Stat 116 Homework 4 Solutions
April 23, 2008
1. The Beta(
α, β
) density is
f
(
x
) =
x
α

1
(1

x
)
β

1
B
(
α,β
)
which is proportional to
g
(
x
) =
x
α

1
(1

x
)
β

1
, and
thus has the same extrema. Since beta densities by construction are smooth, the potential extrema are
x
= 0,
x
= 1, and the roots of the derivative
g
(
x
) = (
α

1)
x
α

2
(1

x
)
β

1

(
β

1)(1

x
)
β

2
x
α

1
.
Setting this equal to zero, we get: (
α

1)
x
α

2
(1

x
)
β

1
= (
β

1)(1

x
)
β

2
x
α

1
, which implies that
(
α

1)(1

x
) = (
β

1)
x
, from which we get
x
=
α

1
α
+
β

2
.
(a)
α >
1
, β >
1: Now
α >
1 means that
g
(0) = 0, and
β >
1 means
g
(1) = 0, so the maximum
density cannot be at
x
= 0 or
x
= 1, so it must occur at
x
=
α

1
α
+
β

2
.
(b)
This condition translates to
α
≤
1
, β
≤
1 and either
α
or
β
or both is strictly less than one. First
suppose
α <
1
, β <
1. Now
α <
1 means that
g
(0) =
∞
, and
β <
1 means
g
(1) =
∞
, so that
means
x
= 0 and
x
= 1 are the maxima, and
x
=
α

1
α
+
β

2
is the minimum, so the density must be U
shaped. Alternatively
α
= 1
, β <
1 means that
g
(
x
) = (1

x
)
β

1
, so
g
(
x
) = (1

β
)(1

x
)
β

2
>
0,
so
g
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 Spring '07
 Staff
 Derivative, Taylor Series, Probability, Probability theory, λxk e−λx dx

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