Higher Dimensional Random Walks
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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 an arbitrary direction (uniformely on the unit sphere $S^1$), 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 set $v_i = v_{i-1} + mu$, where $mu in S^{d-1}$ picked at random at every step and $v_0 = 0$, what is the expected value of $|v_n|$? 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.
(if you could add a reference to the answer as well it would be great! :D )
Thank you very much!
probability approximation random-walk probability-limit-theorems
asked Jan 10 at 23:05
AlfredAlfred
357
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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 an arbitrary direction (uniformely on the unit sphere $S^1$), 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 set $v_i = v_{i-1} + mu$, where $mu in S^{d-1}$ picked at random at every step and $v_0 = 0$, what is the expected value of $|v_n|$? 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.
(if you could add a reference to the answer as well it would be great! :D )
Thank you very much!
probability approximation random-walk probability-limit-theorems
asked Jan 10 at 23:05
AlfredAlfred
357
$endgroup$
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 an arbitrary direction (uniformely on the unit sphere $S^1$), 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 set $v_i = v_{i-1} + mu$, where $mu in S^{d-1}$ picked at random at every step and $v_0 = 0$, what is the expected value of $|v_n|$? 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.
(if you could add a reference to the answer as well it would be great! :D )
Thank you very much!
probability approximation random-walk probability-limit-theorems
asked Jan 10 at 23:05
AlfredAlfred
357
$endgroup$
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 an arbitrary direction (uniformely on the unit sphere $S^1$), 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 set $v_i = v_{i-1} + mu$, where $mu in S^{d-1}$ picked at random at every step and $v_0 = 0$, what is the expected value of $|v_n|$? 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.
(if you could add a reference to the answer as well it would be great! :D )
Thank you very much!
probability approximation random-walk probability-limit-theorems
probability approximation random-walk probability-limit-theorems
asked Jan 10 at 23:05
AlfredAlfred
357
asked Jan 10 at 23:05
AlfredAlfred
357
asked Jan 10 at 23:05
AlfredAlfred
357
asked Jan 10 at 23:05
AlfredAlfred
357
asked Jan 10 at 23:05
AlfredAlfred
357
357
add a comment |
add a comment |
1 Answer
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As described here:
$${displaystyle P(r)={frac {2r}{N}}e^{-r^{2}/N}}$$
answered Jan 10 at 23:08
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
Your answer is just for the case where the steps are all orthogonal between them ( 6 directions: (1,0,0)(0,1,0)(0,0,1) and the negatives for $R^3$ ). I'm looking for the general case, where the steps are allowed to be taken in any random direction. Thank you tho!
$endgroup$
– Alfred
Jan 10 at 23:17
$begingroup$
Actually, if the formula you posted is an upper bound of the one I need, would be great as well, but is it the case?
$endgroup$
– Alfred
Jan 10 at 23:20
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
As described here:
$${displaystyle P(r)={frac {2r}{N}}e^{-r^{2}/N}}$$
answered Jan 10 at 23:08
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
Your answer is just for the case where the steps are all orthogonal between them ( 6 directions: (1,0,0)(0,1,0)(0,0,1) and the negatives for $R^3$ ). I'm looking for the general case, where the steps are allowed to be taken in any random direction. Thank you tho!
$endgroup$
– Alfred
Jan 10 at 23:17
$begingroup$
Actually, if the formula you posted is an upper bound of the one I need, would be great as well, but is it the case?
$endgroup$
– Alfred
Jan 10 at 23:20
add a comment |
$begingroup$
As described here:
$${displaystyle P(r)={frac {2r}{N}}e^{-r^{2}/N}}$$
answered Jan 10 at 23:08
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
Your answer is just for the case where the steps are all orthogonal between them ( 6 directions: (1,0,0)(0,1,0)(0,0,1) and the negatives for $R^3$ ). I'm looking for the general case, where the steps are allowed to be taken in any random direction. Thank you tho!
$endgroup$
– Alfred
Jan 10 at 23:17
$begingroup$
Actually, if the formula you posted is an upper bound of the one I need, would be great as well, but is it the case?
$endgroup$
– Alfred
Jan 10 at 23:20
add a comment |
$begingroup$
As described here:
$${displaystyle P(r)={frac {2r}{N}}e^{-r^{2}/N}}$$
answered Jan 10 at 23:08
David G. StorkDavid G. Stork
11k41432
$endgroup$
As described here:
$${displaystyle P(r)={frac {2r}{N}}e^{-r^{2}/N}}$$
answered Jan 10 at 23:08
David G. StorkDavid G. Stork
11k41432
answered Jan 10 at 23:08
David G. StorkDavid G. Stork
11k41432
answered Jan 10 at 23:08
David G. StorkDavid G. Stork
11k41432
answered Jan 10 at 23:08
David G. StorkDavid G. Stork
11k41432
11k41432
$begingroup$
Your answer is just for the case where the steps are all orthogonal between them ( 6 directions: (1,0,0)(0,1,0)(0,0,1) and the negatives for $R^3$ ). I'm looking for the general case, where the steps are allowed to be taken in any random direction. Thank you tho!
$endgroup$
– Alfred
Jan 10 at 23:17
$begingroup$
Actually, if the formula you posted is an upper bound of the one I need, would be great as well, but is it the case?
$endgroup$
– Alfred
Jan 10 at 23:20
add a comment |
$begingroup$
Your answer is just for the case where the steps are all orthogonal between them ( 6 directions: (1,0,0)(0,1,0)(0,0,1) and the negatives for $R^3$ ). I'm looking for the general case, where the steps are allowed to be taken in any random direction. Thank you tho!
$endgroup$
– Alfred
Jan 10 at 23:17
$begingroup$
Actually, if the formula you posted is an upper bound of the one I need, would be great as well, but is it the case?
$endgroup$
– Alfred
Jan 10 at 23:20
$begingroup$
Your answer is just for the case where the steps are all orthogonal between them ( 6 directions: (1,0,0)(0,1,0)(0,0,1) and the negatives for $R^3$ ). I'm looking for the general case, where the steps are allowed to be taken in any random direction. Thank you tho!
$endgroup$
– Alfred
Jan 10 at 23:17
$begingroup$
Your answer is just for the case where the steps are all orthogonal between them ( 6 directions: (1,0,0)(0,1,0)(0,0,1) and the negatives for $R^3$ ). I'm looking for the general case, where the steps are allowed to be taken in any random direction. Thank you tho!
$endgroup$
– Alfred
Jan 10 at 23:17
$begingroup$
Actually, if the formula you posted is an upper bound of the one I need, would be great as well, but is it the case?
$endgroup$
– Alfred
Jan 10 at 23:20
$begingroup$
Actually, if the formula you posted is an upper bound of the one I need, would be great as well, but is it the case?
$endgroup$
– Alfred
Jan 10 at 23:20
add a comment |
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