.2
0
0
0
2
.1
0
.1
0
3
.1
0
.1
.1
4
0

0
.2 

.1
Solution: a) Check for independence: The row of zeros for x  2 is supposed to fool you - it makes it look
like x and y are independent, but they are not. One thing to look for is proportional rows. The row (.2, .1,
.1, 0) is not proportional to the row (0, .1, .1, 2). If x and y were independent, we would have joint
y
1 2 3 4
Px 
1  .08 .08 .12 .12 
.4
probabilities like
2  0
0
0
0
0 . (Note that the first and third rows are identical and
x
3  .08 .08 .12 .12 
.4


4  .04 .04 .06 .06 
.2
.2 .2 .3 .3
1.0
the first and last rows are proportional) Since the rows are not proportional, x and y are not independent.
b) Compute Covx, y  and Corr x, y  .
y
1
2
x
3
4
P y 
yP y 
y P y 
2
1
 .2

 0
 0

 0
.2
0
.1
0
.2
3
.1
4
0

0
0
.1 .2

.1 .1
.3 .3
Px  xPx  x 2 Px 
.4
0.4
0.4
0
0
0
.4
1.2
3.6
.2
1.0
0.8
2.4
3.2
7.2
0.2 0.4 0.9 1.2
2.7
0.2 0.8 2.7 4.8
8.5
 xPx  2.4 , E x    x
 E  y    yP y   2.7 and E y    y P y   8.5
To summarize -
y
2
.1
E xy  
 Px   1 , 
x
 E x  
2

 .211
 021
xyPxy   
  031

 041
 .3
 0
 .2  .2
 0
0
0
 0

 7.4
 0  0.6  0.9  2.4


 0  1.2  1.6
 0
2
2
 .112   .113  014 
 02 2   02 3  02 4 
 .132   .133  .234 

 04 2   .14 3  .14 4 
2
Px   7.2 ,
 P y   1 ,
251solnK1 3/28/06 (Open this document in 'Page Layout' view!)
 
 xy  Covxy  Exy   x  y  7.40  2.42.7  0.92 ,  x2  E x 2   x2  7.20  2.42  1.44 and
 
 y2  E y 2   y2  8.5  2.72  1.21 . So that  xy 
c) Compute Ex  y  and Var x  y  .
 xy
 x y
0.92

 .70 .
1.44 1.21
Ex  y   Ex  E y    x   y  2.4  2.7  5.1 and
Var x  y    x2   y2  2 xy  Var x   Var  y   2Covx, y   1.44  1.31  20.92   4.49
Downing and Clark, pg. 219, Computational Problem 13: Using the joint probability table below, (a)
check for independence, (b) Compute Covx, y  , Corr x, y  , (c) Compute Ex  y  and Var x  y  .
y
1
x
1
2
3
4
5
6
1
 36
1
 36
1
 36
1
 36
1
 36
1
 36
2
1
36
1
36
1
36
1
36
1
36
1
36
3
4
5
6
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1 
36 
1
36 
1 
36 
1 
36
1 
36 
1 
36 
Solution: a) Check for independence: Of course we have independence here. For example, we have
1
.
Px  3   y  5  Px  3 P y  5  16 16  36
  
y
1
1
2
x
3
4
5
6
P y 
yP y 
1
 36
1
 36
1
 36
1
 36
1
 36
1
 36
y 2 P y 
1
6
1
6
1
6
2
1
36
1
36
1
36
1
36
1
36
1
36
1
6
2
6
4
6
3
4
5
1
36
1
36
1
36
1
36
1
36
1
36
1
6
3
6
9
6
1
36
1
36
1
36
1
36
1
36
1
36
1
6
4
6
16
6
1
36
1
36
1
36
1
36
1
36
1
36
1
6
5
6
25
6
Px  xPx  x 2 Px 
6
1 
36 
1
36 
1 
36 
1 
36
1 
36 
1 
36 
1
6
6
6
36
6
1
6
1
6
1
6
1
6
1
6
1
6
1
1
6
2
6
3
6
4
6
5
6
6
6
21
6
1
6
4
6
9
6
16
6
25
6
36
6
91
6
21
6
91
6
 Px  1 ,   Ex   xPx   3.5 , E x    x Px   15.167 ,
 P y   1 ,   E y    yP y    3.5 and E y    y P y    15.167 .
To summarize -
y
2
21
6
x
21
6
2
2
2
121
6
91
6
251solnK1 3/28/06 (Open this document in 'Page Layout' view!)
E xy  


 1  2  3  4  5  6
 2  4  6
 8  10  12 

1   3  6  9  12  15  18 
xyPxy  


36  4  8  12  16  20  24 
  5  10  15  20  25  30 


  6  12  18  24  30  36 
2
1
21  221  321  421  521  621  21  441  12 .25
36
36
36
 
 xy  Covxy  Exy   x  y  12.25  3.53.5  0 ,  x2  E x 2   x2  15.167  3.52  2.917 and
 
