Mixing
Cramer Rao lower bound in Cauchy distribution
$begingroup$
I need to calculate the Cramer Rao lower bound of variance for the parameter $theta$ of the distribution $$f(x)=frac{1}{pi(1+(x-theta)^2)}$$
How do I proceed I have calculated $$4 Efrac{(X-theta)^2}{1+X^2+theta^2-2Xtheta}$$
Can somebody help
statistics statistical-inference
edited May 22 '18 at 18:14
StubbornAtom
6,16811339
asked Feb 23 '17 at 18:57
UpstartUpstart
1,706617
$endgroup$
add a comment |
$begingroup$
I need to calculate the Cramer Rao lower bound of variance for the parameter $theta$ of the distribution $$f(x)=frac{1}{pi(1+(x-theta)^2)}$$
How do I proceed I have calculated $$4 Efrac{(X-theta)^2}{1+X^2+theta^2-2Xtheta}$$
Can somebody help
statistics statistical-inference
edited May 22 '18 at 18:14
StubbornAtom
6,16811339
asked Feb 23 '17 at 18:57
UpstartUpstart
1,706617
$endgroup$
add a comment |
$begingroup$
I need to calculate the Cramer Rao lower bound of variance for the parameter $theta$ of the distribution $$f(x)=frac{1}{pi(1+(x-theta)^2)}$$
How do I proceed I have calculated $$4 Efrac{(X-theta)^2}{1+X^2+theta^2-2Xtheta}$$
Can somebody help
statistics statistical-inference
edited May 22 '18 at 18:14
StubbornAtom
6,16811339
asked Feb 23 '17 at 18:57
UpstartUpstart
1,706617
$endgroup$
I need to calculate the Cramer Rao lower bound of variance for the parameter $theta$ of the distribution $$f(x)=frac{1}{pi(1+(x-theta)^2)}$$
How do I proceed I have calculated $$4 Efrac{(X-theta)^2}{1+X^2+theta^2-2Xtheta}$$
Can somebody help
statistics statistical-inference
statistics statistical-inference
edited May 22 '18 at 18:14
StubbornAtom
6,16811339
asked Feb 23 '17 at 18:57
UpstartUpstart
1,706617
edited May 22 '18 at 18:14
StubbornAtom
6,16811339
asked Feb 23 '17 at 18:57
UpstartUpstart
1,706617
edited May 22 '18 at 18:14
StubbornAtom
6,16811339
edited May 22 '18 at 18:14
StubbornAtom
6,16811339
edited May 22 '18 at 18:14
StubbornAtom
6,16811339
6,16811339
asked Feb 23 '17 at 18:57
UpstartUpstart
1,706617
asked Feb 23 '17 at 18:57
UpstartUpstart
1,706617
asked Feb 23 '17 at 18:57
UpstartUpstart
1,706617
1,706617
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is easier to work with a single observation $X_1$ to find the information function $I(theta)$.
This is because we have the alternative expression $displaystyle I(theta)=n,text{E}_{theta}left[frac{partial}{partialtheta}ln f_{theta}(X_1)right]^2$.
For $x_1inmathbb R$ and $thetainmathbb R$, one has the density of $X_1$
begin{align}&f_{theta}(x_1)=frac{1}{pi((x_1-theta)^2+1)}\&vdots\&implies left[frac{partial}{partialtheta}ln f_{theta}(x_1)right]^2=4left[frac{x_1-theta}{1+(x_1-theta)^2}right]^2\&implies text{E}_{theta}left[frac{partial}{partialtheta}ln f_{theta}(X_1)right]^2=4text{E}_{theta}left[frac{X_1-theta}{1+(X_1-theta)^2}right]^2tag{1}end{align}
Now, begin{align}text{E}left[frac{X_1-theta}{1+(X_1-theta)^2}right]^2&=frac{1}{pi}int_{mathbb R}left[frac{x-theta}{1+(x-theta)^2}right]^2frac{1}{1+(x-theta)^2},mathrm{d}x\&=frac{1}{pi}int_{mathbb R}frac{(x-theta)^2}{(1+(x-theta)^2)^3},mathrm{d}x\&=frac{2}{pi}int_0^inftyfrac{t^2}{(1+t^2)^3},mathrm{d}t\&=frac{1}{pi}int_0^inftyfrac{sqrt u}{(1+u)^3},mathrm{d}u\&=frac{1}{pi}Bleft(frac{3}{2},frac{3}{2}right)=frac{1}{8}end{align}
That is, from $(1)$, we have $displaystyletext{E}_{theta}left[frac{partial}{partialtheta}ln f_{theta}(X_1)right]^2=4timesfrac{1}{8}=frac{1}{2}quadforall,theta$.
