Inequality about exponential moment of bounded random variable

2

$begingroup$

Let $X$ be a random variable with $mathbb{E}X=0$, $-1 le X le 1$, $text{Var} (X)=sigma ^2$.

I've been trying to prove the Bernstein's inequality, and I have to show the following, but don't know how to deal with those $sigma ^2$ on the RHS.

$$mathbb{E}(e^X)le frac{1}{1+sigma ^2} e^{-sigma ^2}+frac{sigma ^2}{1+sigma ^2} e$$

I wish I can get some help.

probability probability-theory inequality random-variables

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edited Jan 12 at 10:35

saz

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asked Jan 10 at 11:00

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$endgroup$

add a comment | 

2

$begingroup$

Let $X$ be a random variable with $mathbb{E}X=0$, $-1 le X le 1$, $text{Var} (X)=sigma ^2$.

I've been trying to prove the Bernstein's inequality, and I have to show the following, but don't know how to deal with those $sigma ^2$ on the RHS.

$$mathbb{E}(e^X)le frac{1}{1+sigma ^2} e^{-sigma ^2}+frac{sigma ^2}{1+sigma ^2} e$$

I wish I can get some help.

probability probability-theory inequality random-variables

share|cite|improve this question

edited Jan 12 at 10:35

saz

80.3k860125

asked Jan 10 at 11:00

456 123456 123

1616

$endgroup$

add a comment | 

2

2

2

0

$begingroup$

Let $X$ be a random variable with $mathbb{E}X=0$, $-1 le X le 1$, $text{Var} (X)=sigma ^2$.

I've been trying to prove the Bernstein's inequality, and I have to show the following, but don't know how to deal with those $sigma ^2$ on the RHS.

$$mathbb{E}(e^X)le frac{1}{1+sigma ^2} e^{-sigma ^2}+frac{sigma ^2}{1+sigma ^2} e$$

I wish I can get some help.

probability probability-theory inequality random-variables

share|cite|improve this question

edited Jan 12 at 10:35

saz

80.3k860125

asked Jan 10 at 11:00

456 123456 123

1616

$endgroup$

Let $X$ be a random variable with $mathbb{E}X=0$, $-1 le X le 1$, $text{Var} (X)=sigma ^2$.

I've been trying to prove the Bernstein's inequality, and I have to show the following, but don't know how to deal with those $sigma ^2$ on the RHS.

$$mathbb{E}(e^X)le frac{1}{1+sigma ^2} e^{-sigma ^2}+frac{sigma ^2}{1+sigma ^2} e$$

I wish I can get some help.

