Sample median of Cauchy distribution is consistent. How?
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When we use chebyshev's inequality to show whether an estimator is consistent or not, we require the mean square error of the estimator and I do not know sample median's probability distribution. So please advice how this can be shown.
statistics statistical-inference quantile
edited Jun 26 '15 at 5:54
A.D
3,9281235
asked Jun 25 '15 at 13:25
Kiran PrajapatKiran Prajapat
135
$endgroup$
add a comment |
$begingroup$
When we use chebyshev's inequality to show whether an estimator is consistent or not, we require the mean square error of the estimator and I do not know sample median's probability distribution. So please advice how this can be shown.
statistics statistical-inference quantile
edited Jun 26 '15 at 5:54
A.D
3,9281235
asked Jun 25 '15 at 13:25
Kiran PrajapatKiran Prajapat
135
$endgroup$
1
$begingroup$
More generally I think you can say that all empirical quantiles converge to the corresponding true quantiles for any continuous distribution as the sample size tends to infinity.
$endgroup$
– dsaxton
Jun 26 '15 at 3:26
$begingroup$
For every $x$, the median $M_n$ of an i.i.d. sample of size $2n+1$ is such that $P(M_nleqslant x)=P(Y_n^xgeqslant n)$, where $Y_n^x$ is binomial $(2n+1,p(x))$ and $p(x)=P(Xleqslant x)$. If $x<0$, by the law of large numbers, $Y_n^x/(2n+1)to p(x)$ and $p(x)<frac12$ hence $P(Y_n^xgeqslant n)to0$. Likewise, if $x>0$, $P(Y_n^xgeqslant n)to1$. Thus, $M_nto0$ in distribution (and in probability).
$endgroup$
– Did
Jun 26 '15 at 6:07
add a comment |
$begingroup$
When we use chebyshev's inequality to show whether an estimator is consistent or not, we require the mean square error of the estimator and I do not know sample median's probability distribution. So please advice how this can be shown.
statistics statistical-inference quantile
edited Jun 26 '15 at 5:54
A.D
3,9281235
asked Jun 25 '15 at 13:25
Kiran PrajapatKiran Prajapat
135
$endgroup$
When we use chebyshev's inequality to show whether an estimator is consistent or not, we require the mean square error of the estimator and I do not know sample median's probability distribution. So please advice how this can be shown.
statistics statistical-inference quantile
statistics statistical-inference quantile
edited Jun 26 '15 at 5:54
A.D
3,9281235
asked Jun 25 '15 at 13:25
Kiran PrajapatKiran Prajapat
135
edited Jun 26 '15 at 5:54
A.D
3,9281235
asked Jun 25 '15 at 13:25
Kiran PrajapatKiran Prajapat
135
edited Jun 26 '15 at 5:54
A.D
3,9281235
edited Jun 26 '15 at 5:54
A.D
3,9281235
edited Jun 26 '15 at 5:54
A.D
3,9281235
3,9281235
asked Jun 25 '15 at 13:25
Kiran PrajapatKiran Prajapat
135
asked Jun 25 '15 at 13:25
Kiran PrajapatKiran Prajapat
135
asked Jun 25 '15 at 13:25
Kiran PrajapatKiran Prajapat
135
135
1
$begingroup$
More generally I think you can say that all empirical quantiles converge to the corresponding true quantiles for any continuous distribution as the sample size tends to infinity.
$endgroup$
– dsaxton
Jun 26 '15 at 3:26
$begingroup$
For every $x$, the median $M_n$ of an i.i.d. sample of size $2n+1$ is such that $P(M_nleqslant x)=P(Y_n^xgeqslant n)$, where $Y_n^x$ is binomial $(2n+1,p(x))$ and $p(x)=P(Xleqslant x)$. If $x<0$, by the law of large numbers, $Y_n^x/(2n+1)to p(x)$ and $p(x)<frac12$ hence $P(Y_n^xgeqslant n)to0$. Likewise, if $x>0$, $P(Y_n^xgeqslant n)to1$. Thus, $M_nto0$ in distribution (and in probability).