So finally we get $displaystyle I(theta)=frac{n}{2}quadforall,theta$
The Cramer-Rao lower bound for $theta$ is then $1/I(theta)=2/nquadforall,theta$.
answered May 22 '18 at 16:58
StubbornAtomStubbornAtom
6,16811339
$endgroup$
add a comment |
$begingroup$
First of all you should notice that there's no suficient estimator for the center of the bell $theta$. Let's see this.
The likelihood for the Cauchy distribution is
$$L(x;theta) = prod _i^n frac{1}{pileft [ 1+(x_i-theta)^2 right ]}, $$
and its logarithm is
$$ln L(x;theta) = -n ln pi -sum_i^nlnleft [ 1+(x_i-theta)^2 right ].$$
The estimator will maximize the likelihood and if there's a suficient estimator it's derivate can be factoriced i.e.:
$$frac{partial L(x;theta)}{partial theta} = A(theta)left[t(x)-h(theta)-b(theta)right], $$
where $A(theta)$ is a function exclusively from the parameter, $t(x)$ is function exclusively of your data, $h(theta$) is what you want to estimate and $b(theta)$ is a possible bias.
If you derivate you should notice that the Cauchy distribution can not be factorized, but Cramer-Rao lets you find a bound for the variance, that is
$$ Var(t) geq frac{left(frac{partial}{partial theta}(h+b)right)^2}{Eleft[(frac{partial}{partial theta}ln L)^2right]},$$
where the equality hold only if $frac{partial ln L}{partial theta}$ can be factorized.
The most you can do is calculate the bound but it has no analytic closed solution for $theta$
answered Jun 20 '17 at 15:33
DanielDaniel
14
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
It is easier to work with a single observation $X_1$ to find the information function $I(theta)$.
This is because we have the alternative expression $displaystyle I(theta)=n,text{E}_{theta}left[frac{partial}{partialtheta}ln f_{theta}(X_1)right]^2$.
For $x_1inmathbb R$ and $thetainmathbb R$, one has the density of $X_1$
begin{align}&f_{theta}(x_1)=frac{1}{pi((x_1-theta)^2+1)}\&vdots\&implies left[frac{partial}{partialtheta}ln f_{theta}(x_1)right]^2=4left[frac{x_1-theta}{1+(x_1-theta)^2}right]^2\&implies text{E}_{theta}left[frac{partial}{partialtheta}ln f_{theta}(X_1)right]^2=4text{E}_{theta}left[frac{X_1-theta}{1+(X_1-theta)^2}right]^2tag{1}end{align}
Now, begin{align}text{E}left[frac{X_1-theta}{1+(X_1-theta)^2}right]^2&=frac{1}{pi}int_{mathbb R}left[frac{x-theta}{1+(x-theta)^2}right]^2frac{1}{1+(x-theta)^2},mathrm{d}x\&=frac{1}{pi}int_{mathbb R}frac{(x-theta)^2}{(1+(x-theta)^2)^3},mathrm{d}x\&=frac{2}{pi}int_0^inftyfrac{t^2}{(1+t^2)^3},mathrm{d}t\&=frac{1}{pi}int_0^inftyfrac{sqrt u}{(1+u)^3},mathrm{d}u\&=frac{1}{pi}Bleft(frac{3}{2},frac{3}{2}right)=frac{1}{8}end{align}
That is, from $(1)$, we have $displaystyletext{E}_{theta}left[frac{partial}{partialtheta}ln f_{theta}(X_1)right]^2=4timesfrac{1}{8}=frac{1}{2}quadforall,theta$.
So finally we get $displaystyle I(theta)=frac{n}{2}quadforall,theta$
The Cramer-Rao lower bound for $theta$ is then $1/I(theta)=2/nquadforall,theta$.
answered May 22 '18 at 16:58
StubbornAtomStubbornAtom
6,16811339
$endgroup$
add a comment |
$begingroup$
It is easier to work with a single observation $X_1$ to find the information function $I(theta)$.
