Post on 27-May-2020
Expectations
DefinitionLet X be a discrete rv with set of possible values D and pmf p(x).The expected value or mean value of X , denoted by E (X ) orµX , is
E (X ) = µX =∑x∈D
x · p(x)
e.g (Problem 30)A group of individuals who have automobile insurance from acertain company is randomly selected. Let Y be the number ofmoving violations for which the individual was cited during the last3 years. The pmf of Y is
y 0 1 2 3
p(y) 0.60 0.25 0.10 0.05Then the expected value of
moving violations for that group is
µY = E (Y ) = 0 · 0.60 + 1 · 0.25 + 2 · 0.10 + 3 · 0.05 = 0.60
Expectations
DefinitionLet X be a discrete rv with set of possible values D and pmf p(x).The expected value or mean value of X , denoted by E (X ) orµX , is
E (X ) = µX =∑x∈D
x · p(x)
e.g (Problem 30)A group of individuals who have automobile insurance from acertain company is randomly selected. Let Y be the number ofmoving violations for which the individual was cited during the last3 years. The pmf of Y is
y 0 1 2 3
p(y) 0.60 0.25 0.10 0.05Then the expected value of
moving violations for that group is
µY = E (Y ) = 0 · 0.60 + 1 · 0.25 + 2 · 0.10 + 3 · 0.05 = 0.60
Expectations
DefinitionLet X be a discrete rv with set of possible values D and pmf p(x).The expected value or mean value of X , denoted by E (X ) orµX , is
E (X ) = µX =∑x∈D
x · p(x)
e.g (Problem 30)A group of individuals who have automobile insurance from acertain company is randomly selected. Let Y be the number ofmoving violations for which the individual was cited during the last3 years. The pmf of Y is
y 0 1 2 3
p(y) 0.60 0.25 0.10 0.05Then the expected value of
moving violations for that group is
µY = E (Y ) = 0 · 0.60 + 1 · 0.25 + 2 · 0.10 + 3 · 0.05 = 0.60
Expectations
y 0 1 2 3p(y) 0.60 0.25 0.10 0.05
Assume the total number of individuals in that group is 100, then there
are 60 individuals without moving violation, 25 with 1 moving violation,
10 with 2 moving violations and 5 with 3 moving violations.
The population mean is calculated as
µ =0 · 60 + ·1 · 25 + ·2 · 10 + 3 · 5
100= 0.60
µ = 0 · 60
100+ 1 · 25
100+ 2 · 10
100+ 3 · 5
100= 0 · 0.60 + 1 · 0.25 + 2 · 0.10 + 3 · 0.05
= 0.60
The population size is irrevelant if we know the pmf!
Expectations
y 0 1 2 3p(y) 0.60 0.25 0.10 0.05
Assume the total number of individuals in that group is 100, then there
are 60 individuals without moving violation, 25 with 1 moving violation,
10 with 2 moving violations and 5 with 3 moving violations.
The population mean is calculated as
µ =0 · 60 + ·1 · 25 + ·2 · 10 + 3 · 5
100= 0.60
µ = 0 · 60
100+ 1 · 25
100+ 2 · 10
100+ 3 · 5
100= 0 · 0.60 + 1 · 0.25 + 2 · 0.10 + 3 · 0.05
= 0.60
The population size is irrevelant if we know the pmf!
Expectations
y 0 1 2 3p(y) 0.60 0.25 0.10 0.05
Assume the total number of individuals in that group is 100, then there
are 60 individuals without moving violation, 25 with 1 moving violation,
10 with 2 moving violations and 5 with 3 moving violations.
The population mean is calculated as
µ =0 · 60 + ·1 · 25 + ·2 · 10 + 3 · 5
100= 0.60
µ = 0 · 60
100+ 1 · 25
100+ 2 · 10
100+ 3 · 5
100= 0 · 0.60 + 1 · 0.25 + 2 · 0.10 + 3 · 0.05
= 0.60
The population size is irrevelant if we know the pmf!
Expectations
y 0 1 2 3p(y) 0.60 0.25 0.10 0.05
Assume the total number of individuals in that group is 100, then there
are 60 individuals without moving violation, 25 with 1 moving violation,
10 with 2 moving violations and 5 with 3 moving violations.
The population mean is calculated as
µ =0 · 60 + ·1 · 25 + ·2 · 10 + 3 · 5
100= 0.60
µ = 0 · 60
100+ 1 · 25
100+ 2 · 10
100+ 3 · 5
100= 0 · 0.60 + 1 · 0.25 + 2 · 0.10 + 3 · 0.05
= 0.60
The population size is irrevelant if we know the pmf!
