Will simple random walk on $n$-cycle converges to Brownian motion on $S^1$?
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I know that, by Donsker's theorem, simple random walk on $mathbbZ$ will converge to Brownian motion on $mathbbR$. Here, simple random walk means that the Markov chain with probability from $n$ to $n+1$, and from $n$ to $n-1$ are equally $1/2$.
Then, we can consider similar random walk on $n$-cycle. More rigorously, consider $mathbbZ_n$ and suppose probability from $n$ to $n+1$, and from $n$ to $n-1$ are equally $1/2$. Then, as $n$ increases this will also converges to Brownian motion on $S^1$? (Here, Brownian motion on $S^1$ is the Markov process induced by the heat kernel)
If it is true, probably we will need to rescale the original random walk. Since the simple random walk on $mathbbZ_n$ does not consider geometry of $S^1$.
Maybe this is classical question, but I couldn't find good references. Is there any book about "convergence of Random walk to Brownian motion on manifolds(Or, just $S^1$)"?
probability markov-chains brownian-motion markov-process random-walk
asked Aug 12 at 0:05
KGEO
976
 |Â
show 5 more comments
up vote
3
down vote
favorite
I know that, by Donsker's theorem, simple random walk on $mathbbZ$ will converge to Brownian motion on $mathbbR$. Here, simple random walk means that the Markov chain with probability from $n$ to $n+1$, and from $n$ to $n-1$ are equally $1/2$.
Then, we can consider similar random walk on $n$-cycle. More rigorously, consider $mathbbZ_n$ and suppose probability from $n$ to $n+1$, and from $n$ to $n-1$ are equally $1/2$. Then, as $n$ increases this will also converges to Brownian motion on $S^1$? (Here, Brownian motion on $S^1$ is the Markov process induced by the heat kernel)
If it is true, probably we will need to rescale the original random walk. Since the simple random walk on $mathbbZ_n$ does not consider geometry of $S^1$.
Maybe this is classical question, but I couldn't find good references. Is there any book about "convergence of Random walk to Brownian motion on manifolds(Or, just $S^1$)"?
probability markov-chains brownian-motion markov-process random-walk
asked Aug 12 at 0:05
KGEO
976
So you want $n$ going to $infty$; does $k$ go to $infty$ as well? If $k < fracn8$ then this of course reduces of course to the random walk on $mathbbZ$. If $k$ is much larger than $n$ then of course this reduces to the uniform distribution (almost, if $n$ is even parity will be preserved, so for $n$ even uniform on the evens or odds depending on which step of the walk)
– Mike
Aug 14 at 1:52
$k$ means the number of transition, I guess? How about the parabolic scaling $k=n^2$?
– KGEO
Aug 14 at 1:55
Yes I meant $k$ is the number of transitions. But for any even $n$, parity will still be preserved. So $n$ odd?
– Mike
Aug 14 at 1:57
Yes. Let's assume $n$ is odd if necessary. Moreover we can use spectral expansion if $n$ is odd, which is probably good thing..?
– KGEO
Aug 14 at 1:59
For $n$ odd the only eigenvector w normalized eigenvalue that has absolute value exactly 1 is the vector on $mathbbZ_n$ that is 1 everywhere--and the eigenvalue is 1, for $n$ even the eigenvector w normalized eigenvalue exactly -1 is the vector that is -1 on the odds and 1 on the evens
– Mike
Aug 14 at 2:14
 |Â
show 5 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I know that, by Donsker's theorem, simple random walk on $mathbbZ$ will converge to Brownian motion on $mathbbR$. Here, simple random walk means that the Markov chain with probability from $n$ to $n+1$, and from $n$ to $n-1$ are equally $1/2$.
