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Possibility of a point moving out of a shape

  1. Feb 1, 2008 #1
    Dear all:

    The question is like this:

    Given a point in a shape like square or triangle. Let's take square as the simplest one. Say the length of the side of the square is [tex]l[/tex]. If add a noise to the point which will move the point [tex]r[\tex] distance and the direction is arbitrary, what will be the possibility that the point will move out of the square?

    And what will be the possibility if the point is in a triangle? Also what about a cube or tetrahedron?

    Does anyone have any idea to solve the problem please?
    Ideally if there is a equation to model the problem.

  2. jcsd
  3. Feb 1, 2008 #2


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    YOu are talking about "Brownian Motion". As I recall, one can show that, with a step size of r, the point will move on average, a distance [itex]r\sqrt{n}[/itex] from its initial position in n steps. Whether that will move it out of the figure depends on shape of the position. You could probably do an integral over [itex]d\theta[/itex] with [itex]\theta[/itex] going from 0 to [itex]2\pi[/itex] using the distance from the point to boundary at each [itex]\theta[/itex].
  4. Feb 1, 2008 #3


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    I will simplify the shape to a line. If you can solve the problem in 1-D (a line), you can generalize to 2-D (a plane).

    If x is given, then the random part of y is r. Without loss of generality, assume the endpoints of the line are 0 and 1, and [itex]\ell[/itex] = 1. Let y = x + r. Then the probability you have inquired about is given by Pr{y < 0 or y > 1|x} = 1 - Pr{0 < y < 1|x} = 1 - (F(1|x) - F(0|x)) where F is the cumulative distribution (CDF) of y|x, which you need to derive.

    For example, one might assume that given x, r is distributed uniformly over [x-0.1, x+0.1]; show that distribution as U(r) = (r - (x-0.1))/0.2. Then F(y*|x) = Pr{y < y*|x} = Pr{x+r < y*|x} = Pr{r < y*-x|x} = U(y*-x) = (y*-x - (x-0.1))/0.2 = (y*+ 0.1)/0.2.

    When generalizing to a plane, you need to define the distribution of r over a circle around x, instead of an interval (the interval [x-0.1, x+0.1] in my example).
    Last edited: Feb 2, 2008
  5. Feb 2, 2008 #4


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    The expression (y*+ 0.1)/0.2 in my above post should have been (y*-2x + 0.1)/0.2.
  6. Feb 2, 2008 #5


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    I'm not 100% sure what the OP is asking for. If the "noise" is a single shock of size r then you need to calculate the length of the portion of the circle of radius r centred at the point that lies outside of the given set.
    If it is a mathematical brownian motion (http://en.wikipedia.org/wiki/Wiener_process" [Broken]) and want to know the probability of it lying outside the set at a given time T, then it becomes an integral over the normal distribution.
    If you want the probability of it going outside the set at any time before T, then the problem is much harder, and becomes a statement about the distribution of the http://en.wikipedia.org/wiki/Hitting_time" [Broken] of the brownian motion.
    Last edited by a moderator: May 3, 2017
  7. Feb 4, 2008 #6
    I don't think I am talking about Brownian motion as the energy of noise is fixed. Given a length r, the point should only noised by r.

    Thanks EnumaElish, It really helps.
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