Expectations
Expectations
Definition
Let 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 ) or
µX , is
X
E (X ) = µX =
x · p(x)
x∈D
Expectations
Definition
Let 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 ) or
µX , is
X
E (X ) = µX =
x · p(x)
x∈D
e.g (Problem 30)
A group of individuals who have automobile insurance from a
certain company is randomly selected. Let Y be the number of
moving violations for which the individual was cited during the last
3 years. The pmf of Y is
y
0
1
2
3
Then the expected value of
p(y) 0.60 0.25 0.10 0.05
moving violations for that group is
µY = E (Y ) = 0 · 0.60 + 1 · 0.25 + 2 · 0.10 + 3 · 0.05 = 0.60
Expectations
Expectations
y
0
1
2
3
p(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.
Expectations
y
0
1
2
3
p(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
= 0.60
100
Expectations
y
0
1
2
3
p(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
= 0.60
100
60
25
10
5
+1·
+2·
+3·
100
100
100
100
= 0 · 0.60 + 1 · 0.25 + 2 · 0.10 + 3 · 0.05
µ=0·
= 0.60
Expectations
y
0
1
2
3
p(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
= 0.60
100
60
25
10
5
+1·
+2·
+3·
100
100
100
100
= 0 · 0.60 + 1 · 0.25 + 2 · 0.10 + 3 · 0.05
µ=0·
= 0.60
The population size is irrevelant if we know the pmf!
Expectations
Expectations
Examples:
Let X be a Bernoulli rv with pmf


1 − p x = 0
p(x) = p
x =1


0
x 6= 0, or 1
Expectations
Examples:
Let X be a Bernoulli rv with pmf


1 − p x = 0
p(x) = 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
Expectations
Examples:
Let X be a Bernoulli rv with pmf