Expectations
y 0 1 2 3p(y) 0.60 0.25 0.10 0.05
Assume the total number of individuals in that group is 100, then there
are 60 individuals without moving violation, 25 with 1 moving violation,
10 with 2 moving violations and 5 with 3 moving violations.
The population mean is calculated as
µ =0 · 60 + ·1 · 25 + ·2 · 10 + 3 · 5
100= 0.60
µ = 0 · 60
100+ 1 · 25
100+ 2 · 10
100+ 3 · 5
100= 0 · 0.60 + 1 · 0.25 + 2 · 0.10 + 3 · 0.05
= 0.60
The population size is irrevelant if we know the pmf!
Expectations
Examples:Let X be a Bernoulli rv with pmf
p(x) =
1− p x = 0
p x = 1
0 x 6= 0, or 1
Then the expected value for X is
E (X ) = 0 · p(0) + 1 · p(1) = p
We see that the expected value of a Bernoulli rv X is just theprobability that X takes on the value 1.
Expectations
Examples:Let X be a Bernoulli rv with pmf
p(x) =
1− p x = 0
p x = 1
0 x 6= 0, or 1
Then the expected value for X is
E (X ) = 0 · p(0) + 1 · p(1) = p
We see that the expected value of a Bernoulli rv X is just theprobability that X takes on the value 1.
Expectations
Examples:Let X be a Bernoulli rv with pmf
p(x) =
1− p x = 0
p x = 1
0 x 6= 0, or 1
Then the expected value for X is
E (X ) = 0 · p(0) + 1 · p(1) = p
We see that the expected value of a Bernoulli rv X is just theprobability that X takes on the value 1.
Expectations
Examples:Let X be a Bernoulli rv with pmf
p(x) =
1− p x = 0
p x = 1
0 x 6= 0, or 1
Then the expected value for X is
E (X ) = 0 · p(0) + 1 · p(1) = p
We see that the expected value of a Bernoulli rv X is just theprobability that X takes on the value 1.
Expectations
Examples:Consider the cards drawing example again and assume we haveinfinitely many cards this time. Let X = the number of drawingsuntil we get a ♠. If the probability for getting a ♠ is α, then thepmf for X is
p(x) =
{α(1− α)x−1 x = 1, 2, 3, . . .
0 otherwise
The expected value for X is
E (X ) =∑D
x · p(x) =∞∑
x=1
xα(1− α)x−1 = α∞∑
x=1
[− d
dα(1− α)x ]
E (X ) = α{− d
dα[∞∑
x=1
(1− α)x ]} = α{− d
dα(1− αα
)} =1
α
Expectations
Examples:Consider the cards drawing example again and assume we haveinfinitely many cards this time. Let X = the number of drawingsuntil we get a ♠. If the probability for getting a ♠ is α, then thepmf for X is
p(x) =
{α(1− α)x−1 x = 1, 2, 3, . . .
0 otherwise
The expected value for X is
E (X ) =∑D
x · p(x) =∞∑
x=1
xα(1− α)x−1 = α∞∑
x=1
[− d
dα(1− α)x ]
E (X ) = α{− d
dα[∞∑
x=1
(1− α)x ]} = α{− d
dα(1− αα
)} =1
α
Expectations
Examples:Consider the cards drawing example again and assume we haveinfinitely many cards this time. Let X = the number of drawingsuntil we get a ♠. If the probability for getting a ♠ is α, then thepmf for X is
p(x) =
{α(1− α)x−1 x = 1, 2, 3, . . .
0 otherwise
The expected value for X is
E (X ) =∑D
x · p(x) =∞∑
x=1
xα(1− α)x−1 = α
∞∑x=1
[− d
dα(1− α)x ]
E (X ) = α{− d
dα[∞∑
x=1
(1− α)x ]} = α{− d
dα(1− αα
)} =1
α
Expectations
Examples:Consider the cards drawing example again and assume we haveinfinitely many cards this time. Let X = the number of drawingsuntil we get a ♠. If the probability for getting a ♠ is α, then thepmf for X is
p(x) =
{α(1− α)x−1 x = 1, 2, 3, . . .
