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COMP9020 23T1
Week 10
Expected Values, Course Review
Textbook (R & W) - Ch. 9, Sec. 9.2-9.4
Problem set week 10
1
Random Variables
Definition
An (integer) random variable is a function from Ω to Z.
In other words, it associates a number value with every outcome.
Random variables are often denoted by X ,Y ,Z , . . .
2
Example
Random variable Xs
def
= sum of rolling two dice
Ω = {(1, 1), (1, 2), . . . , (6, 6)}
Xs((1, 1)) = 2 Xs((1, 2)) = 3 = Xs((2, 1)) . . .
Example
9.3.3 Buy one lottery ticket for $1. The only prize is $1M.
Ω = {win, lose} XL(win) = $999, 999 XL(lose) = −$1
3
Expectation
Definition
The expected value (often called “expectation” or “average”) of
a random variable X is
E (X ) =
∑
k∈Z
P(X = k) · k
4
Example
The expected sum when rolling two dice is
E (Xs) =
1
36
· 2 + 2
36
· 3 + . . .+ 6
36
· 7 + . . .+ 1
36
· 12 = 7
Example
9.3.3 Buy one lottery ticket for $1. The only prize is $1M. Each
ticket has probability 6 · 10−7 of winning.
E (XL) = 6 · 10−7 · $999, 999 + (1− 6 · 10−7) · −$1 = −$0.4
NB
Expectation is a truly universal concept; it is the basis of all
decision making, of estimating gains and losses, in all actions
under risk. Historically, a rudimentary concept of expected value
arose long before the notion of probability.
5
Theorem (linearity of expected value)
E (X + Y ) = E (X ) + E (Y )
E (c · X ) = c · E (X )
Example
The expected sum when rolling two dice can be computed as
E (Xs) = E (X1) + E (X2) = 3.5 + 3.5 = 7
since E (Xi ) =
1
6 · 1 + 16 · 2 + . . .+ 16 · 6, for each die Xi
6
Example
E (Sn), where Sn
def
= |no. of heads in n tosses|
‘hard way’
E (Sn) =
∑n
k=0 P(Sn = k) · k =
∑n
k=0
1
2n
(n
k
) · k
since there are
(n
k
)
sequences of n tosses with k heads,
and each sequence has the probability 12n
= 12n
∑n
k=1
n
k
(n−1
k−1
)
k = n2n
∑n−1
k=0
(n−1
k
)
= n2n · 2n−1 = n2
using the ‘binomial identity’
∑n
k=0
(n
k
)
= 2n
‘easy way’
E (Sn) = E (S
1
1 + . . .+ S
n
1 ) =
∑
i=1...n E (S
i
1) = nE (S1) = n · 12
Note: Sn
def
= |heads in n tosses| while each S i1 def= |heads in 1 toss|
7
Example
E (Sn), where Sn
def
= |no. of heads in n tosses|
‘hard way’
E (Sn) =
∑n
k=0 P(Sn = k) · k =
∑n
k=0
1
2n
(n
k
) · k
since there are
(n
k
)
sequences of n tosses with k heads,
and each sequence has the probability 12n
= 12n
∑n
k=1
n
k
(n−1
k−1
)
k = n2n
∑n−1
k=0
(n−1
k
)
= n2n · 2n−1 = n2
using the ‘binomial identity’
∑n
k=0
(n
k
)
= 2n
‘easy way’
E (Sn) = E (S
1
1 + . . .+ S
n
1 ) =
∑
i=1...n E (S
i
1) = nE (S1) = n · 12
Note: Sn
def
= |heads in n tosses| while each S i1 def= |heads in 1 toss|
8
Example
E (Sn), where Sn
def
= |no. of heads in n tosses|
‘hard way’
E (Sn) =
∑n
k=0 P(Sn = k) · k =
∑n
k=0
1
2n
(n
k
) · k
since there are
(n
k
)
sequences of n tosses with k heads,
and each sequence has the probability 12n
= 12n
∑n
k=1
n
k
(n−1
k−1
)
k = n2n
∑n−1
k=0
(n−1
k
)
= n2n · 2n−1 = n2
using the ‘binomial identity’
∑n
k=0
(n
k
)
= 2n
‘easy way’
E (Sn) = E (S
1
1 + . . .+ S
n
1 ) =
∑
i=1...n E (S
i
1) = nE (S1) = n · 12
Note: Sn
def
= |heads in n tosses| while each S i1 def= |heads in 1 toss|
9
NB
If X1,X2, . . . ,Xn are independent, identically distributed random
variables, then E (X1 + X2 + . . .+ Xn) happens to be the same as
E (nX1), but these are very different random variables.