Then, we can consider similar random walk on $n$-cycle. More rigorously, consider $mathbbZ_n$ and suppose probability from $n$ to $n+1$, and from $n$ to $n-1$ are equally $1/2$. Then, as $n$ increases this will also converges to Brownian motion on $S^1$? (Here, Brownian motion on $S^1$ is the Markov process induced by the heat kernel)
If it is true, probably we will need to rescale the original random walk. Since the simple random walk on $mathbbZ_n$ does not consider geometry of $S^1$.
Maybe this is classical question, but I couldn't find good references. Is there any book about "convergence of Random walk to Brownian motion on manifolds(Or, just $S^1$)"?
probability markov-chains brownian-motion markov-process random-walk
asked Aug 12 at 0:05
KGEO
976
I know that, by Donsker's theorem, simple random walk on $mathbbZ$ will converge to Brownian motion on $mathbbR$. Here, simple random walk means that the Markov chain with probability from $n$ to $n+1$, and from $n$ to $n-1$ are equally $1/2$.
Then, we can consider similar random walk on $n$-cycle. More rigorously, consider $mathbbZ_n$ and suppose probability from $n$ to $n+1$, and from $n$ to $n-1$ are equally $1/2$. Then, as $n$ increases this will also converges to Brownian motion on $S^1$? (Here, Brownian motion on $S^1$ is the Markov process induced by the heat kernel)
If it is true, probably we will need to rescale the original random walk. Since the simple random walk on $mathbbZ_n$ does not consider geometry of $S^1$.
Maybe this is classical question, but I couldn't find good references. Is there any book about "convergence of Random walk to Brownian motion on manifolds(Or, just $S^1$)"?
probability markov-chains brownian-motion markov-process random-walk
asked Aug 12 at 0:05
KGEO
976
edited Aug 14 at 1:50
asked Aug 12 at 0:05
KGEO
976
asked Aug 12 at 0:05
KGEO
976
asked Aug 12 at 0:05
KGEO
976
976
So you want $n$ going to $infty$; does $k$ go to $infty$ as well? If $k < fracn8$ then this of course reduces of course to the random walk on $mathbbZ$. If $k$ is much larger than $n$ then of course this reduces to the uniform distribution (almost, if $n$ is even parity will be preserved, so for $n$ even uniform on the evens or odds depending on which step of the walk)
– Mike
Aug 14 at 1:52
$k$ means the number of transition, I guess? How about the parabolic scaling $k=n^2$?
– KGEO
Aug 14 at 1:55
Yes I meant $k$ is the number of transitions. But for any even $n$, parity will still be preserved. So $n$ odd?
– Mike
Aug 14 at 1:57
Yes. Let's assume $n$ is odd if necessary. Moreover we can use spectral expansion if $n$ is odd, which is probably good thing..?
– KGEO
Aug 14 at 1:59
For $n$ odd the only eigenvector w normalized eigenvalue that has absolute value exactly 1 is the vector on $mathbbZ_n$ that is 1 everywhere--and the eigenvalue is 1, for $n$ even the eigenvector w normalized eigenvalue exactly -1 is the vector that is -1 on the odds and 1 on the evens
– Mike
Aug 14 at 2:14
 |Â
show 5 more comments
So you want $n$ going to $infty$; does $k$ go to $infty$ as well? If $k < fracn8$ then this of course reduces of course to the random walk on $mathbbZ$. If $k$ is much larger than $n$ then of course this reduces to the uniform distribution (almost, if $n$ is even parity will be preserved, so for $n$ even uniform on the evens or odds depending on which step of the walk)
– Mike
Aug 14 at 1:52
$k$ means the number of transition, I guess? How about the parabolic scaling $k=n^2$?
– KGEO
Aug 14 at 1:55
Yes I meant $k$ is the number of transitions. But for any even $n$, parity will still be preserved. So $n$ odd?
– Mike
Aug 14 at 1:57
Yes. Let's assume $n$ is odd if necessary. Moreover we can use spectral expansion if $n$ is odd, which is probably good thing..?
