251solnJ1 10/16/00 (Open this document in 'Page Layout' view!)
J. Random Variables.
1. Definitions.
4.3.
2. Probability Distribution of a Discrete Random Variable.
4.11-4.14, 4.16.
3. Expected Value (Expectation) of a Discrete Random Variable.
4.22a, 4.25a, 4.27. J1.
4. Variance of a Discrete Random Variable.
4.22, 4.25, 4.26. J2-J4.
5. Summary
J6, J7
Exercise 4.3: a, b and e are discrete. c, d and f are usually considered continuous.
Exercise 4.11: a) By the extended addition rule for mutually exclusive events, we can add together the
probabilities that we get the various values. If 2, 3, 5, 8, and 10 are the only values of x , the total
P2  P3  P5  P8  P10   1 . Since P2  P3  P8  P10   .15  .10  .25  .25  .75 ,
P8  1  .75  .25
b) Px  2  x  10   P2  P10   .15  .25  .40
c) F 8  Px  8  P2  P3  P5  P8  1  P10   1  .25  .75 . Question: What is Px  8 ?
Exercise 4.12:
x Px 
a)
0
.1
1
2
.3
.3
b)
x Px 
 2 .25
1
.50
0
.25
x Px 
4  .3
c)
9
.4
20
.3
d)
x
2
Px 
.15
3
.15
5
.45
To be valid, these must have positive probabilities
3
.2
1.00
.4
6
.35
.9
1.10
and the probabilities must add to 1. Only b) is valid.
Exercise 4.13: If we write down the pattern, the value of x and its probability, we get
pattern x probabilit y
HHH
3
1
HHT
2
1
HTH
2
1
HTT
1
1
THH
2
1
THT
1
1
TTH
1
1
TTT
0
1
8
8
8
8
The probabilities, of course, add to 1, so that, if we add together the combinations
8
8
8
8
that will give us values of 0, 1, 2 or 3, we find P1 
1
8
, P2  3 8 , P3  3 8 and P4 
1
8
.
Exercise 4.16: a) We can consider relative frequencies estimates of probabilities. Of course, they should be
all between zero and one and add to 1.
c) Px  30   P31  P32   P33  .1273  .1455  .1273  .4000 , Px  40   0 ,
Px  30   P20   P21  P22   P23   P24   P25   P26   P27   P28   P29 
 .0182  .0364  .0545  .0727  .0364  .0182  .0727  .0182  .1090  .0727  .5090
d) Px  25   x  26   Px  25   Px  26   .0182  .0727  .0909
251solnJ1 10/16/00
Exercise 4.22:
x
1
2
4
10
From this table

Px 
x P x 
.2
0.2
.4
0.8
.2
0.8
.2
2.0
1.0
3.8
Px   1.00 (valid distribution!), 
x 2 P x 
0.2
1.6
3.2
20.0
25.0
 E x  
x   
 x    P x 
-2.8
-1.8
0.2
6.2
-0.56
-0.72
0.04
1.24
0.00
2
x Px   25 .0 ,
 xPx  3.8 , 
 x   2 P x 
1.568
1.296
0.008
7.688
10.560
 x   Px   0.00 (a check for accuracy!) and  x   2 Px  10.560 . Remember! You do not need
the last two columns if you use the computational formula.
xPx   3.8
a)   E x  

 
b) Computational Formula (the easy way!): Var x    x2  E x 2   2 
 25 .0  3.82  10 .56

x
2
Px    2
  x    Px  10.560 .
Definitional Formula (the hard way): Varx    x2  E x   2 
2
c)   10.56  3.2496
d) It is the average value that would be attained if the experiment described by the distribution were
repeated many, many times.
e) No!!!
f) Depends on the distribution - If E x  is the same as a value that could be assumed by x , it is possible.
Exercise 4.25:
Px 
x P x 
x 2 P x 
P y 
y P y 
y 2 P y 
y
.3
0.0
0.0
0
.1
0.0
0.0
.4
0.4
0.4
1
.8
0.8
0.8
.3
0.6
1.2
2
.1
0.2
0.4
1.0
1.0
1.6
1.0
1.0
1.2
a) What did you guess? Did you notice that these are symmetrical, unimodal distributions around 1?
b) Since x has more probability away from the mean, we would expect it to be more variable.
c) Since these have identical means   Ex   1.0 , variability is shown by the variance.
x
0
1
2
Varx   x2 
 xPx  
2
x
 1.6  1.02  0.6 , Var y    y2 
 yPy  
2
y
 1.2  1.02  0.2 .
Exercise 4.27: The spread is used to adjust your chance of winning so that the probability is .5. The
probability of three wins in a row is .53  .125 . Your probability of losing is thus 1 - .125 = .875. That is,
the bookie has a .125 chance of losing $5 and a .875 chance of winning $1. From his viewpoint, the mean
x
1
and variance are:
5
Px 
xPx  x 2 Px 
.875
0.875
.875
.125  0.625
3.125
1.000
4  0.25   3.9375 .
2
0.250
4.000
So the mean is $0.25 and the variance is
251solnJ1 10/16/00
Exercise 4.26: For Firm A we can compute:
x Px  xPx 
x 2 Px 
0
500
1000
1500
2000
2500
3000
3500
4000
4500
5000
.01
0
0
 x  2450
.01
5
2500
Var x    x2  E x 2   x2 
x 2 Px    x2
.01
10
10000
 6440000  2450 2  437500
.02
30
45000
 x  661 .44
.35
700
1400000
You should be also able to compute for Firm B :
.30
750
1875000
 y  2450
.25
750
2250000
.02
70
245000
Var  y    y2  E y 2   y2 
y 2 P y    y2
.01
40
160000
 6495000  2450 2  492500
.01
45
202500
 y  701 .78
.01
50
250000
1.00 2450 64400000
Thus Firm B has a larger variance, standard deviation and coefficient of variation and thus faces greater
risk.
Problem J1:
 

