Re: Weekend Puzzle

Andor wrote:

This reminds me of another probability puzzle (due to
Mandelbrot, if I remember correctly). Imagine an archer standing
on the y axis and facing an infinitely long wall along the x
axis (Mandelbrot is a mathematician:-). The archer is spun
around blindfolded, which effectively points him in a direction
uniformly distributed between -90° and +90° (while still facing
the x axis from his position). He then shoots his arrow.

Call the archer's position y, the entry point x, the angle a and the
path length p. We have a = atan x/y = asec p/y.

1. What is the expected x coordinate of the point of entry of the
arrow (remembering that the archer is standing on the y axis)?

So P(x) = P(a) da/dx = 1/(2 pi) y/(x^2 + y^2)
and E(x) = int -inf..inf x P(x) dx,
which has principal value zero since the integrand is odd.

2. What is the expected length of the flight path of the arrow?

Here, P(p) = 1/(2 pi) y/p 1/sqrt(p^2 - y^2)
and E(p) = 2 int y..inf p P(p) dp = inf
because of the pole at p = y.


Martin

--
Quidquid latine scriptum sit, altum viditur.
.

Relevant Pages

  • Re: Weekend Puzzle
    ... Mandelbrot, if I remember correctly). ... Imagine an archer standing ... axis. ...
    (comp.dsp)
  • Re: Weekend Puzzle
    ... arrow (remembering that the archer is standing on the y axis)? ...
    (comp.dsp)