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.