10
Example
You face a quiz consisting of six true/false questions, and your
plan is to guess the answer to each question (randomly, with
probability 0.5 of being right). There are no negative marks, and
answering four or more questions correctly suffices to pass.
What is the probability of passing and what is the expected score?
To pass you would need four, five or six correct guesses. Therefore,
p(pass) =
(6
4
)
+
(6
5
)
+
(6
6
)
64
=
15 + 6 + 1
64
≈ 34%
The expected score from a single question is 0.5, as there is no
penalty for errors. For six questions the expected value is 6 ·0.5 = 3
11
Example
You face a quiz consisting of six true/false questions, and your
plan is to guess the answer to each question (randomly, with
probability 0.5 of being right). There are no negative marks, and
answering four or more questions correctly suffices to pass.
What is the probability of passing and what is the expected score?
To pass you would need four, five or six correct guesses. Therefore,
p(pass) =
(6
4
)
+
(6
5
)
+
(6
6
)
64
=
15 + 6 + 1
64
≈ 34%
The expected score from a single question is 0.5, as there is no
penalty for errors. For six questions the expected value is 6 ·0.5 = 3
12
Exercise
9.3.7
An urn has m + n = 10 marbles, m ≥ 0 red and n ≥ 0 blue.
7 marbles selected at random without replacement.
What is the expected number of red marbles drawn?(m
0
)(n
7
)(10
7
) · 0 + (m1)(n6)(10
7
) · 1 + (m2)(n5)(10
7
) · 2 + . . .+ (m7)(n0)(10
7
) · 7
e.g. (5
2
)(5
5
)(10
7
) · 2 + (53)(54)(10
7
) · 3 + (54)(53)(10
7
) · 4 + (55)(52)(10
7
) · 5
=
10
120
· 2 + 50
120
· 3 + 50
120
· 4 + 10
120
· 5 = 420
120
= 3.5
13
Exercise
9.3.7
An urn has m + n = 10 marbles, m ≥ 0 red and n ≥ 0 blue.
7 marbles selected at random without replacement.
What is the expected number of red marbles drawn?(m
0
)(n
7
)(10
7
) · 0 + (m1)(n6)(10
7
) · 1 + (m2)(n5)(10
7
) · 2 + . . .+ (m7)(n0)(10
7
) · 7
e.g. (5
2
)(5
5
)(10
7
) · 2 + (53)(54)(10
7
) · 3 + (54)(53)(10
7
) · 4 + (55)(52)(10
7
) · 5
=
10
120
· 2 + 50
120
· 3 + 50
120
· 4 + 10
120
· 5 = 420
120
= 3.5
14
Example
Find the average waiting time for the first head, with no upper
bound on the ‘duration’ (one allows for all possible sequences of
tosses, regardless of how many times tails occur initially).
A = E (Xw ) =
∑∞
k=1 k · P(Xw = k) =
∑∞
k=1 k
1
2k
= 1
21
+ 2
22
+ 3
23
+ . . .
This can be evaluated by breaking the sum into a sequence of
geometric progressions
1
2
+
2
22
+
3
23
+ . . .
=
(
1
2
+
1
22
+
1
23
+ . . .
)
+
(
1
22
+
1
23
+ . . .
)
+
(
1
23
+ . . .
)
+ . . .
= 1 +
1
2
+
1
22
+ . . . = 2
15
Example
Find the average waiting time for the first head, with no upper
bound on the ‘duration’ (one allows for all possible sequences of
tosses, regardless of how many times tails occur initially).