0 otherwise
The expected value for X is
E (X ) =∑D
x · p(x) =∞∑
x=1
xα(1− α)x−1 = α
∞∑x=1
[− d
dα(1− α)x ]
E (X ) = α{− d
dα[∞∑
x=1
(1− α)x ]} = α{− d
dα(1− αα
)} =1
α
Expectations
Examples 3.20Let X be the number of interviews a student has prior to getting ajob. The pmf for X is
p(x) =
{kx2 x = 1, 2, 3, . . .
0 otherwise
where k is chosen so that∑∞
x=1(k/x2) = 1. (It can be showedthat
∑∞x=1(1/x2) <∞, which implies that such a k exists.)
The expected value of X is
µ = E (X ) =∞∑
x=1
x · k
x2= k
∞∑x=1
1
x=∞!
The expected value is NOT finite!Heavy Tail: distribution with a large amount of probability farfrom µ
Expectations
Examples 3.20Let X be the number of interviews a student has prior to getting ajob. The pmf for X is
p(x) =
{kx2 x = 1, 2, 3, . . .
0 otherwise
where k is chosen so that∑∞
x=1(k/x2) = 1. (It can be showedthat
∑∞x=1(1/x2) <∞, which implies that such a k exists.)
The expected value of X is
µ = E (X ) =∞∑
x=1
x · k
x2= k
∞∑x=1
1
x=∞!
The expected value is NOT finite!
Heavy Tail: distribution with a large amount of probability farfrom µ
Expectations
Examples 3.20Let X be the number of interviews a student has prior to getting ajob. The pmf for X is
p(x) =
{kx2 x = 1, 2, 3, . . .
0 otherwise
where k is chosen so that∑∞
x=1(k/x2) = 1. (It can be showedthat
∑∞x=1(1/x2) <∞, which implies that such a k exists.)
The expected value of X is
µ = E (X ) =∞∑
x=1
x · k
x2= k
∞∑x=1
1
x=∞!
The expected value is NOT finite!Heavy Tail:
distribution with a large amount of probability farfrom µ
Expectations
Examples 3.20Let X be the number of interviews a student has prior to getting ajob. The pmf for X is
p(x) =
{kx2 x = 1, 2, 3, . . .
0 otherwise
where k is chosen so that∑∞
x=1(k/x2) = 1. (It can be showedthat
∑∞x=1(1/x2) <∞, which implies that such a k exists.)
The expected value of X is
µ = E (X ) =∞∑
x=1
x · k
x2= k
∞∑x=1
1
x=∞!
The expected value is NOT finite!Heavy Tail: distribution with a large amount of probability farfrom µ
Expectations
Example (Problem 38)Let X = the outcome when a fair die is rolled once. If before thedie is rolled you are offered either 1
3.5 dollars or 1X dollars, would
you accept the guaranteed amount or would you gamble?x 1 2 3 4 5 6
p(x) 16
16
16
16
16
16
1x 1 1
213
14
15
16
Then the expected dollars from gambling is
E (1
X) =
6∑x=1
1
x· p(
1
x)
= 1 · 16
+1
2· 16
+ · · ·+ 1
6· 16
=49
120<
1
3.5
Expectations
Example (Problem 38)Let X = the outcome when a fair die is rolled once. If before thedie is rolled you are offered either 1
3.5 dollars or 1X dollars, would
you accept the guaranteed amount or would you gamble?
x 1 2 3 4 5 6
p(x) 16
16
16
16
16
16
1x 1 1
213
14
15
16
Then the expected dollars from gambling is
E (1
X) =
6∑x=1
1
x· p(
1
x)
= 1 · 16
+1
2· 16
+ · · ·+ 1
6· 16
=49
120<
1
3.5
Expectations
Example (Problem 38)Let X = the outcome when a fair die is rolled once. If before thedie is rolled you are offered either 1
3.5 dollars or 1X dollars, would
you accept the guaranteed amount or would you gamble?x 1 2 3 4 5 6
p(x) 16
16
16
16
16
16
1x 1 1
213
14
15
16
Then the expected dollars from gambling is
E (1
X) =
6∑x=1
1
x· p(
1
x)
= 1 · 16
+1
2· 16
+ · · ·+ 1
6· 16
=49
120<
1
3.5
Expectations
Example (Problem 38)Let X = the outcome when a fair die is rolled once. If before thedie is rolled you are offered either 1
3.5 dollars or 1X dollars, would