 

Assume that values of x as shown below represent sales of your product.
Sales
Probability
Except for part a do the parts below in two ways - first,
x
p(x)
by figuring out each value of the function of x and its
0
.50
probability, and, second, by using a formula for a function
1
.10
of x .
2
.10
a. What is Ex ,E x 2 ?
3
.10
b. If costs are C1  10 x  10 , what is E C1  ?
4
.20
c. If costs are C2  0.5x 2  10x  10 , What is E C 2  ?
 
d. If the product sells for $25 per unit, and R represents revenue, what is E R  ?
e. If Costs are described by C1 , and  represents profit, what is E   ?
f. If Costs are described by C 2 , and  represents profit, what is E   ?
Solution: To do these directly , make a table by first computing the values of the functions of x for each
value of x and then multiplying each value by the probability of the corresponding value of x.
x
0
1
2
3
4
px 
xpx 
.50
.10
.10
.10
.20
1.00
0
.10
.20
.30
.80
1.40
x
x2
x 2 px 
0
1
4
9
16
0
0.10
0.40
0.90
3.20
4.60
E x2
C1
10
20
30
40
50
 
 
a. From the table: Ex  1.40,E x 2  4.60 .
b. Using the formula Eax  b  aEx   b,
 10 Ex   10  101.4  10  24 .
C1 px 
5
2
3
4
10
24
E C1 
C2
10.0
19.5
28.0
35.5
42.0
EC1   E10 x  10 
C 2 px 
5.00
1.95
2.80
3.55
8.40
21.70
E C 2 
R
0
25
50
75
100
Rpx 
0
2.5
5.0
7.5
20.0
35.0
E R 
251solnJ1 10/16/00
c. Using the formula E ax  aEx  , E .5x 2  .5E x 2 . Then, using the fact that you can
   
add and subtract means, EC   EC  .5x   EC   E.5x   EC   .5Ex 
2
2
 24  .54.60   21.70 .
d. The formula for revenue is
1
2
1
2
1
R  25x , so using the formula Eax  aEx  ,
ER   E 25 x   25 E x   251.4  35 .
e. Since we can add or subtract means, and   R  C , E    E R   E C1   35  24  11 .
f. E    E R   E C2   35  21 .70  13.30 .
Problem J2: If y  200x  50, x  3 and  x2  10 , what are 
y
and  y2 ?
Solution: The formula says that if y  ax  b, y  a x  b and  y2  a 2 x2 , so, if
a  200 and b  50 , then  y  200 x  50  2003  50  650 , and
 y2  2002  x2  2002 10  400000 . Thus  y  400000  632.46 .
Problem J3:.I have measured the temperature at which successful outcomes of a biochemical process
occur. It turns out that the sample mean is 72 degrees Fahrenheit and the sample standard deviation is 10
degrees Fahrenheit. My European affiliate needs this data. What are the mean and standard deviation in
Celsius? Remember that if x f is the temperature in Fahrenheit and xc is the temperature in Celsius,
x f  9 5 xc  32
 
Solution: x f  72 ,Var x f  10 2  100 . If we solve the equation above for Celsius temperature, we get
xc  5 9 x f  1609 . Using the formula y  ax  b , and remembering E  y   aEx   b and
Var y   a 2Varx  , let y  xc , x  x f , a  5 9 , and b   1609 .
Also, remember that we can use these formulas for sample means and variances, so that
2500
2
2
.
xc  ax f  b  5 9 72  1609  22.22 and Varxc   5 9  Var x f  5 9  10 2 
81
 