This is because we have the alternative expression $displaystyle I(theta)=n,text{E}_{theta}left[frac{partial}{partialtheta}ln f_{theta}(X_1)right]^2$.
For $x_1inmathbb R$ and $thetainmathbb R$, one has the density of $X_1$
begin{align}&f_{theta}(x_1)=frac{1}{pi((x_1-theta)^2+1)}\&vdots\&implies left[frac{partial}{partialtheta}ln f_{theta}(x_1)right]^2=4left[frac{x_1-theta}{1+(x_1-theta)^2}right]^2\&implies text{E}_{theta}left[frac{partial}{partialtheta}ln f_{theta}(X_1)right]^2=4text{E}_{theta}left[frac{X_1-theta}{1+(X_1-theta)^2}right]^2tag{1}end{align}
Now, begin{align}text{E}left[frac{X_1-theta}{1+(X_1-theta)^2}right]^2&=frac{1}{pi}int_{mathbb R}left[frac{x-theta}{1+(x-theta)^2}right]^2frac{1}{1+(x-theta)^2},mathrm{d}x\&=frac{1}{pi}int_{mathbb R}frac{(x-theta)^2}{(1+(x-theta)^2)^3},mathrm{d}x\&=frac{2}{pi}int_0^inftyfrac{t^2}{(1+t^2)^3},mathrm{d}t\&=frac{1}{pi}int_0^inftyfrac{sqrt u}{(1+u)^3},mathrm{d}u\&=frac{1}{pi}Bleft(frac{3}{2},frac{3}{2}right)=frac{1}{8}end{align}
That is, from $(1)$, we have $displaystyletext{E}_{theta}left[frac{partial}{partialtheta}ln f_{theta}(X_1)right]^2=4timesfrac{1}{8}=frac{1}{2}quadforall,theta$.
So finally we get $displaystyle I(theta)=frac{n}{2}quadforall,theta$
The Cramer-Rao lower bound for $theta$ is then $1/I(theta)=2/nquadforall,theta$.
answered May 22 '18 at 16:58
StubbornAtomStubbornAtom
6,16811339
$endgroup$
add a comment |
$begingroup$
It is easier to work with a single observation $X_1$ to find the information function $I(theta)$.
This is because we have the alternative expression $displaystyle I(theta)=n,text{E}_{theta}left[frac{partial}{partialtheta}ln f_{theta}(X_1)right]^2$.
For $x_1inmathbb R$ and $thetainmathbb R$, one has the density of $X_1$
begin{align}&f_{theta}(x_1)=frac{1}{pi((x_1-theta)^2+1)}\&vdots\&implies left[frac{partial}{partialtheta}ln f_{theta}(x_1)right]^2=4left[frac{x_1-theta}{1+(x_1-theta)^2}right]^2\&implies text{E}_{theta}left[frac{partial}{partialtheta}ln f_{theta}(X_1)right]^2=4text{E}_{theta}left[frac{X_1-theta}{1+(X_1-theta)^2}right]^2tag{1}end{align}
Now, begin{align}text{E}left[frac{X_1-theta}{1+(X_1-theta)^2}right]^2&=frac{1}{pi}int_{mathbb R}left[frac{x-theta}{1+(x-theta)^2}right]^2frac{1}{1+(x-theta)^2},mathrm{d}x\&=frac{1}{pi}int_{mathbb R}frac{(x-theta)^2}{(1+(x-theta)^2)^3},mathrm{d}x\&=frac{2}{pi}int_0^inftyfrac{t^2}{(1+t^2)^3},mathrm{d}t\&=frac{1}{pi}int_0^inftyfrac{sqrt u}{(1+u)^3},mathrm{d}u\&=frac{1}{pi}Bleft(frac{3}{2},frac{3}{2}right)=frac{1}{8}end{align}
That is, from $(1)$, we have $displaystyletext{E}_{theta}left[frac{partial}{partialtheta}ln f_{theta}(X_1)right]^2=4timesfrac{1}{8}=frac{1}{2}quadforall,theta$.