1 − p x = 0
p(x) = 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 the
probability that X takes on the value 1.
Expectations
Expectations
Examples:
Consider the cards drawing example again and assume we have
infinitely many cards this time. Let X = the number of drawings
until we get a ♠. If the probability for getting a ♠ is α, then the
pmf for X is
(
α(1 − α)x−1 x = 1, 2, 3, . . .
p(x) =
0
otherwise
Expectations
Examples:
Consider the cards drawing example again and assume we have
infinitely many cards this time. Let X = the number of drawings
until we get a ♠. If the probability for getting a ♠ is α, then the
pmf for X is
(
α(1 − α)x−1 x = 1, 2, 3, . . .
p(x) =
0
otherwise
The expected value for X is
E (X ) =
X
D
x · p(x) =
∞
X
x=1
xα(1 − α)x−1 = α
∞
X
d
[− (1 − α)x ]
dα
x=1
Expectations
Examples:
Consider the cards drawing example again and assume we have
infinitely many cards this time. Let X = the number of drawings
until we get a ♠. If the probability for getting a ♠ is α, then the
pmf for X is
(
α(1 − α)x−1 x = 1, 2, 3, . . .
p(x) =
0
otherwise
The expected value for X is
E (X ) =
X
x · p(x) =
D
∞
X
x=1
∞
E (X ) = α{−
xα(1 − α)x−1 = α
∞
X
d
[− (1 − α)x ]
dα
x=1
d X
d 1−α
1
[ (1 − α)x ]} = α{− (
)} =
dα
dα α
α
x=1
Expectations
Expectations
Examples 3.20
Let X be the number of interviews a student has prior to getting a
job. The pmf for X is
(
k
x = 1, 2, 3, . . .
2
p(x) = x
0 otherwise
P
2
wherePk is chosen so that ∞
x=1 (k/x ) = 1. (It can be showed
∞
2
that x=1 (1/x ) < ∞, which implies that such a k exists.)
The expected value of X is
µ = E (X ) =
∞
X
x=1
∞
X1
k
= ∞!
x· 2 =k
x
x
The expected value is NOT finite!
x=1
Expectations
Examples 3.20
Let X be the number of interviews a student has prior to getting a
job. The pmf for X is
(
k
x = 1, 2, 3, . . .
2
p(x) = x
0 otherwise
P
2
wherePk is chosen so that ∞
x=1 (k/x ) = 1. (It can be showed
∞
2
that x=1 (1/x ) < ∞, which implies that such a k exists.)
The expected value of X is
µ = E (X ) =
∞
X
x=1
∞
X1
k
= ∞!
x· 2 =k
x
x
The expected value is NOT finite!
Heavy Tail:
x=1
Expectations
Examples 3.20
Let X be the number of interviews a student has prior to getting a
job. The pmf for X is
(
k
x = 1, 2, 3, . . .
2
p(x) = x
0 otherwise
P
2
wherePk is chosen so that ∞
x=1 (k/x ) = 1. (It can be showed
∞
2
that x=1 (1/x ) < ∞, which implies that such a k exists.)
The expected value of X is
µ = E (X ) =
∞
X
x=1
∞
X1
k
= ∞!
x· 2 =k
x
x
x=1
The expected value is NOT finite!
Heavy Tail: distribution with a large amount of probability far
from µ
Expectations
Expectations
Example (Problem 38)
Let X = the outcome when a fair die is rolled once. If before the
1
die is rolled you are offered either 3.5
dollars or X1 dollars, would
you accept the guaranteed amount or would you gamble?
Expectations
Example (Problem 38)
Let X = the outcome when a fair die is rolled once. If before the
1
die is rolled you are offered either 3.5
dollars or X1 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
Expectations
Example (Problem 38)
Let X = the outcome when a fair die is rolled once. If before the
1
die is rolled you are offered either 3.5
dollars or X1 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
1
1 12 13 14 15 16
x
Expectations
Example (Problem 38)
Let X = the outcome when a fair die is rolled once. If before the
1
die is rolled you are offered either 3.5
dollars or X1 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
1
1 12 13 14 15 16
x
Then the expected dollars from gambling is
6
E(
X1
1
1
)=
· p( )
X
x
x
x=1
1 1 1
1 1
+ · + ··· + ·
6 2 6
6 6
49
1
=
<
120
3.5
=1·
Expectations
Expectations
Proposition
If the rv X has a set of possible values D and pmf p(x), then the
expected value of any function h(X ), denoted by E [h(X )] or µhX ,
is computed by
X
E [h(X )] =
h(x) · p(x)
D
Expectations
Expectations
Example 3.23
A computer store has purchased three computers of a certain type
at $500 apiece. It will sell them for $1000 apiece. The
manufacturer has agreed to repurchase any computers still unsold
after a specified period at $200 apiece.
Expectations
Example 3.23
A computer store has purchased three computers of a certain type
at $500 apiece. It will sell them for $1000 apiece. The
manufacturer has agreed to repurchase any computers still unsold
after a specified period at $200 apiece.
Let X denote the number of 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 profit associated with selling X units,
Expectations
Example 3.23
A computer store has purchased three computers of a certain type
at $500 apiece. It will sell them for $1000 apiece. The
manufacturer has agreed to repurchase any computers still unsold
after a specified period at $200 apiece.
Let X denote the number of 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 profit associated with selling X units, then
h(X ) = revenue − cost
= 1000X + 200(3 − X ) − 1500 = 800X − 900.
Expectations
Example 3.23
A computer store has purchased three computers of a certain type
at $500 apiece. It will sell them for $1000 apiece. The
manufacturer has agreed to repurchase any computers still unsold
after a specified period at $200 apiece.
Let X denote the number of 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 profit associated with selling X units, then
h(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
Expectations
Proposition
E (aX + b) = a · E (X ) + b
(Or, using alternative notation, µaX +b = a · µX + 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
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
Expectations
Definition
Let X have pmf p(x) and expected value µ. Then the variance of
X, denoted by V (X ) or σX2 , or just σX2 , is
V (X ) =
X
(x − µ)2 · p(x) = E [(X − µ)2 ]
D
The stand deviation (SD) of X is
q
σX = σX2
Expectations
Expectations
Example:
For the previous example, the pmf is given as
0
1
2
3
x
p(x) 0.1 0.2 0.3 0.4
Expectations
Example:
For the previous example, the pmf is given as
0
1
2
3
x
p(x) 0.1 0.2 0.3 0.4
then the variance of X is
2
V (X ) = σ =
3
X
x=0
2
(x − 2)2 · p(x)
= (0 − 2) (0.1) + (1 − 2)2 (0.2) + (2 − 2)2 (0.3) + (3 − 2)2 (0.4)
=1
Expectations
Expectations
Recall that for sample variance s 2 , we have
Sxx
=
s2 =
n−1
P
xi )2
n
P
xi2 − (
n−1
Expectations
Recall that for sample variance s 2 , we have
Sxx
=
s2 =
n−1
P
xi )2
n
P
xi2 − (
n−1
Proposition
X
V (X ) = σ 2 = [
x 2 · p(x)] − µ2 = E (X 2 ) − [E (X )]2
D
Expectations
Recall that for sample variance s 2 , we have
Sxx
=
s2 =
n−1
P
xi )2
n
P
xi2 − (
n−1
Proposition
X
V (X ) = σ 2 = [
x 2 · p(x)] − µ2 = E (X 2 ) − [E (X )]2
D
e.g. 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
V (X ) = E (X 2 ) − [E (X )]2 = 12 · 0.2 + 22 · 0.3 + 32 · 0.4 − (2)2 = 1
Expectations
Expectations
Proposition
If h(X ) is a function of a rv X , then
X
2
V [h(X )] = σh(X
{h(x)−E [h(X )]}2 ·p(x) = E [h(X )2 ]−{E [h(X )]}2
) =
D
If h(X ) is linear, i.e. h(X ) = aX + b for some nonrandom constant
a and b, then
2
2
2
V (aX + b) = σaX
+b = a · σX and σaX +b =| a | ·σX
In particular,
σaX =| a | ·σX , σX +b = σX
Expectations
Expectations
Example 3.23 continued
A computer store has purchased three computers of a certain type
at $500 apiece. It will sell them for $1000 apiece. The
manufacturer has agreed to repurchase any computers still unsold
after a specified period at $200 apiece. Let X denote the number
of 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 profit
associated with selling X units, then
h(X ) = revenue − cost
= 1000X + 200(3 − X ) − 1500 = 800X − 900.
Expectations
Example 3.23 continued
A computer store has purchased three computers of a certain type
at $500 apiece. It will sell them for $1000 apiece. The
manufacturer has agreed to repurchase any computers still unsold
after a specified period at $200 apiece. Let X denote the number
of 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 profit
associated with selling X units, then
h(X ) = revenue − cost
= 1000X + 200(3 − X ) − 1500 = 800X − 900.
The variance of h(X ) is
V [h(X )] = V [800X − 900]
= 8002 V [X ]
= 640, 000
And the SD is σh(X ) =
p
V [h(X )] = 800.