probability probability-theory inequality random-variables

probability probability-theory inequality random-variables

share|cite|improve this question

edited Jan 12 at 10:35

saz

80.3k860125

asked Jan 10 at 11:00

456 123456 123

1616

share|cite|improve this question

edited Jan 12 at 10:35

saz

80.3k860125

asked Jan 10 at 11:00

456 123456 123

1616

share|cite|improve this question

share|cite|improve this question

edited Jan 12 at 10:35

saz

80.3k860125

edited Jan 12 at 10:35

saz

80.3k860125

edited Jan 12 at 10:35

saz

80.3k860125

80.3k860125

asked Jan 10 at 11:00

456 123456 123

1616

asked Jan 10 at 11:00

456 123456 123

1616

asked Jan 10 at 11:00

456 123456 123

1616

1616

add a comment | 

add a comment | 

1 Answer
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0

$begingroup$

Let us define
$$
F(x) = e^x-Ax^2-Bx.
$$ where
$$
A=frac{e-(sigma^2+2)e^{-sigma^2}}{(1+sigma^2)^2},
$$ and
$$
B=frac{2sigma^2e-(sigma^4+2sigma^2-1)e^{-sigma^2}}{(1+sigma^2)^2}.
$$The constants $A,B$ are chosen so that
$$
F(-sigma^2) = F(1), quad F'(-sigma^2)=0, quad F''(-sigma^2)<0.
$$ Since $xmapsto e^x$ is convex and $F''(-sigma^2)<0$, the equation $F'(x) = e^x -2Ax-B=0$ has only $2$ roots $x=-sigma^2$ and $x=x_0>-sigma^2$. Since $$F'(x)begin{cases}>0,quad xin(-infty,-sigma^2)\<0,quad xin (-sigma^2,x_0)\>0,quad xin (x_0,infty)end{cases},$$ it follows that
$$
F(x) le F(-sigma^2)=F(1)quadforall xin [-1,1].
$$ Hence by taking expectation on $F(X)$, we have
$$
E[e^X-AX^2-BX]=E[e^X]-Asigma^2 le F(1),
$$ or equivalently
$$
E[e^X]le Asigma^2 +F(1)= A(sigma^2-1)-B+e.
$$ After some calculation, we get
$$
A(sigma^2-1)-B+e=frac{e^{-sigma^2}}{1+sigma^2}+frac{sigma^2 e}{1+sigma^2}
$$ and the inequality follows.

share|cite|improve this answer

edited Jan 17 at 18:50

answered Jan 17 at 17:51

SongSong

11.7k628

$endgroup$

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

0

$begingroup$

Let us define
$$
F(x) = e^x-Ax^2-Bx.
$$ where
$$
A=frac{e-(sigma^2+2)e^{-sigma^2}}{(1+sigma^2)^2},
$$ and
$$
B=frac{2sigma^2e-(sigma^4+2sigma^2-1)e^{-sigma^2}}{(1+sigma^2)^2}.
$$The constants $A,B$ are chosen so that
$$
F(-sigma^2) = F(1), quad F'(-sigma^2)=0, quad F''(-sigma^2)<0.
$$ Since $xmapsto e^x$ is convex and $F''(-sigma^2)<0$, the equation $F'(x) = e^x -2Ax-B=0$ has only $2$ roots $x=-sigma^2$ and $x=x_0>-sigma^2$. Since $$F'(x)begin{cases}>0,quad xin(-infty,-sigma^2)\<0,quad xin (-sigma^2,x_0)\>0,quad xin (x_0,infty)end{cases},$$ it follows that
$$
F(x) le F(-sigma^2)=F(1)quadforall xin [-1,1].
$$ Hence by taking expectation on $F(X)$, we have
$$
E[e^X-AX^2-BX]=E[e^X]-Asigma^2 le F(1),
$$ or equivalently
$$
E[e^X]le Asigma^2 +F(1)= A(sigma^2-1)-B+e.
$$ After some calculation, we get
$$
A(sigma^2-1)-B+e=frac{e^{-sigma^2}}{1+sigma^2}+frac{sigma^2 e}{1+sigma^2}
$$ and the inequality follows.

share|cite|improve this answer

edited Jan 17 at 18:50

answered Jan 17 at 17:51

SongSong

11.7k628

$endgroup$

add a comment | 

0

$begingroup$

Let us define
$$
F(x) = e^x-Ax^2-Bx.
$$ where
$$
A=frac{e-(sigma^2+2)e^{-sigma^2}}{(1+sigma^2)^2},
$$ and
$$
B=frac{2sigma^2e-(sigma^4+2sigma^2-1)e^{-sigma^2}}{(1+sigma^2)^2}.
$$The constants $A,B$ are chosen so that
$$
F(-sigma^2) = F(1), quad F'(-sigma^2)=0, quad F''(-sigma^2)<0.
$$ Since $xmapsto e^x$ is convex and $F''(-sigma^2)<0$, the equation $F'(x) = e^x -2Ax-B=0$ has only $2$ roots $x=-sigma^2$ and $x=x_0>-sigma^2$. Since $$F'(x)begin{cases}>0,quad xin(-infty,-sigma^2)\<0,quad xin (-sigma^2,x_0)\>0,quad xin (x_0,infty)end{cases},$$ it follows that
$$
F(x) le F(-sigma^2)=F(1)quadforall xin [-1,1].
$$ Hence by taking expectation on $F(X)$, we have
$$
E[e^X-AX^2-BX]=E[e^X]-Asigma^2 le F(1),
$$ or equivalently
$$
E[e^X]le Asigma^2 +F(1)= A(sigma^2-1)-B+e.
$$ After some calculation, we get
$$
A(sigma^2-1)-B+e=frac{e^{-sigma^2}}{1+sigma^2}+frac{sigma^2 e}{1+sigma^2}
$$ and the inequality follows.