$endgroup$
– Did
Jun 26 '15 at 6:07
add a comment |
1
$begingroup$
More generally I think you can say that all empirical quantiles converge to the corresponding true quantiles for any continuous distribution as the sample size tends to infinity.
$endgroup$
– dsaxton
Jun 26 '15 at 3:26
$begingroup$
For every $x$, the median $M_n$ of an i.i.d. sample of size $2n+1$ is such that $P(M_nleqslant x)=P(Y_n^xgeqslant n)$, where $Y_n^x$ is binomial $(2n+1,p(x))$ and $p(x)=P(Xleqslant x)$. If $x<0$, by the law of large numbers, $Y_n^x/(2n+1)to p(x)$ and $p(x)<frac12$ hence $P(Y_n^xgeqslant n)to0$. Likewise, if $x>0$, $P(Y_n^xgeqslant n)to1$. Thus, $M_nto0$ in distribution (and in probability).
$endgroup$
– Did
Jun 26 '15 at 6:07
1
1
$begingroup$
More generally I think you can say that all empirical quantiles converge to the corresponding true quantiles for any continuous distribution as the sample size tends to infinity.
$endgroup$
– dsaxton
Jun 26 '15 at 3:26
$begingroup$
More generally I think you can say that all empirical quantiles converge to the corresponding true quantiles for any continuous distribution as the sample size tends to infinity.
$endgroup$
– dsaxton
Jun 26 '15 at 3:26
$begingroup$
For every $x$, the median $M_n$ of an i.i.d. sample of size $2n+1$ is such that $P(M_nleqslant x)=P(Y_n^xgeqslant n)$, where $Y_n^x$ is binomial $(2n+1,p(x))$ and $p(x)=P(Xleqslant x)$. If $x<0$, by the law of large numbers, $Y_n^x/(2n+1)to p(x)$ and $p(x)<frac12$ hence $P(Y_n^xgeqslant n)to0$. Likewise, if $x>0$, $P(Y_n^xgeqslant n)to1$. Thus, $M_nto0$ in distribution (and in probability).
$endgroup$
– Did
Jun 26 '15 at 6:07
$begingroup$
For every $x$, the median $M_n$ of an i.i.d. sample of size $2n+1$ is such that $P(M_nleqslant x)=P(Y_n^xgeqslant n)$, where $Y_n^x$ is binomial $(2n+1,p(x))$ and $p(x)=P(Xleqslant x)$. If $x<0$, by the law of large numbers, $Y_n^x/(2n+1)to p(x)$ and $p(x)<frac12$ hence $P(Y_n^xgeqslant n)to0$. Likewise, if $x>0$, $P(Y_n^xgeqslant n)to1$. Thus, $M_nto0$ in distribution (and in probability).
$endgroup$
– Did
Jun 26 '15 at 6:07
add a comment |
1 Answer
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Partial answer
For uneven samples $n=2k+1$ the distribution of the median of a distribution with probability density function $f(x)$ and cumulative density function F(x) can be expressed as:
$$f_{text{median n= 2k+1}}(x) = frac{(2k+1)!}{k! k!} F(x)^kf(x) (1-F(x))^k$$
which relates to the sampling of a number at $x$ (the middle term $f(x)$), having $k$ samples that are below $x$ (the first term $F(x)^k$), having $k$ samples above $x$ (the last term $F(x)^k$), and the number of ways that these numbers can be arranged (the factor at the beginning $(2k+1)!/(k!k!)$).
