概率论论文英文版
月亮洞-
Mathematical booster - probability theory
Abstract:
with
the
rapid
development
and
application
of
probability
theory,
probability theory
gradually applied into the various fields, in all
kinds of trades. In
the
proof of some inequalities in mathematics, it is
difficult to solve the problem if use
the algebraic method, however, the use
of probability method can easily
prove
this,
probability
theory
can
also
be
applied
to
solve
some
problems
in
mathematical
analysis.
Non
negative
to
prove
inequality
by
nature
and
the
variance
probability,
limit the
central limit theorem, and clarify the key using a
probabilistic method is a
stochastic
model, according to different math questions, then
using theorem proving,
showing the
probability method widely and the superiority in
application from.
Keywords:
probability theory, mathematical problem solving,
application,
The text
2
Ef
(
)
f
2
< br>(
x
)
p
(
x
)
dx
application
of
probability
theory
1
b
2
in
the proof of inequality
f
(
x
)
dx
=
a
1.1
with
probability
theory
proof
of <
/p>
b
a
integral inequality
2
2
Eg
(
)
g
1
(
x
)
p
(
x
)
dx
=
Example 1
(triangle inequality)
1
b
2
Let
f (x) and G (x) is [a, the continuous
g
(
x
)
dx<
/p>
positive function on b],
then
b
a
a
{
[f
(x)
g(x)]
dx}
[
f
(
x
)
< br>dx
]
a
a
b
2
1
2
b
2
1<
/p>
2
E
[
f
(
)
g
(
)]
f
(
< br>x
)
g
(
x
)
p
(
x
)
dx
<
/p>
[
g
(
x
)
dx
]
a
b
2
1
2
1
b
f
(
x
)
g
(
x
)
dx
b
a
a
Because it is the continuous
positive function
on[a,b],so by Cauchy
Schwartz inequality:
Proof: probability
distribution and probability
density
function
of
random
variables
respectively:
<
br>)] <
br>,
<
br>) (
0,x
F(x)=
E
[
f
(
)
g
(
)]
[
Ef
(
)]
[
Eg
(
x
Thus:
2
1
2
2
1
2
x
a
x
[
a
,
b
],
b
a
1,x>b.
E
[
f
(
)
g
(
)]
2
Ef
2
(
Eg
2
)
2
E
[
f
(
)
g
(
)]
[
Ef
(
)]
[
Eg
(
)]
2
1
2
2
1
2
1
,
x
[
a
,<
/p>
b
]
b
a
F
（
x
）
0
，
x
[
a
,
b
]
Then
2
The
corresponding
values,
namely
triangle
inequality.
1.2
using
the
probability
method
to
prove
algebraic inequality.
E
[
f
(
)
g
(
)]
[
f
(
x
)
g
(
x
)]
2
p
(
x
)
dx
Example 2
2
：
(mean
inequality)if there is n
1<
/p>
b
[
f
(
x
)
g
(
x
)]
dx
a
b
a
positive, there are: <
/p>
n
n
n
n
a
a
a
1
n
1
1
2
n
n
a
1
2
i<
/p>
n
a
i
,
i
1
i
1
i
1
a
i
(
i
1
,
2
,
<
/p>
,
n
),
Prove:The
probability
distribution
of
random
variables
as
p
(
a
i
)
1
n
.
Among
them
a
i
0
,
(
i
1
,
2
,
,
n
)
By:
E
2
(
)
E
2
(
)
That is
1
n
2
n
n
a
< br>2
i
i
1
a
i
i
1
Make
f<
/p>
(
x
)
ln
x
,
Among
them
ln
x
is
the
Concave
function,
and
because
the
deduction
of Jansen
inequality:
Set
f
=
f(
x)
，
x
∈
(
a
，
b)
is
a
concave
function
of
E
ξ
and
Ef(
ξ
)
existed,then E( f(
ξ
) )
≤
f( E
ξ
).
then:
E
< br>(ln
)
< br>ln
E
(
)
(1)
Then:
1
n
n
a
1
n
ln
i
ln
a
i
,
i
1
n
< br>i
1
That is:
n
a
1
n
1
a
2
a
n
< br>
n
a
i
i
1
In
the
formula
(1),
if
the
Probability
distribution of
random variables
is:
P
(
1
a
)<
/p>
1
,
(
i
1
,
2
,
,
n
),
< br>i
n
there
are
1
n
1
1
n
1
n
ln
ln
1
a
i
n
,
That is:
i
i
1
a
i
n
n
a
n
1
1
a
2
a
n
i<
/p>
1
a
i
n
n
a
1
n
1
n
2
n
1
1
a
2
a
n
<
/p>
n
a
i
i
1
n
a
i
i
1
i
1
a
i
and application of relevant theory.
2.1About
(Weierstrass
inequality)
inference
proof.
Example
3<
/p>
3
:
(corollary
of
the
Weierstrass
inequality)
n
Set
a
k
0
,
(
k
1
,
< br>2
,
,
n
),
and
a
k
1
k
1<
/p>
2
,
Th
en<
/p>
n
(
1
a
1
k
)
.
k
1
2
Prove:
set
event
A
1
,
A
2
,
A
n
are
independent
events,and
P
(
A
k
< br>)
a
k
,
(
k
1
,
2
,
,
n
),
Because
n
n
n
P
(<
/p>
A
k
)
1
P
(
1
a
k
)
1
(
1
a
k
)<
/p>
.
k
1
k
1
k
1
n
n
n
and
P
(
A
k
)
P
(
A
k
)
1<
/p>
a
k
,
k
1
k
k
1
n
n
So there i
s:
1
-
(
1
-
a
k
)<
/p>
a
k
.
That is:
k
1
k
1
n
n
(<
/p>
1
a
k
)
1
a
1
k
k
1
k
1
2
Probability
method
to
prove
inequality
is
generally relatively simple, relatively
algebra.
integral
and
differential
methods
used
should
be
concise and to the point than the image, to
many
problems,
even
to
prove
inequality
is
very complex can also use probability
method
is simple.