Inequality about exponential moment of bounded random variable
$begingroup$
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
edited Jan 12 at 10:35
saz
82.2k862131
asked Jan 10 at 11:00
456 123456 123
1616
$endgroup$
add a comment |
$begingroup$
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
edited Jan 12 at 10:35
saz
82.2k862131
asked Jan 10 at 11:00
456 123456 123
1616
$endgroup$
add a comment |
$begingroup$
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
edited Jan 12 at 10:35
saz
82.2k862131
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
edited Jan 12 at 10:35
saz
82.2k862131
asked Jan 10 at 11:00
456 123456 123
1616
edited Jan 12 at 10:35
saz
82.2k862131
asked Jan 10 at 11:00
456 123456 123
1616
edited Jan 12 at 10:35
saz
82.2k862131
edited Jan 12 at 10:35
saz
82.2k862131
edited Jan 12 at 10:35
saz
82.2k862131
82.2k862131
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
1
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$begingroup$
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.
answered Jan 17 at 17:51
SongSong
18.6k21651
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Jan 17 at 17:51
SongSong
18.6k21651
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 17 at 17:51
SongSong
18.6k21651
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 17 at 17:51
SongSong
18.6k21651
$endgroup$
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.
answered Jan 17 at 17:51
SongSong
18.6k21651
edited Jan 17 at 18:50
answered Jan 17 at 17:51
SongSong
18.6k21651
answered Jan 17 at 17:51
SongSong
18.6k21651
answered Jan 17 at 17:51
SongSong
18.6k21651
18.6k21651
add a comment |
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