share|cite|improve this answer

edited Jan 17 at 18:50

answered Jan 17 at 17:51

SongSong

11.7k628

$endgroup$

add a comment | 

0

0

0

$begingroup$

Let us define
$$
F(x) = e^x-Ax^2-Bx.
$$ where
$$
A=frac{e-(sigma^2+2)e^{-sigma^2}}{(1+sigma^2)^2},
$$ and
$$
B=frac{2sigma^2e-(sigma^4+2sigma^2-1)e^{-sigma^2}}{(1+sigma^2)^2}.
$$The constants $A,B$ are chosen so that
$$
F(-sigma^2) = F(1), quad F'(-sigma^2)=0, quad F''(-sigma^2)<0.
$$ Since $xmapsto e^x$ is convex and $F''(-sigma^2)<0$, the equation $F'(x) = e^x -2Ax-B=0$ has only $2$ roots $x=-sigma^2$ and $x=x_0>-sigma^2$. Since $$F'(x)begin{cases}>0,quad xin(-infty,-sigma^2)\<0,quad xin (-sigma^2,x_0)\>0,quad xin (x_0,infty)end{cases},$$ it follows that
$$
F(x) le F(-sigma^2)=F(1)quadforall xin [-1,1].
$$ Hence by taking expectation on $F(X)$, we have
$$
E[e^X-AX^2-BX]=E[e^X]-Asigma^2 le F(1),
$$ or equivalently
$$
E[e^X]le Asigma^2 +F(1)= A(sigma^2-1)-B+e.
$$ After some calculation, we get
$$
A(sigma^2-1)-B+e=frac{e^{-sigma^2}}{1+sigma^2}+frac{sigma^2 e}{1+sigma^2}
$$ and the inequality follows.

share|cite|improve this answer

edited Jan 17 at 18:50

answered Jan 17 at 17:51

SongSong

11.7k628

$endgroup$

Let us define
$$
F(x) = e^x-Ax^2-Bx.
$$ where
$$
A=frac{e-(sigma^2+2)e^{-sigma^2}}{(1+sigma^2)^2},
$$ and
$$
B=frac{2sigma^2e-(sigma^4+2sigma^2-1)e^{-sigma^2}}{(1+sigma^2)^2}.
$$The constants $A,B$ are chosen so that
$$
F(-sigma^2) = F(1), quad F'(-sigma^2)=0, quad F''(-sigma^2)<0.
$$ Since $xmapsto e^x$ is convex and $F''(-sigma^2)<0$, the equation $F'(x) = e^x -2Ax-B=0$ has only $2$ roots $x=-sigma^2$ and $x=x_0>-sigma^2$. Since $$F'(x)begin{cases}>0,quad xin(-infty,-sigma^2)\<0,quad xin (-sigma^2,x_0)\>0,quad xin (x_0,infty)end{cases},$$ it follows that
$$
F(x) le F(-sigma^2)=F(1)quadforall xin [-1,1].
$$ Hence by taking expectation on $F(X)$, we have
$$
E[e^X-AX^2-BX]=E[e^X]-Asigma^2 le F(1),
$$ or equivalently
$$
E[e^X]le Asigma^2 +F(1)= A(sigma^2-1)-B+e.
$$ After some calculation, we get
$$
A(sigma^2-1)-B+e=frac{e^{-sigma^2}}{1+sigma^2}+frac{sigma^2 e}{1+sigma^2}
$$ and the inequality follows.

share|cite|improve this answer

edited Jan 17 at 18:50

answered Jan 17 at 17:51

SongSong

11.7k628

share|cite|improve this answer

share|cite|improve this answer

edited Jan 17 at 18:50

edited Jan 17 at 18:50

edited Jan 17 at 18:50

answered Jan 17 at 17:51

SongSong

11.7k628

answered Jan 17 at 17:51

SongSong

11.7k628

answered Jan 17 at 17:51

SongSong

11.7k628

11.7k628

add a comment | 

add a comment | 

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