So finally we get $displaystyle I(theta)=frac{n}{2}quadforall,theta$
The Cramer-Rao lower bound for $theta$ is then $1/I(theta)=2/nquadforall,theta$.
answered May 22 '18 at 16:58
StubbornAtomStubbornAtom
6,16811339
$endgroup$
It is easier to work with a single observation $X_1$ to find the information function $I(theta)$.
This is because we have the alternative expression $displaystyle I(theta)=n,text{E}_{theta}left[frac{partial}{partialtheta}ln f_{theta}(X_1)right]^2$.
For $x_1inmathbb R$ and $thetainmathbb R$, one has the density of $X_1$
begin{align}&f_{theta}(x_1)=frac{1}{pi((x_1-theta)^2+1)}\&vdots\&implies left[frac{partial}{partialtheta}ln f_{theta}(x_1)right]^2=4left[frac{x_1-theta}{1+(x_1-theta)^2}right]^2\&implies text{E}_{theta}left[frac{partial}{partialtheta}ln f_{theta}(X_1)right]^2=4text{E}_{theta}left[frac{X_1-theta}{1+(X_1-theta)^2}right]^2tag{1}end{align}
Now, begin{align}text{E}left[frac{X_1-theta}{1+(X_1-theta)^2}right]^2&=frac{1}{pi}int_{mathbb R}left[frac{x-theta}{1+(x-theta)^2}right]^2frac{1}{1+(x-theta)^2},mathrm{d}x\&=frac{1}{pi}int_{mathbb R}frac{(x-theta)^2}{(1+(x-theta)^2)^3},mathrm{d}x\&=frac{2}{pi}int_0^inftyfrac{t^2}{(1+t^2)^3},mathrm{d}t\&=frac{1}{pi}int_0^inftyfrac{sqrt u}{(1+u)^3},mathrm{d}u\&=frac{1}{pi}Bleft(frac{3}{2},frac{3}{2}right)=frac{1}{8}end{align}
That is, from $(1)$, we have $displaystyletext{E}_{theta}left[frac{partial}{partialtheta}ln f_{theta}(X_1)right]^2=4timesfrac{1}{8}=frac{1}{2}quadforall,theta$.
So finally we get $displaystyle I(theta)=frac{n}{2}quadforall,theta$
The Cramer-Rao lower bound for $theta$ is then $1/I(theta)=2/nquadforall,theta$.
answered May 22 '18 at 16:58
StubbornAtomStubbornAtom
6,16811339
answered May 22 '18 at 16:58
StubbornAtomStubbornAtom
6,16811339
answered May 22 '18 at 16:58
StubbornAtomStubbornAtom
6,16811339
answered May 22 '18 at 16:58
StubbornAtomStubbornAtom
6,16811339
6,16811339
add a comment |
add a comment |
$begingroup$
First of all you should notice that there's no suficient estimator for the center of the bell $theta$. Let's see this.
The likelihood for the Cauchy distribution is
$$L(x;theta) = prod _i^n frac{1}{pileft [ 1+(x_i-theta)^2 right ]}, $$
and its logarithm is
$$ln L(x;theta) = -n ln pi -sum_i^nlnleft [ 1+(x_i-theta)^2 right ].$$
The estimator will maximize the likelihood and if there's a suficient estimator it's derivate can be factoriced i.e.:
$$frac{partial L(x;theta)}{partial theta} = A(theta)left[t(x)-h(theta)-b(theta)right], $$
where $A(theta)$ is a function exclusively from the parameter, $t(x)$ is function exclusively of your data, $h(theta$) is what you want to estimate and $b(theta)$ is a possible bias.
If you derivate you should notice that the Cauchy distribution can not be factorized, but Cramer-Rao lets you find a bound for the variance, that is
$$ Var(t) geq frac{left(frac{partial}{partial theta}(h+b)right)^2}{Eleft[(frac{partial}{partial theta}ln L)^2right]},$$
where the equality hold only if $frac{partial ln L}{partial theta}$ can be factorized.
The most you can do is calculate the bound but it has no analytic closed solution for $theta$
answered Jun 20 '17 at 15:33
DanielDaniel
14
$endgroup$
add a comment |
$begingroup$
First of all you should notice that there's no suficient estimator for the center of the bell $theta$. Let's see this.
