5.6 Expected Values
35
However you rationalize it though, not only are lottery players guaranteed to
lose in the long run, the odds are that they will lose in the short run too. The
Kansas State instant lottery of Review Exercises 4 Problem 19, returns about
47 cents in the dollar to the players. Wasserstein and Boyer [1990] worked out
the chances of winning more than you lose after playing this game
n
times.
The chances are best after 10 games at which point roughly one person in 6
will be ahead of the game.
But after 100 games, only one person in 50 will
have made a profit.
Exercises on Section 5.6.1
1.Suppose a random variableXhas probability functionx2357pr(x)0.20.10.30.4FindμX, the expected value ofX.2.ComputeμX=E(X) whereX∼Binomial(n= 2, p= 0.3).5.6.2Population standard deviation
2]Just as with “population mean” and “sample mean”, we use the wordpopulationin the term “population standard deviation” to distinguish it from “samplestandard deviation”, which is the standard deviation of a batch of numbers.Thepopulationstandard deviation is the standard deviation of a distribution.It tells you about the spread of the distribution, or equivalently, how variablethe random quantity is.To compute sd(X), we need to be able to compute E[(Xμ)2]. How are weto do this? For discrete distributions, just as we compute E(
36
Discrete Random Variables
[This is part of a general pattern, E(
X
2
) = Σ
x
2
i
pr(
x
i
), E(
e
X
) = Σ
e
x
i
pr(
x
i
), or
generally for any function
h
(
X
) we compute E[
h
(
X
)] = Σ
h
(
x
i
)pr(
x
i
).]
We now illustrate the use of the formula for E[(
X

μ
)
2
].
Example 5.6.3
In Example 5.6.1,
X
has probability function
x
0
1
2
3
pr(
x
)
1
8
5
8
1
8
1
8
and we calculated that
μ
= E(
X
) = 1
.
25. Thus,
E[(
X

μ
)
2
] =
X
(
x
i

μ
)
2
pr(
x
i
)
= (0

1
.
25)
2
1
8
+ (1

1
.
25)
2
5
8
+ (2

1
.
25)
2
1
8
+ (3

1
.
25)
2
1
8
= 0
.
6875
and sd(
X
) =
p
E[(
X

μ
)
2
] =
√
0
.
6875 = 0
.
8292.
Example 5.6.4
If
X
∼
Binomial(
n
= 3
, p
= 0
.
1) then
X
can take values
0
,
1
,
2
,
3 and we calculated in Example 5.6.2 that
μ
= E(
X
) = 0
.
3. Then
E[(
X

μ
)
2
] =
X
(
x
i

μ
)
2
pr(
x
i
)
= (0

0
.
3)
2
pr(0) + (1

0
.
3)
2
pr(1) + (2

0
.
3)
2
pr(2)
+ (3

0
.
3)
2
pr(3)
= 0
.
27
.
Thus, sd(
X
) =
p
E[(
X

μ
)
2
] =
√
0
.
27 = 0
.
5196.
[Note that the formulae for pr(0), pr(1), etc. are given in Example 5.6.2.]
Exercises 5.6.2
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 Probability theory, Binomial distribution, Discrete probability distribution, Hypergeometric