– KGEO
Aug 14 at 1:59
For $n$ odd the only eigenvector w normalized eigenvalue that has absolute value exactly 1 is the vector on $mathbbZ_n$ that is 1 everywhere--and the eigenvalue is 1, for $n$ even the eigenvector w normalized eigenvalue exactly -1 is the vector that is -1 on the odds and 1 on the evens
– Mike
Aug 14 at 2:14
So you want $n$ going to $infty$; does $k$ go to $infty$ as well? If $k < fracn8$ then this of course reduces of course to the random walk on $mathbbZ$. If $k$ is much larger than $n$ then of course this reduces to the uniform distribution (almost, if $n$ is even parity will be preserved, so for $n$ even uniform on the evens or odds depending on which step of the walk)
– Mike
Aug 14 at 1:52
So you want $n$ going to $infty$; does $k$ go to $infty$ as well? If $k < fracn8$ then this of course reduces of course to the random walk on $mathbbZ$. If $k$ is much larger than $n$ then of course this reduces to the uniform distribution (almost, if $n$ is even parity will be preserved, so for $n$ even uniform on the evens or odds depending on which step of the walk)
– Mike
Aug 14 at 1:52
$k$ means the number of transition, I guess? How about the parabolic scaling $k=n^2$?
– KGEO
Aug 14 at 1:55
$k$ means the number of transition, I guess? How about the parabolic scaling $k=n^2$?
– KGEO
Aug 14 at 1:55
Yes I meant $k$ is the number of transitions. But for any even $n$, parity will still be preserved. So $n$ odd?
– Mike
Aug 14 at 1:57
Yes I meant $k$ is the number of transitions. But for any even $n$, parity will still be preserved. So $n$ odd?
– Mike
Aug 14 at 1:57
Yes. Let's assume $n$ is odd if necessary. Moreover we can use spectral expansion if $n$ is odd, which is probably good thing..?
– KGEO
Aug 14 at 1:59
Yes. Let's assume $n$ is odd if necessary. Moreover we can use spectral expansion if $n$ is odd, which is probably good thing..?
– KGEO
Aug 14 at 1:59
For $n$ odd the only eigenvector w normalized eigenvalue that has absolute value exactly 1 is the vector on $mathbbZ_n$ that is 1 everywhere--and the eigenvalue is 1, for $n$ even the eigenvector w normalized eigenvalue exactly -1 is the vector that is -1 on the odds and 1 on the evens
– Mike
Aug 14 at 2:14
For $n$ odd the only eigenvector w normalized eigenvalue that has absolute value exactly 1 is the vector on $mathbbZ_n$ that is 1 everywhere--and the eigenvalue is 1, for $n$ even the eigenvector w normalized eigenvalue exactly -1 is the vector that is -1 on the odds and 1 on the evens
– Mike
Aug 14 at 2:14
 |Â
show 5 more comments
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So you want $n$ going to $infty$; does $k$ go to $infty$ as well? If $k < fracn8$ then this of course reduces of course to the random walk on $mathbbZ$. If $k$ is much larger than $n$ then of course this reduces to the uniform distribution (almost, if $n$ is even parity will be preserved, so for $n$ even uniform on the evens or odds depending on which step of the walk)
– Mike
Aug 14 at 1:52
$k$ means the number of transition, I guess? How about the parabolic scaling $k=n^2$?
– KGEO
Aug 14 at 1:55
Yes I meant $k$ is the number of transitions. But for any even $n$, parity will still be preserved. So $n$ odd?
– Mike
Aug 14 at 1:57
Yes. Let's assume $n$ is odd if necessary. Moreover we can use spectral expansion if $n$ is odd, which is probably good thing..?
– KGEO
Aug 14 at 1:59
For $n$ odd the only eigenvector w normalized eigenvalue that has absolute value exactly 1 is the vector on $mathbbZ_n$ that is 1 everywhere--and the eigenvalue is 1, for $n$ even the eigenvector w normalized eigenvalue exactly -1 is the vector that is -1 on the odds and 1 on the evens
– Mike
Aug 14 at 2:14