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Solutions to Homework Set No. 4 – Probability Theory (235A), Fall 2011
1.
A function
ϕ
: (
a,b
)
→
R
is called
convex
if for any
x,y
∈
(
a,b
) and
α
∈
[0
,
1] we
have
ϕ
(
αx
+ (1

α
)
y
)
≤
αϕ
(
x
) + (1

α
)
ϕ
(
y
)
.
(a) Prove that an equivalent condition for
ϕ
to be convex is that for any
x < z < y
in (
a,b
)
we have
ϕ
(
z
)

ϕ
(
x
)
z

x
≤
ϕ
(
y
)

ϕ
(
z
)
y

z
.
(1)
Deduce using the mean value theorem that if
ϕ
is twice continuously diﬀerentiable and
satisﬁes
ϕ
00
≥
0 then it is convex.
Solution.
Let
x
≤
z
≤
y
and
α
= (
y

z
)
/
(
y

x
) so that
z
=
αx
+ (1

α
)
y
and
α
∈
[0
,
1]. If
ϕ
is convex, then
ϕ
(
z
)
≤
±
y

z
y

x
²
ϕ
(
x
) +
±
z

x
y

x
²
ϕ
(
y
)
,
subtracting
ϕ
(
z
)(
z

x
)
/
(
y

x
) from both sides,
ϕ
(
z
)

ϕ
(
x
)
z

x
≤
ϕ
(
y
)

ϕ
(
z
)
y

z
for any
x
≤
z
≤
y
Reverse process gives the original condition. For the second part of question, use the mean
value theorem for [
x,z
] and [
z,y
] and
ϕ
00
≥
0 to give (1).
(b) Prove
Jensen’s inequality
, which says that if
X
is a random variable such that
P
(
X
∈
(
a,b
)) = 1 and
ϕ
: (
a,b
)
→
R
is convex, then
ϕ
(
E
X
)
≤
E
(
ϕ
(
X
))
.
Hint.
Start by proving the following property of a convex function: If
ϕ
is convex then at
any point
x
0
∈
(
a,b
),
ϕ
has a
supporting line
, that is, a linear function
y
(
x
) =
ax
+
b
such that
y
(
x
0
) =
ϕ
(
x
0
) and such that
ϕ
(
x
)
≥
y
(
x
) for all
x
∈
(
a,b
) (to prove its existence,
use the characterization of convexity from part (a) to show that the leftsided derivative of
ϕ
at
x
0
is less than or equal to the rightsided derivative at
x
0
; the supporting line is a line
1
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View Full Documentpassing through the point (
x
0
,ϕ
(
x
0
)) whose slope lies between these two numbers). Now
take the supporting line function at
x
0
=
E
X
and see what happens.
Solution.
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 Fall '11
 DanRomik
 Probability

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