Mixing
Random Walks on high dimensional spaces
$begingroup$
I've read on a paper that, in the two dimensional case, if you start from the origin and take steps of length one in arbitrary directions (uniformely on the unit sphere $S^1$, not left-right-up-down), the expected distance after $n$ steps from the starting point is approximated by $sqrt{npi}/2$
(source: https://pdfs.semanticscholar.org/6685/1166d588821456477f2007a37bf0428a2cf2.pdf).
I was wondering if there was a similar formula for higher dimensional random walks, which means:
Starting from the origin in $mathbb{R}^d$ if I take $n$ steps in random directions (which doesn't have to be aligned to any axes, can be any uniformly chosen random direction taken from the sphere $S^{d-1}$), what is the expected value of the distance where I end up from the origin?
e.g. how distant is the point from the origin after having taken $n$ steps in random directions?
I don't need a precise formula, everything that just gives an idea on how large the expected distance is works just fine. Could also be a loose upper bound.
(if you could add a reference to the answer as well it would be great! :D )
EDIT: A friend suggested me that the answer should be in the heat equation, which means that I only need to integrate the d-dimensional gaussian. Right?
Thank you very much!
pr.probability st.statistics random-walks
asked Jan 11 at 9:55
AlfredAlfred
1188
$endgroup$
add a comment |
$begingroup$
I've read on a paper that, in the two dimensional case, if you start from the origin and take steps of length one in arbitrary directions (uniformely on the unit sphere $S^1$, not left-right-up-down), the expected distance after $n$ steps from the starting point is approximated by $sqrt{npi}/2$
(source: https://pdfs.semanticscholar.org/6685/1166d588821456477f2007a37bf0428a2cf2.pdf).
I was wondering if there was a similar formula for higher dimensional random walks, which means:
Starting from the origin in $mathbb{R}^d$ if I take $n$ steps in random directions (which doesn't have to be aligned to any axes, can be any uniformly chosen random direction taken from the sphere $S^{d-1}$), what is the expected value of the distance where I end up from the origin?
e.g. how distant is the point from the origin after having taken $n$ steps in random directions?
I don't need a precise formula, everything that just gives an idea on how large the expected distance is works just fine. Could also be a loose upper bound.
(if you could add a reference to the answer as well it would be great! :D )
EDIT: A friend suggested me that the answer should be in the heat equation, which means that I only need to integrate the d-dimensional gaussian. Right?
Thank you very much!
pr.probability st.statistics random-walks
asked Jan 11 at 9:55
AlfredAlfred
1188
$endgroup$
add a comment |
$begingroup$
I've read on a paper that, in the two dimensional case, if you start from the origin and take steps of length one in arbitrary directions (uniformely on the unit sphere $S^1$, not left-right-up-down), the expected distance after $n$ steps from the starting point is approximated by $sqrt{npi}/2$
(source: https://pdfs.semanticscholar.org/6685/1166d588821456477f2007a37bf0428a2cf2.pdf).
I was wondering if there was a similar formula for higher dimensional random walks, which means:
Starting from the origin in $mathbb{R}^d$ if I take $n$ steps in random directions (which doesn't have to be aligned to any axes, can be any uniformly chosen random direction taken from the sphere $S^{d-1}$), what is the expected value of the distance where I end up from the origin?
e.g. how distant is the point from the origin after having taken $n$ steps in random directions?
I don't need a precise formula, everything that just gives an idea on how large the expected distance is works just fine. Could also be a loose upper bound.
(if you could add a reference to the answer as well it would be great! :D )
EDIT: A friend suggested me that the answer should be in the heat equation, which means that I only need to integrate the d-dimensional gaussian. Right?
Thank you very much!
pr.probability st.statistics random-walks
asked Jan 11 at 9:55
AlfredAlfred
1188
$endgroup$
I've read on a paper that, in the two dimensional case, if you start from the origin and take steps of length one in arbitrary directions (uniformely on the unit sphere $S^1$, not left-right-up-down), the expected distance after $n$ steps from the starting point is approximated by $sqrt{npi}/2$
(source: https://pdfs.semanticscholar.org/6685/1166d588821456477f2007a37bf0428a2cf2.pdf).
I was wondering if there was a similar formula for higher dimensional random walks, which means:
Starting from the origin in $mathbb{R}^d$ if I take $n$ steps in random directions (which doesn't have to be aligned to any axes, can be any uniformly chosen random direction taken from the sphere $S^{d-1}$), what is the expected value of the distance where I end up from the origin?
e.g. how distant is the point from the origin after having taken $n$ steps in random directions?