The likelihood for the Cauchy distribution is
$$L(x;theta) = prod _i^n frac{1}{pileft [ 1+(x_i-theta)^2 right ]}, $$
and its logarithm is
$$ln L(x;theta) = -n ln pi -sum_i^nlnleft [ 1+(x_i-theta)^2 right ].$$
The estimator will maximize the likelihood and if there's a suficient estimator it's derivate can be factoriced i.e.:
$$frac{partial L(x;theta)}{partial theta} = A(theta)left[t(x)-h(theta)-b(theta)right], $$
where $A(theta)$ is a function exclusively from the parameter, $t(x)$ is function exclusively of your data, $h(theta$) is what you want to estimate and $b(theta)$ is a possible bias.
If you derivate you should notice that the Cauchy distribution can not be factorized, but Cramer-Rao lets you find a bound for the variance, that is
$$ Var(t) geq frac{left(frac{partial}{partial theta}(h+b)right)^2}{Eleft[(frac{partial}{partial theta}ln L)^2right]},$$
where the equality hold only if $frac{partial ln L}{partial theta}$ can be factorized.
The most you can do is calculate the bound but it has no analytic closed solution for $theta$
answered Jun 20 '17 at 15:33
DanielDaniel
14
$endgroup$
add a comment |
$begingroup$
First of all you should notice that there's no suficient estimator for the center of the bell $theta$. Let's see this.
The likelihood for the Cauchy distribution is
$$L(x;theta) = prod _i^n frac{1}{pileft [ 1+(x_i-theta)^2 right ]}, $$
and its logarithm is
$$ln L(x;theta) = -n ln pi -sum_i^nlnleft [ 1+(x_i-theta)^2 right ].$$
The estimator will maximize the likelihood and if there's a suficient estimator it's derivate can be factoriced i.e.:
$$frac{partial L(x;theta)}{partial theta} = A(theta)left[t(x)-h(theta)-b(theta)right], $$
where $A(theta)$ is a function exclusively from the parameter, $t(x)$ is function exclusively of your data, $h(theta$) is what you want to estimate and $b(theta)$ is a possible bias.
If you derivate you should notice that the Cauchy distribution can not be factorized, but Cramer-Rao lets you find a bound for the variance, that is
$$ Var(t) geq frac{left(frac{partial}{partial theta}(h+b)right)^2}{Eleft[(frac{partial}{partial theta}ln L)^2right]},$$
where the equality hold only if $frac{partial ln L}{partial theta}$ can be factorized.
The most you can do is calculate the bound but it has no analytic closed solution for $theta$
answered Jun 20 '17 at 15:33
DanielDaniel
14
$endgroup$
First of all you should notice that there's no suficient estimator for the center of the bell $theta$. Let's see this.
The likelihood for the Cauchy distribution is
$$L(x;theta) = prod _i^n frac{1}{pileft [ 1+(x_i-theta)^2 right ]}, $$
and its logarithm is
$$ln L(x;theta) = -n ln pi -sum_i^nlnleft [ 1+(x_i-theta)^2 right ].$$
The estimator will maximize the likelihood and if there's a suficient estimator it's derivate can be factoriced i.e.:
$$frac{partial L(x;theta)}{partial theta} = A(theta)left[t(x)-h(theta)-b(theta)right], $$
where $A(theta)$ is a function exclusively from the parameter, $t(x)$ is function exclusively of your data, $h(theta$) is what you want to estimate and $b(theta)$ is a possible bias.
If you derivate you should notice that the Cauchy distribution can not be factorized, but Cramer-Rao lets you find a bound for the variance, that is
$$ Var(t) geq frac{left(frac{partial}{partial theta}(h+b)right)^2}{Eleft[(frac{partial}{partial theta}ln L)^2right]},$$
where the equality hold only if $frac{partial ln L}{partial theta}$ can be factorized.
The most you can do is calculate the bound but it has no analytic closed solution for $theta$
answered Jun 20 '17 at 15:33
DanielDaniel
14
edited Jun 20 '17 at 17:29
answered Jun 20 '17 at 15:33
DanielDaniel
14
answered Jun 20 '17 at 15:33
DanielDaniel
14
answered Jun 20 '17 at 15:33
DanielDaniel
14
14
add a comment |
add a comment |
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