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Robin Chapman
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The bivariate distribution formed by two independent normalized Gaussians is rotationally symmetric (think about the usual argument for evaluating the probability integral). The quotient of two random variables $X$ and $Y$ is the tangent of the angle between $(0,0)$ and $(X,Y)$ with the $x$-axis. If one has a rotationally symmetric distribution for $X$ and $Y$ (with no point mass at the origin) then $Y/X$ is a tangent of a uniformly distributed angle. This is the Cauchy distribution.

Added Your example with the Brownian motion states in effect that if $P$ is the first point that the motion hits the $x$-axis then the angle between the line from $P$ to the starting point and the $y$-axis is uniformly distributed between $-\pi$ and $\pi$. I can't see any reason why this should be so, but perhaps someone (unlike me) who actually knows something about Brownian motion might know why.

The bivariate distribution formed by two independent normalized Gaussians is rotationally symmetric (think about the usual argument for evaluating the probability integral). The quotient of two random variables $X$ and $Y$ is the tangent of the angle between $(0,0)$ and $(X,Y)$ with the $x$-axis. If one has a rotationally symmetric distribution for $X$ and $Y$ (with no point mass at the origin) then $Y/X$ is a tangent of a uniformly distributed angle. This is the Cauchy distribution.

The bivariate distribution formed by two independent normalized Gaussians is rotationally symmetric (think about the usual argument for evaluating the probability integral). The quotient of two random variables $X$ and $Y$ is the tangent of the angle between $(0,0)$ and $(X,Y)$ with the $x$-axis. If one has a rotationally symmetric distribution for $X$ and $Y$ (with no point mass at the origin) then $Y/X$ is a tangent of a uniformly distributed angle. This is the Cauchy distribution.

Added Your example with the Brownian motion states in effect that if $P$ is the first point that the motion hits the $x$-axis then the angle between the line from $P$ to the starting point and the $y$-axis is uniformly distributed between $-\pi$ and $\pi$. I can't see any reason why this should be so, but perhaps someone (unlike me) who actually knows something about Brownian motion might know why.

removed dubious statement
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Robin Chapman
  • 20.8k
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  • 81

The bivariate distribution formed by two independent normalized Gaussians is rotationally symmetric (think about the usual argument for evaluating the probability integral). The quotient of two random variables $X$ and $Y$ is the tangent of the angle between $(0,0)$ and $(X,Y)$ with the $x$-axis. If one has a rotationally symmetric distribution for $X$ and $Y$ (with no point mass at the origin) then $Y/X$ is a tangent of a uniformly distributed angle. This is the Cauchy distribution.

Surely at least your first Brownian motion example is an example of rotational symmetry.

The bivariate distribution formed by two independent normalized Gaussians is rotationally symmetric (think about the usual argument for evaluating the probability integral). The quotient of two random variables $X$ and $Y$ is the tangent of the angle between $(0,0)$ and $(X,Y)$ with the $x$-axis. If one has a rotationally symmetric distribution for $X$ and $Y$ (with no point mass at the origin) then $Y/X$ is a tangent of a uniformly distributed angle. This is the Cauchy distribution.

Surely at least your first Brownian motion example is an example of rotational symmetry.

The bivariate distribution formed by two independent normalized Gaussians is rotationally symmetric (think about the usual argument for evaluating the probability integral). The quotient of two random variables $X$ and $Y$ is the tangent of the angle between $(0,0)$ and $(X,Y)$ with the $x$-axis. If one has a rotationally symmetric distribution for $X$ and $Y$ (with no point mass at the origin) then $Y/X$ is a tangent of a uniformly distributed angle. This is the Cauchy distribution.

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Robin Chapman
  • 20.8k
  • 2
  • 66
  • 81

The bivariate distribution formed by two independent normalized Gaussians is rotationally symmetric (think about the usual argument for evaluating the probability integral). The quotient of two random variables $X$ and $Y$ is the tangent of the angle between $(0,0)$ and $(X,Y)$ with the $x$-axis. If one has a rotationally symmetric distribution for $X$ and $Y$ (with no point mass at the origin) then $Y/X$ is a tangent of a uniformly distributed angle. This is the Cauchy distribution.

Surely at least your first Brownian motion example is an example of rotational symmetry.