I don't need a precise formula, everything that just gives an idea on how large the expected distance is works just fine. Could also be a loose upper bound.
(if you could add a reference to the answer as well it would be great! :D )
EDIT: A friend suggested me that the answer should be in the heat equation, which means that I only need to integrate the d-dimensional gaussian. Right?
Thank you very much!
pr.probability st.statistics random-walks
pr.probability st.statistics random-walks
asked Jan 11 at 9:55
AlfredAlfred
1188
asked Jan 11 at 9:55
AlfredAlfred
1188
edited Jan 11 at 12:02
Alfred
asked Jan 11 at 9:55
AlfredAlfred
1188
asked Jan 11 at 9:55
AlfredAlfred
1188
asked Jan 11 at 9:55
AlfredAlfred
1188
1188
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $X_1,X_2,dots$ be iid random vectors each uniformly distributed on $S^{d-1}$. Let $S_n:=sum_1^n X_i$. By the symmetry, $EX_1=0$. Also, $1=|X_1|^2=sum_{j=1}^d X_{1j}^2$, where $X_1=(X_{11},dots,X_{1d})$. Since the $X_{1j}$'s are exchangeable and $1=E|X_1|^2=sum_{j=1}^d EX_{1j}^2$, we conclude that $EX_{1j}^2=frac1d$ for all $j$. Also, for any distinct $j,k=1,dots,d$, the pair $(X_{1j},X_{1k})$ equals $(-X_{1j},X_{1k})$ in distribution, whence $EX_{1j}X_{1k}=0$. So, $EX_1=0$ and the covariance matrix of $X_1$ is $frac1d,I_d$, where $I_d$ is the $dtimes d$ identity matrix.
So, by the multivariate central limit theorem, $frac1{sqrt n},S_n$ converges in distribution (as $ntoinfty$) to the random vector $frac1{sqrt d}Z$, where $Z=(Z_1,dots,Z_d)$ and $Z_1,dots,Z_d$ are iid $N(0,1)$. Therefore and because of the uniform integrability of $frac1{sqrt n},|S_n|$ (provided by the observation that $E|S_n|^2=n$, made in the answer by Pierre PC), we have
begin{equation}
Efrac1{sqrt n},|S_n|toell_d:=frac1{sqrt d}Esqrt{sum_1^d Z_j^2},
end{equation}
and
hence
begin{equation}
E|S_n|sim ell_dsqrt n. tag{*}
end{equation}
Since the distribution of $sum_1^d Z_j^2$ is $chi^2_d=text{Gamma}(frac d2,2)$, we see that
begin{equation}
ell_d=sqrt{frac2d},frac{Gamma((d+1)/2)}{Gamma(d/2)}.
end{equation}
In particular, for $d=2$ (*) becomes
begin{equation}
E|S_n|sim tfrac12,sqrt{pi n},
end{equation}
which is what you read in that paper.
answered Jan 11 at 13:45
Iosif PinelisIosif Pinelis
18.9k22159
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Alfred
Jan 11 at 15:11
add a comment |
$begingroup$
Let $(X_k)_{kgeq1}$ be a sequence of points chosen independently and uniformly on the $(d-1)$-sphere. Set $S_n=X_1+cdots+X_n$ and $rho_n=|S_n|$.
It is clear that
$$ rho_{n+1}^2 = rho_{n}^2 + 2langle S_n,X_{n+1} rangle + 1. $$
Now because $mathbb E[langle S_n,X_{n+1}rangle|X_1,cdots, X_n] = langle S_n,mathbb E[X_{n+1}]rangle = 0$, we get
$$ mathbb E[rho_{n+1}^2] = mathbb E[rho_{n}^2] + 2mathbb E[langle S_n,X_{n+1} rangle] + 1 = mathbb E[rho_{n}^2] + 1 = cdots = n+1. $$
Hence, for instance, $mathbb Erho_nleqsqrt{mathbb Erho_n^2}leqsqrt n$.
answered Jan 11 at 12:44
Pierre PCPierre PC
1718
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Alfred
Jan 11 at 15:11
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Let $X_1,X_2,dots$ be iid random vectors each uniformly distributed on $S^{d-1}$. Let $S_n:=sum_1^n X_i$. By the symmetry, $EX_1=0$. Also, $1=|X_1|^2=sum_{j=1}^d X_{1j}^2$, where $X_1=(X_{11},dots,X_{1d})$. Since the $X_{1j}$'s are exchangeable and $1=E|X_1|^2=sum_{j=1}^d EX_{1j}^2$, we conclude that $EX_{1j}^2=frac1d$ for all $j$. Also, for any distinct $j,k=1,dots,d$, the pair $(X_{1j},X_{1k})$ equals $(-X_{1j},X_{1k})$ in distribution, whence $EX_{1j}X_{1k}=0$. So, $EX_1=0$ and the covariance matrix of $X_1$ is $frac1d,I_d$, where $I_d$ is the $dtimes d$ identity matrix.