5 9 2 10 2  5 9 10   5.556 .
So that the standard deviation of x c is
Problem J4:.A package of a product sells for $21.00/lb. plus a $1.00 packing cost, so that A 2 lb. package
would sell for $21.00(2)+$1.00 = $43.00. If the mean package size is 37 oz. and the standard deviation of
package size is 3 oz. , what are the mean and standard deviation of package price?
Solution: If we choose to work in pounds, let x be the weight in pounds,  x  3716 and  x  316 . The
cost is y  21x  1 . Using the formula y  ax  b , and letting a  21 and b  1 , we can use our formulas
 37 
for the means and variances of y .  y  E  y   aEx   b  a x  b  21   1  49 .56 and
 16 
2
 3
  15 .5039 . So that  y  15.5039  3.9375 .
 16 
 y2  Var  y   a 2Var x   a 2 x2  212 
We could also take the square root of the variance equation and say  y  a 2 x2  a  x
 3
 21   15 .5039 .
 16 
251solnJ1 10/16/00
Problem J5: You are investing in a project with the following distribution of profits(in millions of dollars):
x
P(x)
a. What is the mean and variance of the profits?
10
.500
b. Assume that you must pay me 50% of the profits plus a
20
.250
finder's fee of $1(million). What is the mean and variance of
30
.125
your profit?
40
.125
c. Use your answer in part b. to find the mean and variance
of your profit if I raise my finder's fee to $2(million)?
Solution: a)
x
10
20
30
40
Px  xPx  x 2 Px 
.500
5.00
50 .0
.250
5.00 100 .0
.125
3.75 112 .5
.125
5.00 200 .0
1.000 18 .75 462 .5
From the table at left
 xPx  18.75
Var x     E x      x Px   
x 
2
x
2
2
x
2
2
x
 462 .5  18 .75 2  110 .9375
 x  110 .9375  10 .53
b) Since x represents the profits, the formula for my share is 0.5x  1 . So your share is
y  x  0.5x  1  0.5x  1 . Using the formula y  ax  b , and letting a  0.5 and b  1 , we can use
our formulas for the means and variances of y .  y  E y   aEx  b  a x  b  0.518.75  1  8.375
and  y2  Var y   a 2Varx  a 2 x2  0.52 110.9375  27.7344 . (or  y  4.892 )
c) Just as if y  ax  b ,  y  E y   aEx  b  a x  b and  y2  Var  y   a 2Var x   a 2 x2 , so if
w  cy  d , w  Ew  cE y   d  c x  d and  w2  Var w  c 2Var  y   c 2 y2 . Our new return is
w  1y  1 , so that c  1 and d  1 . Therefore w  Ew  c x  d  18.375   1  7.375 and
 w2  Varw  c2 y2  12 27.7344  27.7344.
Problem J6: Assume that x is a standardized random variable. If y  5x  3 , find the mean and variance
of y .
Solution: By definition a standardized random variable has a mean of zero and a standard deviation of 1.
 x  0 and  x  1 . y  5x  3 . Using the formula y  ax  b , and letting a  5 and b  3 , we can use
our formulas for the means and variances of y .  y  E y   aEx  b  a x  b  50  3  3 and
 y2  Var y   a 2Varx  a 2 x2  52 12  25 .
251solnJ1 10/16/00
Problem J7: Consider the random variable x in the table below. Make a probability table for a
standardized random variable z , created by subtracting
x
the mean (which is 4) and dividing by the standard deviation
P x 
0
1/16
(which is 2). Show that the mean and standard deviation of
2
1/4
z are 0 and 1.
4
3/8
6
1/4
8
1/16
x x4

Solution: We standardize a variable by using the formula z 
, so that x  2 is replaced by

2
2 24
z

 1 . First we check the mean and variance of x .

2
x P x  xPx  x 2 P x 
From the table at left
0 1
0.00
0.00
16
x 
xPx   4.00
1
2
0.50
1.00
4
Var x    x2  E x 2   x2 
x 2 Px    x2
3
4
1.50
6.00
8
 20 .00  4.00 2  4.00
1
6
1.50
9.00
4
 x  4.00  2.00
8 1
0.50
4.00
16
1.00 4.00 20 .00
Now we write out the variable z with its probabilities and compute its mean and standard deviation.
z Pz 
zP z  z 2 Pz 
From the table at left
2 1
 0.125
0.25
16
 z  zP z   0.00
1 1
 0.250
0.25
4
Var z    z2  E z 2   z2 
z 2 Pz    z2
3
0
0.000
0.00
8
 1.00  0.00 2  1.00
1
1
0.250
0.25
4
 x  1.00  1.00
2 1
0.125
0.25
16
1.00
0.000
1.00

 

 