For the Cauchy distribution this becomes:
$$f_{text{Cauchy median $n=2k+1$}}(x) = frac{(2k+1)!}{k! k!} left( frac{1}{pi gamma left[1 + left(frac{x-x_0}{gamma} right)^2 right] } right) times left( 0.25 - frac{text{tan}^{-1}left(frac{x-x_0}{gamma} right)}{pi^2} vphantom{frac{1}{pi gamma left[1 + left(frac{x-x_0}{gamma} right)^2 right] }}right)^k $$
The variance does seem to be finite for $k geq 2$ (at least checking in wolfram alpha). I can imagine it might be possible to obtain an explicit expression. It seems like the indefinite integral can be expressed in terms of inverse tangent, logarithm and polylogarithm functions. But it is not a very pretty looking expression.
answered Jan 12 at 17:46
Martijn WeteringsMartijn Weterings
17010
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$begingroup$
Partial answer
For uneven samples $n=2k+1$ the distribution of the median of a distribution with probability density function $f(x)$ and cumulative density function F(x) can be expressed as:
$$f_{text{median n= 2k+1}}(x) = frac{(2k+1)!}{k! k!} F(x)^kf(x) (1-F(x))^k$$
which relates to the sampling of a number at $x$ (the middle term $f(x)$), having $k$ samples that are below $x$ (the first term $F(x)^k$), having $k$ samples above $x$ (the last term $F(x)^k$), and the number of ways that these numbers can be arranged (the factor at the beginning $(2k+1)!/(k!k!)$).
For the Cauchy distribution this becomes:
$$f_{text{Cauchy median $n=2k+1$}}(x) = frac{(2k+1)!}{k! k!} left( frac{1}{pi gamma left[1 + left(frac{x-x_0}{gamma} right)^2 right] } right) times left( 0.25 - frac{text{tan}^{-1}left(frac{x-x_0}{gamma} right)}{pi^2} vphantom{frac{1}{pi gamma left[1 + left(frac{x-x_0}{gamma} right)^2 right] }}right)^k $$
The variance does seem to be finite for $k geq 2$ (at least checking in wolfram alpha). I can imagine it might be possible to obtain an explicit expression. It seems like the indefinite integral can be expressed in terms of inverse tangent, logarithm and polylogarithm functions. But it is not a very pretty looking expression.
answered Jan 12 at 17:46
Martijn WeteringsMartijn Weterings
17010
$endgroup$
add a comment |
$begingroup$
Partial answer
For uneven samples $n=2k+1$ the distribution of the median of a distribution with probability density function $f(x)$ and cumulative density function F(x) can be expressed as:
$$f_{text{median n= 2k+1}}(x) = frac{(2k+1)!}{k! k!} F(x)^kf(x) (1-F(x))^k$$
which relates to the sampling of a number at $x$ (the middle term $f(x)$), having $k$ samples that are below $x$ (the first term $F(x)^k$), having $k$ samples above $x$ (the last term $F(x)^k$), and the number of ways that these numbers can be arranged (the factor at the beginning $(2k+1)!/(k!k!)$).
For the Cauchy distribution this becomes:
$$f_{text{Cauchy median $n=2k+1$}}(x) = frac{(2k+1)!}{k! k!} left( frac{1}{pi gamma left[1 + left(frac{x-x_0}{gamma} right)^2 right] } right) times left( 0.25 - frac{text{tan}^{-1}left(frac{x-x_0}{gamma} right)}{pi^2} vphantom{frac{1}{pi gamma left[1 + left(frac{x-x_0}{gamma} right)^2 right] }}right)^k $$
The variance does seem to be finite for $k geq 2$ (at least checking in wolfram alpha). I can imagine it might be possible to obtain an explicit expression. It seems like the indefinite integral can be expressed in terms of inverse tangent, logarithm and polylogarithm functions. But it is not a very pretty looking expression.
answered Jan 12 at 17:46
Martijn WeteringsMartijn Weterings
17010
$endgroup$
add a comment |
$begingroup$
Partial answer
For uneven samples $n=2k+1$ the distribution of the median of a distribution with probability density function $f(x)$ and cumulative density function F(x) can be expressed as:
$$f_{text{median n= 2k+1}}(x) = frac{(2k+1)!}{k! k!} F(x)^kf(x) (1-F(x))^k$$
which relates to the sampling of a number at $x$ (the middle term $f(x)$), having $k$ samples that are below $x$ (the first term $F(x)^k$), having $k$ samples above $x$ (the last term $F(x)^k$), and the number of ways that these numbers can be arranged (the factor at the beginning $(2k+1)!/(k!k!)$).
