Proof that if you take enough steps of equal magnitude on a plane, you'll always end up at the starting point
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Is the claim in the title true? If yes, is the following sufficient to justify it intuitively?
Assume for simplicity that the steps are performed with magnitude 1 in a Cartesian plane. To construct a random step, first randomly pick some angle $theta$. For said $theta$, let the walk be a translation of distance 1 either at an angle $theta$ to the $+x$ direction, or at an angle $theta$ to the $-x$ direction (i.e. in the oppositive direction). Because the steps are random, the probability of either is $0.5$. The central limit theorem implies that if said angle $theta$ is picked a sufficiently large number, the number of steps going either way will cancel out. Since every step can be randomly constructed using the method above, therefore eventually all the steps will cancel out.
One obvious flaw of this justification is that for every random $theta$ picked, the probability that picking it again is infinitesimally close to $0$. Will this affect the validity of the justification?
Thanks.
probability proof-verification random-walk central-limit-theorem
asked Dec 21 '18 at 16:55
36843684
1317
$endgroup$
add a comment |
$begingroup$
Is the claim in the title true? If yes, is the following sufficient to justify it intuitively?
Assume for simplicity that the steps are performed with magnitude 1 in a Cartesian plane. To construct a random step, first randomly pick some angle $theta$. For said $theta$, let the walk be a translation of distance 1 either at an angle $theta$ to the $+x$ direction, or at an angle $theta$ to the $-x$ direction (i.e. in the oppositive direction). Because the steps are random, the probability of either is $0.5$. The central limit theorem implies that if said angle $theta$ is picked a sufficiently large number, the number of steps going either way will cancel out. Since every step can be randomly constructed using the method above, therefore eventually all the steps will cancel out.
One obvious flaw of this justification is that for every random $theta$ picked, the probability that picking it again is infinitesimally close to $0$. Will this affect the validity of the justification?
Thanks.
probability proof-verification random-walk central-limit-theorem
asked Dec 21 '18 at 16:55
36843684
1317
$endgroup$
$begingroup$
As stated, the result is obviously false. At the very least, you need to add "uniformly at random" in there somewhere, and be clear about what you mean by "uniform". And no, your proof doesn't work, and no proof of this form can possibly work. In particular, if it did work, then it would equally apply in 3-space, where the result is false, even if you restrict to a lattice. The first answer here might help.
$endgroup$
– user3482749
Dec 21 '18 at 18:06
add a comment |
$begingroup$
Is the claim in the title true? If yes, is the following sufficient to justify it intuitively?
Assume for simplicity that the steps are performed with magnitude 1 in a Cartesian plane. To construct a random step, first randomly pick some angle $theta$. For said $theta$, let the walk be a translation of distance 1 either at an angle $theta$ to the $+x$ direction, or at an angle $theta$ to the $-x$ direction (i.e. in the oppositive direction). Because the steps are random, the probability of either is $0.5$. The central limit theorem implies that if said angle $theta$ is picked a sufficiently large number, the number of steps going either way will cancel out. Since every step can be randomly constructed using the method above, therefore eventually all the steps will cancel out.
One obvious flaw of this justification is that for every random $theta$ picked, the probability that picking it again is infinitesimally close to $0$. Will this affect the validity of the justification?
Thanks.
probability proof-verification random-walk central-limit-theorem
asked Dec 21 '18 at 16:55
36843684
1317
$endgroup$
Is the claim in the title true? If yes, is the following sufficient to justify it intuitively?
Assume for simplicity that the steps are performed with magnitude 1 in a Cartesian plane. To construct a random step, first randomly pick some angle $theta$. For said $theta$, let the walk be a translation of distance 1 either at an angle $theta$ to the $+x$ direction, or at an angle $theta$ to the $-x$ direction (i.e. in the oppositive direction). Because the steps are random, the probability of either is $0.5$. The central limit theorem implies that if said angle $theta$ is picked a sufficiently large number, the number of steps going either way will cancel out. Since every step can be randomly constructed using the method above, therefore eventually all the steps will cancel out.
One obvious flaw of this justification is that for every random $theta$ picked, the probability that picking it again is infinitesimally close to $0$. Will this affect the validity of the justification?
Thanks.
probability proof-verification random-walk central-limit-theorem
probability proof-verification random-walk central-limit-theorem
asked Dec 21 '18 at 16:55
36843684
1317
asked Dec 21 '18 at 16:55
36843684
1317
asked Dec 21 '18 at 16:55
36843684
1317
asked Dec 21 '18 at 16:55
36843684
1317
asked Dec 21 '18 at 16:55
36843684
1317
1317
$begingroup$
As stated, the result is obviously false. At the very least, you need to add "uniformly at random" in there somewhere, and be clear about what you mean by "uniform". And no, your proof doesn't work, and no proof of this form can possibly work. In particular, if it did work, then it would equally apply in 3-space, where the result is false, even if you restrict to a lattice. The first answer here might help.
$endgroup$
– user3482749
Dec 21 '18 at 18:06
add a comment |
$begingroup$
As stated, the result is obviously false. At the very least, you need to add "uniformly at random" in there somewhere, and be clear about what you mean by "uniform". And no, your proof doesn't work, and no proof of this form can possibly work. In particular, if it did work, then it would equally apply in 3-space, where the result is false, even if you restrict to a lattice. The first answer here might help.
$endgroup$
– user3482749
Dec 21 '18 at 18:06
$begingroup$
As stated, the result is obviously false. At the very least, you need to add "uniformly at random" in there somewhere, and be clear about what you mean by "uniform". And no, your proof doesn't work, and no proof of this form can possibly work. In particular, if it did work, then it would equally apply in 3-space, where the result is false, even if you restrict to a lattice. The first answer here might help.
$endgroup$
– user3482749
Dec 21 '18 at 18:06
$begingroup$
As stated, the result is obviously false. At the very least, you need to add "uniformly at random" in there somewhere, and be clear about what you mean by "uniform". And no, your proof doesn't work, and no proof of this form can possibly work. In particular, if it did work, then it would equally apply in 3-space, where the result is false, even if you restrict to a lattice. The first answer here might help.
$endgroup$
– user3482749
Dec 21 '18 at 18:06
add a comment |
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$begingroup$
As stated, the result is obviously false. At the very least, you need to add "uniformly at random" in there somewhere, and be clear about what you mean by "uniform". And no, your proof doesn't work, and no proof of this form can possibly work. In particular, if it did work, then it would equally apply in 3-space, where the result is false, even if you restrict to a lattice. The first answer here might help.
$endgroup$
– user3482749
Dec 21 '18 at 18:06