you accept the guaranteed amount or would you gamble?x 1 2 3 4 5 6
p(x) 16
16
16
16
16
16
1x 1 1
213
14
15
16
Then the expected dollars from gambling is
E (1
X) =
6∑x=1
1
x· p(
1
x)
= 1 · 16
+1
2· 16
+ · · ·+ 1
6· 16
=49
120<
1
3.5
Expectations
Example (Problem 38)Let X = the outcome when a fair die is rolled once. If before thedie is rolled you are offered either 1
3.5 dollars or 1X dollars, would
you accept the guaranteed amount or would you gamble?x 1 2 3 4 5 6
p(x) 16
16
16
16
16
16
1x 1 1
213
14
15
16
Then the expected dollars from gambling is
E (1
X) =
6∑x=1
1
x· p(
1
x)
= 1 · 16
+1
2· 16
+ · · ·+ 1
6· 16
=49
120<
1
3.5
Expectations
Proposition
If the rv X has a set of possible values D and pmf p(x), then theexpected value of any function h(X ), denoted by E [h(X )] or µhX
,is computed by
E [h(X )] =∑D
h(x) · p(x)
Expectations
Proposition
If the rv X has a set of possible values D and pmf p(x), then theexpected value of any function h(X ), denoted by E [h(X )] or µhX
,is computed by
E [h(X )] =∑D
h(x) · p(x)
Expectations
Example 3.23A computer store has purchased three computers of a certain typeat $500 apiece. It will sell them for $1000 apiece. Themanufacturer has agreed to repurchase any computers still unsoldafter a specified period at $200 apiece.Let X denote the number of computers sold, and suppose thatp(0) = 0.1, p(1) = 0.2, p(2) = 0.3, p(3) = 0.4.Let h(X ) denote the profit associated with selling X units, thenh(X ) = revenue − cost= 1000X + 200(3− X )− 1500 = 800X − 900.The expected profit is
E [h(X )] = h(0) · p(0) + h(1) · p(1) + h(2) · p(2) + h(3) · p(3)
= (−900)(0.1) + (−100)(0.2) + (700)(0.3) + (1500)(0.4)
= 700
Expectations
Example 3.23A computer store has purchased three computers of a certain typeat $500 apiece. It will sell them for $1000 apiece. Themanufacturer has agreed to repurchase any computers still unsoldafter a specified period at $200 apiece.
Let X denote the number of computers sold, and suppose thatp(0) = 0.1, p(1) = 0.2, p(2) = 0.3, p(3) = 0.4.Let h(X ) denote the profit associated with selling X units, thenh(X ) = revenue − cost= 1000X + 200(3− X )− 1500 = 800X − 900.The expected profit is
E [h(X )] = h(0) · p(0) + h(1) · p(1) + h(2) · p(2) + h(3) · p(3)
= (−900)(0.1) + (−100)(0.2) + (700)(0.3) + (1500)(0.4)
= 700
Expectations
Example 3.23A computer store has purchased three computers of a certain typeat $500 apiece. It will sell them for $1000 apiece. Themanufacturer has agreed to repurchase any computers still unsoldafter a specified period at $200 apiece.Let X denote the number of computers sold, and suppose thatp(0) = 0.1, p(1) = 0.2, p(2) = 0.3, p(3) = 0.4.Let h(X ) denote the profit associated with selling X units,
thenh(X ) = revenue − cost= 1000X + 200(3− X )− 1500 = 800X − 900.The expected profit is
E [h(X )] = h(0) · p(0) + h(1) · p(1) + h(2) · p(2) + h(3) · p(3)
= (−900)(0.1) + (−100)(0.2) + (700)(0.3) + (1500)(0.4)
= 700
Expectations
Example 3.23A computer store has purchased three computers of a certain typeat $500 apiece. It will sell them for $1000 apiece. Themanufacturer has agreed to repurchase any computers still unsoldafter a specified period at $200 apiece.Let X denote the number of computers sold, and suppose thatp(0) = 0.1, p(1) = 0.2, p(2) = 0.3, p(3) = 0.4.Let h(X ) denote the profit associated with selling X units, thenh(X ) = revenue − cost= 1000X + 200(3− X )− 1500 = 800X − 900.