For the Cauchy distribution this becomes:
$$f_{text{Cauchy median $n=2k+1$}}(x) = frac{(2k+1)!}{k! k!} left( frac{1}{pi gamma left[1 + left(frac{x-x_0}{gamma} right)^2 right] } right) times left( 0.25 - frac{text{tan}^{-1}left(frac{x-x_0}{gamma} right)}{pi^2} vphantom{frac{1}{pi gamma left[1 + left(frac{x-x_0}{gamma} right)^2 right] }}right)^k $$
The variance does seem to be finite for $k geq 2$ (at least checking in wolfram alpha). I can imagine it might be possible to obtain an explicit expression. It seems like the indefinite integral can be expressed in terms of inverse tangent, logarithm and polylogarithm functions. But it is not a very pretty looking expression.
answered Jan 12 at 17:46
Martijn WeteringsMartijn Weterings
17010
$endgroup$
Partial answer
For uneven samples $n=2k+1$ the distribution of the median of a distribution with probability density function $f(x)$ and cumulative density function F(x) can be expressed as:
$$f_{text{median n= 2k+1}}(x) = frac{(2k+1)!}{k! k!} F(x)^kf(x) (1-F(x))^k$$
which relates to the sampling of a number at $x$ (the middle term $f(x)$), having $k$ samples that are below $x$ (the first term $F(x)^k$), having $k$ samples above $x$ (the last term $F(x)^k$), and the number of ways that these numbers can be arranged (the factor at the beginning $(2k+1)!/(k!k!)$).
For the Cauchy distribution this becomes:
$$f_{text{Cauchy median $n=2k+1$}}(x) = frac{(2k+1)!}{k! k!} left( frac{1}{pi gamma left[1 + left(frac{x-x_0}{gamma} right)^2 right] } right) times left( 0.25 - frac{text{tan}^{-1}left(frac{x-x_0}{gamma} right)}{pi^2} vphantom{frac{1}{pi gamma left[1 + left(frac{x-x_0}{gamma} right)^2 right] }}right)^k $$
The variance does seem to be finite for $k geq 2$ (at least checking in wolfram alpha). I can imagine it might be possible to obtain an explicit expression. It seems like the indefinite integral can be expressed in terms of inverse tangent, logarithm and polylogarithm functions. But it is not a very pretty looking expression.
answered Jan 12 at 17:46
Martijn WeteringsMartijn Weterings
17010
edited Jan 12 at 17:51
answered Jan 12 at 17:46
Martijn WeteringsMartijn Weterings
17010
answered Jan 12 at 17:46
Martijn WeteringsMartijn Weterings
17010
answered Jan 12 at 17:46
Martijn WeteringsMartijn Weterings
17010
17010
add a comment |
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$begingroup$
More generally I think you can say that all empirical quantiles converge to the corresponding true quantiles for any continuous distribution as the sample size tends to infinity.
$endgroup$
– dsaxton
Jun 26 '15 at 3:26
$begingroup$
For every $x$, the median $M_n$ of an i.i.d. sample of size $2n+1$ is such that $P(M_nleqslant x)=P(Y_n^xgeqslant n)$, where $Y_n^x$ is binomial $(2n+1,p(x))$ and $p(x)=P(Xleqslant x)$. If $x<0$, by the law of large numbers, $Y_n^x/(2n+1)to p(x)$ and $p(x)<frac12$ hence $P(Y_n^xgeqslant n)to0$. Likewise, if $x>0$, $P(Y_n^xgeqslant n)to1$. Thus, $M_nto0$ in distribution (and in probability).
$endgroup$
– Did
Jun 26 '15 at 6:07