 y2  E y 2   y2  15.167  3.52  2.917 . So that  xy 
 xy

0.
 0 . We didn't have
7.917 7.917
to go to all this work to show this. It was explained in class that if x and y are independent  xy  0 and
 x y
 xy  0 .
c) Compute Ex  y  and Var x  y  .
Ex  y   Ex  E y    x   y  3.5  3.5  7.0 and
Var x  y    x2   y2  2 xy  Varx   Var y   2Covx, y   2.917  2.917  20  5.833
251solnK1 3/28/06 (Open this document in 'Page Layout' view!)
Downing and Clark (formerly pg. 348 now posted at end of 251hwkadd) Old Computational Problem
1: This is obviously sample data, so we compute only a sample covariance and correlation.
b) Compute Covx, y  and Corr x, y  .
c) Compute the mean of x  y  and Var x  y  .
x 34 26 9 30 47 10 34 34 45 10 47 37 47 8 45
y 6 57 89 60 95 42 31 28 90 25 45 23 52 95 48
Solution: This is sample data, so we compute a sample covariance and correlation.
b) Compute Covx, y  and Corr x, y  . Only the x and y columns are given. If we use computational
formulas, we must start with the table below.
xy
y2
x2
y
x
obs
1
34
6
1156
36
204
2
3
4
26
9
30
57
89
60
676
81
900
3249
7921
3600
1482
801
1800
5
47
10
95
2209
100
9025
1764
4465
420
6
42
961
784
1054
952
90
1156
1156
2025
8100
4080
10
47
25
45
100
2209
625
2025
250
2115
12
37
23
1369
529
736
13
47
52
2209
2704
2444
14
15
8
45
95
48
64
2025
9025
2304
760
2160
Total 463 786 17435
52652
23808
9
34
34
45
10
11
7
8
31
28
 x  463 ,  x
x
s x2
 x  463  30.867
n
x

15
2
 nx 2
n 1
 224 .5524
y
s 2y 

17435  1530 .867 2
14
 y  786  52.400
n
y
15
2
 ny 2
n 1
 818 .9714

52652  1552 .400 2
14
 y  786 ,  y  52652 and  xy  23808
 xy  nx y  23808  1530.867 52.400 
The formula for the sample covariance is s 
2
 17435 ,
2
xy

453 .2
 32 .3714
14
.
So rxy 
sxy
sx s y

n 1
 32 .3714
224 .5514 818 .9714
14
 0.07548 .
c) Compute the mean of x  y  and Var x  y  .
x  y  is x  y   x  y  30.867  52.400  83.267
The sample variance of x  y  , s x2 y is calculated by the familiar formula, Var x  y  
 Var x   Var  y   2Covx, y  , which, for a sample, becomes s x2 y  s x2  s 2y  2s xy
 224 .5524  818 .9714  2 32.3714   973 .7810 .
The sample mean of
251solnK1 3/28/06 (Open this document in 'Page Layout' view!)
Minitab Computation of the Above Problem.
————— 3/28/2006 9:03:25 PM ————————————————————
Welcome to Minitab, press F1 for help.
Results for: 251K1s1.MTW
MTB > WSave "C:\Documents and Settings\rbove\My Documents\Minitab\251K1s1.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\251K1s1.MTW'
MTB > exec '251samcov.txt'
Executing from file: 251samcov.txt
Executing from file: var973.txt
Data Display
svar
3143.73
#Sample variance before dividing by n-1
Descriptive Statistics: C1
Variable
C1
N
15
N*
0
Mean
30.87
SE Mean
3.87
StDev
14.99
Minimum
8.00
Q1
10.00
Median
34.00
Q3
45.00
Maximum
47.00
Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
C1
34
26
9
30
47
10
34
34
45
10
47
37
47
8
45
C2
1156
676
81
900
2209
100
1156
1156
2025
100
2209
1369
2209
64
2025
# x and x squared.
Data Display
sum
sumsq
count
smean
svar
sdev
DF
sterr
463.000
17435.0
15.0000
30.8667
224.552
14.9851
14.0000
3.86913
Executing from file: var973.txt
Data Display
svar
11465.6
# Sample variance of y before division by n-1.
Descriptive Statistics: C1
Variable
C1
N
15
N*
0
Mean
52.40
SE Mean
7.39
StDev
28.62
Minimum
6.00
Q1
28.00
Median
48.00
Q3
89.00
Maximum
95.00
251solnK1 3/28/06 (Open this document in 'Page Layout' view!)
Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
C1
6
57
89
60
95
42
31
28
90
25
45
23
52
95
48
C2
36
3249
7921
3600
9025
1764
961
784
8100
625
2025
529
2704
9025
2304
# y and y squared.
Data Display
sum
sumsq
count
smean
svar
sdev
DF
sterr
786.000
52652.0
15.0000
52.4000
818.971
28.6177
14.0000
7.38905
Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
x
34
26
9
30
47
10
34
34
45
10
47
37
47
8
45
xsq
1156
676
81
900
2209
100
1156
1156
2025
100
2209
1369
2209
64
2025
y
6
57
89
60
95
42
31
28
90
25
45
23
52
95
48
ysq
36
3249
7921
3600
9025
1764
961
784
8100
625
2025
529
2704
9025
2304
xy
204
1482
801
1800
4465
420
1054
952
4050
250
2115
851
2444
760
2160
# The worksheet.
Data Display
sumx
sumx2
sumy
sumy2
sumxy
n
xbar
ybar
svarx
svary
scovxy
sx
sy
rxy
rxy2
463.000
17435.0
786.000
52652.0
23808.0
15.0000
30.8667
52.4000
224.552
818.971
-32.3714
14.9851
28.6177
-0.0754864
0.00569820
#
#
#
#
#
Sum
Sum
Sum
Sum
Sum
of
of
of
of
of
#
#
#
#
#
#
#
#
#
Sample
Sample
Sample
Sample
Sample
Sample
Sample
Sample
Sample
x
x squared
y
y squared
xy
mean of x
mean of y
variance of x
variance of y
covariance of x and y
standard deviation of x
standard deviation of y
correlation of x and y
correlation of x and y squared.