The expected profit is
E [h(X )] = h(0) · p(0) + h(1) · p(1) + h(2) · p(2) + h(3) · p(3)
= (−900)(0.1) + (−100)(0.2) + (700)(0.3) + (1500)(0.4)
= 700
Expectations
Example 3.23A computer store has purchased three computers of a certain typeat $500 apiece. It will sell them for $1000 apiece. Themanufacturer has agreed to repurchase any computers still unsoldafter a specified period at $200 apiece.Let X denote the number of computers sold, and suppose thatp(0) = 0.1, p(1) = 0.2, p(2) = 0.3, p(3) = 0.4.Let h(X ) denote the profit associated with selling X units, thenh(X ) = revenue − cost= 1000X + 200(3− X )− 1500 = 800X − 900.The expected profit is
E [h(X )] = h(0) · p(0) + h(1) · p(1) + h(2) · p(2) + h(3) · p(3)
= (−900)(0.1) + (−100)(0.2) + (700)(0.3) + (1500)(0.4)
= 700
Expectations
Proposition
E (aX + b) = a · E (X ) + b
(Or, using alternative notation, µaX+b = a · µX + b.)
e.g. for the previous example,
E [h(X )] = E (800X − 900) = 800 · E (X )− 900 = 700
Corollary
1. For any constant a, E (aX ) = a · E (X ).2. For any constant b, E (X + b) = E (X ) + b.
Expectations
Proposition
E (aX + b) = a · E (X ) + b
(Or, using alternative notation, µaX+b = a · µX + b.)
e.g. for the previous example,
E [h(X )] = E (800X − 900) = 800 · E (X )− 900 = 700
Corollary
1. For any constant a, E (aX ) = a · E (X ).2. For any constant b, E (X + b) = E (X ) + b.
Expectations
Proposition
E (aX + b) = a · E (X ) + b
(Or, using alternative notation, µaX+b = a · µX + b.)
e.g. for the previous example,
E [h(X )] = E (800X − 900) = 800 · E (X )− 900 = 700
Corollary
1. For any constant a, E (aX ) = a · E (X ).2. For any constant b, E (X + b) = E (X ) + b.
Expectations
Proposition
E (aX + b) = a · E (X ) + b
(Or, using alternative notation, µaX+b = a · µX + b.)
e.g. for the previous example,
E [h(X )] = E (800X − 900) = 800 · E (X )− 900 = 700
Corollary
1. For any constant a, E (aX ) = a · E (X ).2. For any constant b, E (X + b) = E (X ) + b.
Expectations
DefinitionLet X have pmf p(x) and expected value µ. Then the variance ofX, denoted by V (X ) or σ2
X , or just σ2X , is
V (X ) =∑D
(x − µ)2 · p(x) = E [(X − µ)2]
The stand deviation (SD) of X is
σX =√σ2
X
Expectations
DefinitionLet X have pmf p(x) and expected value µ. Then the variance ofX, denoted by V (X ) or σ2
X , or just σ2X , is
V (X ) =∑D
(x − µ)2 · p(x) = E [(X − µ)2]
The stand deviation (SD) of X is
σX =√σ2
X
Expectations
Example:For the previous example, the pmf is given as
x 0 1 2 3
p(x) 0.1 0.2 0.3 0.4then the variance of X is
V (X ) = σ2 =3∑
x=0
(x − 2)2 · p(x)
= (0− 2)2(0.1) + (1− 2)2(0.2) + (2− 2)2(0.3) + (3− 2)2(0.4)
= 1
Expectations
Example:For the previous example, the pmf is given as
x 0 1 2 3
p(x) 0.1 0.2 0.3 0.4
then the variance of X is
V (X ) = σ2 =3∑
x=0
(x − 2)2 · p(x)
= (0− 2)2(0.1) + (1− 2)2(0.2) + (2− 2)2(0.3) + (3− 2)2(0.4)
= 1
Expectations
Example:For the previous example, the pmf is given as
x 0 1 2 3
p(x) 0.1 0.2 0.3 0.4then the variance of X is
V (X ) = σ2 =3∑
x=0
(x − 2)2 · p(x)
= (0− 2)2(0.1) + (1− 2)2(0.2) + (2− 2)2(0.3) + (3− 2)2(0.4)
= 1
Expectations
Recall that for sample variance s2, we have
s2 =Sxx
n − 1=
∑x2i −
(∑
xi )2
n
n − 1
Proposition
V (X ) = σ2 = [∑D
x2 · p(x)]− µ2 = E (X 2)− [E (X )]2
e.g. for the previous example, the pmf is given asx 0 1 2 3
p(x) 0.1 0.2 0.3 0.4ThenV (X ) = E (X 2)− [E (X )]2 = 12 · 0.2 + 22 · 0.3 + 32 · 0.4− (2)2 = 1
Expectations