So, by the multivariate central limit theorem, $frac1{sqrt n},S_n$ converges in distribution (as $ntoinfty$) to the random vector $frac1{sqrt d}Z$, where $Z=(Z_1,dots,Z_d)$ and $Z_1,dots,Z_d$ are iid $N(0,1)$. Therefore and because of the uniform integrability of $frac1{sqrt n},|S_n|$ (provided by the observation that $E|S_n|^2=n$, made in the answer by Pierre PC), we have
begin{equation}
Efrac1{sqrt n},|S_n|toell_d:=frac1{sqrt d}Esqrt{sum_1^d Z_j^2},
end{equation}
and
hence
begin{equation}
E|S_n|sim ell_dsqrt n. tag{*}
end{equation}
Since the distribution of $sum_1^d Z_j^2$ is $chi^2_d=text{Gamma}(frac d2,2)$, we see that
begin{equation}
ell_d=sqrt{frac2d},frac{Gamma((d+1)/2)}{Gamma(d/2)}.
end{equation}
In particular, for $d=2$ (*) becomes
begin{equation}
E|S_n|sim tfrac12,sqrt{pi n},
end{equation}
which is what you read in that paper.
answered Jan 11 at 13:45
Iosif PinelisIosif Pinelis
18.9k22159
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Alfred
Jan 11 at 15:11
add a comment |
$begingroup$
Let $X_1,X_2,dots$ be iid random vectors each uniformly distributed on $S^{d-1}$. Let $S_n:=sum_1^n X_i$. By the symmetry, $EX_1=0$. Also, $1=|X_1|^2=sum_{j=1}^d X_{1j}^2$, where $X_1=(X_{11},dots,X_{1d})$. Since the $X_{1j}$'s are exchangeable and $1=E|X_1|^2=sum_{j=1}^d EX_{1j}^2$, we conclude that $EX_{1j}^2=frac1d$ for all $j$. Also, for any distinct $j,k=1,dots,d$, the pair $(X_{1j},X_{1k})$ equals $(-X_{1j},X_{1k})$ in distribution, whence $EX_{1j}X_{1k}=0$. So, $EX_1=0$ and the covariance matrix of $X_1$ is $frac1d,I_d$, where $I_d$ is the $dtimes d$ identity matrix.
So, by the multivariate central limit theorem, $frac1{sqrt n},S_n$ converges in distribution (as $ntoinfty$) to the random vector $frac1{sqrt d}Z$, where $Z=(Z_1,dots,Z_d)$ and $Z_1,dots,Z_d$ are iid $N(0,1)$. Therefore and because of the uniform integrability of $frac1{sqrt n},|S_n|$ (provided by the observation that $E|S_n|^2=n$, made in the answer by Pierre PC), we have
begin{equation}
Efrac1{sqrt n},|S_n|toell_d:=frac1{sqrt d}Esqrt{sum_1^d Z_j^2},
end{equation}
and
hence
begin{equation}
E|S_n|sim ell_dsqrt n. tag{*}
end{equation}
Since the distribution of $sum_1^d Z_j^2$ is $chi^2_d=text{Gamma}(frac d2,2)$, we see that
begin{equation}
ell_d=sqrt{frac2d},frac{Gamma((d+1)/2)}{Gamma(d/2)}.
end{equation}
In particular, for $d=2$ (*) becomes
begin{equation}
E|S_n|sim tfrac12,sqrt{pi n},
end{equation}
which is what you read in that paper.
answered Jan 11 at 13:45
Iosif PinelisIosif Pinelis
18.9k22159
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Alfred
Jan 11 at 15:11
add a comment |
$begingroup$
Let $X_1,X_2,dots$ be iid random vectors each uniformly distributed on $S^{d-1}$. Let $S_n:=sum_1^n X_i$. By the symmetry, $EX_1=0$. Also, $1=|X_1|^2=sum_{j=1}^d X_{1j}^2$, where $X_1=(X_{11},dots,X_{1d})$. Since the $X_{1j}$'s are exchangeable and $1=E|X_1|^2=sum_{j=1}^d EX_{1j}^2$, we conclude that $EX_{1j}^2=frac1d$ for all $j$. Also, for any distinct $j,k=1,dots,d$, the pair $(X_{1j},X_{1k})$ equals $(-X_{1j},X_{1k})$ in distribution, whence $EX_{1j}X_{1k}=0$. So, $EX_1=0$ and the covariance matrix of $X_1$ is $frac1d,I_d$, where $I_d$ is the $dtimes d$ identity matrix.
