Statistics 403 Problem Set 2
Due in lab on Friday, September 24th
1. We are given the following probability distribution, where
p
is an unspecified proba
bility:
x
1
0
1
P
(
X
=
x
)
0.3
p
0
.
7

p
(a) Find the value of
p
that maximizes the variance of this distribution.
Solution:
The expected value is
(0
.
3
× 
1) + (
p
×
0) + ((0
.
7

p
)
×
1) = 0
.
4

p.
The direct way to get the variance is from the definition:
0
.
3
·
(

1

(0
.
4

p
))
2
+
p
·
(0

(0
.
4

p
))
2
+ (0
.
7

p
)
·
(1

(0
.
4

p
))
2
If you expand this out and simplify you will get

p
2

0
.
2
p
+ 0
.
84
You can also use the identity var(
X
) =
E
(
X
2
)

(
EX
)
2
:
E
(
X
2
) = (

1)
2
·
0
.
3 + 0
2
·
p
+ 1
2
·
(0
.
7

p
) = 1

p
Thus the variance is
1

p

(0
.
4

p
)
2
=

p
2

0
.
2
p
+ 0
.
84
This is a concave parabola with a global maximum value at
p
=

0
.
1. Since a
negative number is not possible as a probability, we can conclude that there is
no global maximum within the allowed range, which is the interval (0
,
0
.
7). Thus
we only need to check the endpoints of this interval, which gives us values 0
.
84
(when
p
= 0) and 0.21 (when
p
= 0
.
7). So 0.21 is the minimum possible variance.
Note that the minimum variance occurs when the point
x
= 1 is eliminated from
the sample space. So we see that the variance is minimized when we squeeze as
much probability mass as possible into a small range.
1
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(b) Find the value of
p
that minimizes the variance of this distribution.
Solution:
Following the work from part (a), 0
.
84 is the maximum possible vari
ance.
This occurs when when
p
= 0, which eliminates
x
= 0 from the sample
space. Thus the variance is maximized when we stretch the probability mass over
as wide a range as possible.
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 Winter '12
 KerbyShedden
 Statistics, Probability, Variance, Probability theory, probability density function

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