Recall that for sample variance s2, we have
s2 =Sxx
n − 1=
∑x2i −
(∑
xi )2
n
n − 1
Proposition
V (X ) = σ2 = [∑D
x2 · p(x)]− µ2 = E (X 2)− [E (X )]2
e.g. for the previous example, the pmf is given asx 0 1 2 3
p(x) 0.1 0.2 0.3 0.4ThenV (X ) = E (X 2)− [E (X )]2 = 12 · 0.2 + 22 · 0.3 + 32 · 0.4− (2)2 = 1
Expectations
Recall that for sample variance s2, we have
s2 =Sxx
n − 1=
∑x2i −
(∑
xi )2
n
n − 1
Proposition
V (X ) = σ2 = [∑D
x2 · p(x)]− µ2 = E (X 2)− [E (X )]2
e.g. for the previous example, the pmf is given asx 0 1 2 3
p(x) 0.1 0.2 0.3 0.4ThenV (X ) = E (X 2)− [E (X )]2 = 12 · 0.2 + 22 · 0.3 + 32 · 0.4− (2)2 = 1
Expectations
Recall that for sample variance s2, we have
s2 =Sxx
n − 1=
∑x2i −
(∑
xi )2
n
n − 1
Proposition
V (X ) = σ2 = [∑D
x2 · p(x)]− µ2 = E (X 2)− [E (X )]2
e.g. for the previous example, the pmf is given asx 0 1 2 3
p(x) 0.1 0.2 0.3 0.4ThenV (X ) = E (X 2)− [E (X )]2 = 12 · 0.2 + 22 · 0.3 + 32 · 0.4− (2)2 = 1
Expectations
Proposition
If h(X ) is a function of a rv X , then
V [h(X )] = σ2h(X ) =
∑D
{h(x)−E [h(X )]}2·p(x) = E [h(X )2]−{E [h(X )]}2
If h(X ) is linear, i.e. h(X ) = aX + b for some nonrandom constanta and b, then
V (aX + b) = σ2aX+b = a2 · σ2
X and σaX+b =| a | ·σX
In particular,σaX =| a | ·σX , σX+b = σX
Expectations
Proposition
If h(X ) is a function of a rv X , then
V [h(X )] = σ2h(X ) =
∑D
{h(x)−E [h(X )]}2·p(x) = E [h(X )2]−{E [h(X )]}2
If h(X ) is linear, i.e. h(X ) = aX + b for some nonrandom constanta and b, then
V (aX + b) = σ2aX+b = a2 · σ2
X and σaX+b =| a | ·σX
In particular,σaX =| a | ·σX , σX+b = σX
Expectations
Example 3.23 continuedA computer store has purchased three computers of a certain typeat $500 apiece. It will sell them for $1000 apiece. Themanufacturer has agreed to repurchase any computers still unsoldafter a specified period at $200 apiece. Let X denote the numberof computers sold, and suppose that p(0) = 0.1,p(1) = 0.2, p(2) = 0.3, p(3) = 0.4. Let h(X ) denote the profitassociated with selling X units, thenh(X ) = revenue − cost= 1000X + 200(3− X )− 1500 = 800X − 900.The variance of h(X ) is
V [h(X )] = V [800X − 900]
= 8002V [X ]
= 640, 000
And the SD is σh(X ) =√
V [h(X )] = 800.
Expectations
Example 3.23 continuedA computer store has purchased three computers of a certain typeat $500 apiece. It will sell them for $1000 apiece. Themanufacturer has agreed to repurchase any computers still unsoldafter a specified period at $200 apiece. Let X denote the numberof computers sold, and suppose that p(0) = 0.1,p(1) = 0.2, p(2) = 0.3, p(3) = 0.4. Let h(X ) denote the profitassociated with selling X units, thenh(X ) = revenue − cost= 1000X + 200(3− X )− 1500 = 800X − 900.
The variance of h(X ) is
V [h(X )] = V [800X − 900]
= 8002V [X ]
= 640, 000
And the SD is σh(X ) =√
V [h(X )] = 800.
Expectations
Example 3.23 continuedA computer store has purchased three computers of a certain typeat $500 apiece. It will sell them for $1000 apiece. Themanufacturer has agreed to repurchase any computers still unsoldafter a specified period at $200 apiece. Let X denote the numberof computers sold, and suppose that p(0) = 0.1,p(1) = 0.2, p(2) = 0.3, p(3) = 0.4. Let h(X ) denote the profitassociated with selling X units, thenh(X ) = revenue − cost= 1000X + 200(3− X )− 1500 = 800X − 900.The variance of h(X ) is
V [h(X )] = V [800X − 900]
= 8002V [X ]
= 640, 000
And the SD is σh(X ) =√
V [h(X )] = 800.