So, by the multivariate central limit theorem, $frac1{sqrt n},S_n$ converges in distribution (as $ntoinfty$) to the random vector $frac1{sqrt d}Z$, where $Z=(Z_1,dots,Z_d)$ and $Z_1,dots,Z_d$ are iid $N(0,1)$. Therefore and because of the uniform integrability of $frac1{sqrt n},|S_n|$ (provided by the observation that $E|S_n|^2=n$, made in the answer by Pierre PC), we have
begin{equation}
Efrac1{sqrt n},|S_n|toell_d:=frac1{sqrt d}Esqrt{sum_1^d Z_j^2},
end{equation}
and
hence
begin{equation}
E|S_n|sim ell_dsqrt n. tag{*}
end{equation}
Since the distribution of $sum_1^d Z_j^2$ is $chi^2_d=text{Gamma}(frac d2,2)$, we see that
begin{equation}
ell_d=sqrt{frac2d},frac{Gamma((d+1)/2)}{Gamma(d/2)}.
end{equation}
In particular, for $d=2$ (*) becomes
begin{equation}
E|S_n|sim tfrac12,sqrt{pi n},
end{equation}
which is what you read in that paper.
answered Jan 11 at 13:45
Iosif PinelisIosif Pinelis
18.9k22159
$endgroup$
Let $X_1,X_2,dots$ be iid random vectors each uniformly distributed on $S^{d-1}$. Let $S_n:=sum_1^n X_i$. By the symmetry, $EX_1=0$. Also, $1=|X_1|^2=sum_{j=1}^d X_{1j}^2$, where $X_1=(X_{11},dots,X_{1d})$. Since the $X_{1j}$'s are exchangeable and $1=E|X_1|^2=sum_{j=1}^d EX_{1j}^2$, we conclude that $EX_{1j}^2=frac1d$ for all $j$. Also, for any distinct $j,k=1,dots,d$, the pair $(X_{1j},X_{1k})$ equals $(-X_{1j},X_{1k})$ in distribution, whence $EX_{1j}X_{1k}=0$. So, $EX_1=0$ and the covariance matrix of $X_1$ is $frac1d,I_d$, where $I_d$ is the $dtimes d$ identity matrix.
So, by the multivariate central limit theorem, $frac1{sqrt n},S_n$ converges in distribution (as $ntoinfty$) to the random vector $frac1{sqrt d}Z$, where $Z=(Z_1,dots,Z_d)$ and $Z_1,dots,Z_d$ are iid $N(0,1)$. Therefore and because of the uniform integrability of $frac1{sqrt n},|S_n|$ (provided by the observation that $E|S_n|^2=n$, made in the answer by Pierre PC), we have
begin{equation}
Efrac1{sqrt n},|S_n|toell_d:=frac1{sqrt d}Esqrt{sum_1^d Z_j^2},
end{equation}
and
hence
begin{equation}
E|S_n|sim ell_dsqrt n. tag{*}
end{equation}
Since the distribution of $sum_1^d Z_j^2$ is $chi^2_d=text{Gamma}(frac d2,2)$, we see that
begin{equation}
ell_d=sqrt{frac2d},frac{Gamma((d+1)/2)}{Gamma(d/2)}.
end{equation}
In particular, for $d=2$ (*) becomes
begin{equation}
E|S_n|sim tfrac12,sqrt{pi n},
end{equation}
which is what you read in that paper.
answered Jan 11 at 13:45
Iosif PinelisIosif Pinelis
18.9k22159
answered Jan 11 at 13:45
Iosif PinelisIosif Pinelis
18.9k22159
answered Jan 11 at 13:45
Iosif PinelisIosif Pinelis
18.9k22159
answered Jan 11 at 13:45
Iosif PinelisIosif Pinelis
18.9k22159
18.9k22159
$begingroup$
Thank you very much!
$endgroup$
– Alfred
Jan 11 at 15:11
add a comment |
$begingroup$
Thank you very much!
$endgroup$
– Alfred
Jan 11 at 15:11
$begingroup$
Thank you very much!
$endgroup$
– Alfred
Jan 11 at 15:11
$begingroup$
Thank you very much!
$endgroup$
– Alfred
Jan 11 at 15:11
add a comment |
$begingroup$
Let $(X_k)_{kgeq1}$ be a sequence of points chosen independently and uniformly on the $(d-1)$-sphere. Set $S_n=X_1+cdots+X_n$ and $rho_n=|S_n|$.
It is clear that
$$ rho_{n+1}^2 = rho_{n}^2 + 2langle S_n,X_{n+1} rangle + 1. $$
Now because $mathbb E[langle S_n,X_{n+1}rangle|X_1,cdots, X_n] = langle S_n,mathbb E[X_{n+1}]rangle = 0$, we get
$$ mathbb E[rho_{n+1}^2] = mathbb E[rho_{n}^2] + 2mathbb E[langle S_n,X_{n+1} rangle] + 1 = mathbb E[rho_{n}^2] + 1 = cdots = n+1. $$
Hence, for instance, $mathbb Erho_nleqsqrt{mathbb Erho_n^2}leqsqrt n$.
answered Jan 11 at 12:44
Pierre PCPierre PC
1718
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Alfred
Jan 11 at 15:11
add a comment |
$begingroup$
Let $(X_k)_{kgeq1}$ be a sequence of points chosen independently and uniformly on the $(d-1)$-sphere. Set $S_n=X_1+cdots+X_n$ and $rho_n=|S_n|$.
It is clear that
$$ rho_{n+1}^2 = rho_{n}^2 + 2langle S_n,X_{n+1} rangle + 1. $$
Now because $mathbb E[langle S_n,X_{n+1}rangle|X_1,cdots, X_n] = langle S_n,mathbb E[X_{n+1}]rangle = 0$, we get
$$ mathbb E[rho_{n+1}^2] = mathbb E[rho_{n}^2] + 2mathbb E[langle S_n,X_{n+1} rangle] + 1 = mathbb E[rho_{n}^2] + 1 = cdots = n+1. $$
Hence, for instance, $mathbb Erho_nleqsqrt{mathbb Erho_n^2}leqsqrt n$.
answered Jan 11 at 12:44
Pierre PCPierre PC
1718
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Alfred
Jan 11 at 15:11
add a comment |
$begingroup$
Let $(X_k)_{kgeq1}$ be a sequence of points chosen independently and uniformly on the $(d-1)$-sphere. Set $S_n=X_1+cdots+X_n$ and $rho_n=|S_n|$.
It is clear that
$$ rho_{n+1}^2 = rho_{n}^2 + 2langle S_n,X_{n+1} rangle + 1. $$
Now because $mathbb E[langle S_n,X_{n+1}rangle|X_1,cdots, X_n] = langle S_n,mathbb E[X_{n+1}]rangle = 0$, we get
$$ mathbb E[rho_{n+1}^2] = mathbb E[rho_{n}^2] + 2mathbb E[langle S_n,X_{n+1} rangle] + 1 = mathbb E[rho_{n}^2] + 1 = cdots = n+1. $$
Hence, for instance, $mathbb Erho_nleqsqrt{mathbb Erho_n^2}leqsqrt n$.
answered Jan 11 at 12:44
Pierre PCPierre PC
1718
$endgroup$
Let $(X_k)_{kgeq1}$ be a sequence of points chosen independently and uniformly on the $(d-1)$-sphere. Set $S_n=X_1+cdots+X_n$ and $rho_n=|S_n|$.
It is clear that
$$ rho_{n+1}^2 = rho_{n}^2 + 2langle S_n,X_{n+1} rangle + 1. $$
Now because $mathbb E[langle S_n,X_{n+1}rangle|X_1,cdots, X_n] = langle S_n,mathbb E[X_{n+1}]rangle = 0$, we get
$$ mathbb E[rho_{n+1}^2] = mathbb E[rho_{n}^2] + 2mathbb E[langle S_n,X_{n+1} rangle] + 1 = mathbb E[rho_{n}^2] + 1 = cdots = n+1. $$
Hence, for instance, $mathbb Erho_nleqsqrt{mathbb Erho_n^2}leqsqrt n$.
answered Jan 11 at 12:44
Pierre PCPierre PC
1718
answered Jan 11 at 12:44
Pierre PCPierre PC
1718
answered Jan 11 at 12:44
Pierre PCPierre PC
1718
answered Jan 11 at 12:44
Pierre PCPierre PC
1718
1718
$begingroup$
Thank you very much!
$endgroup$
– Alfred
Jan 11 at 15:11
add a comment |
$begingroup$
Thank you very much!
$endgroup$
– Alfred
Jan 11 at 15:11
$begingroup$
Thank you very much!
$endgroup$
– Alfred
Jan 11 at 15:11
$begingroup$
Thank you very much!
$endgroup$
– Alfred
Jan 11 at 15:11
add a comment |
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