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## Homework Statement

Consider a two-ended laser spinner; that is a pen-like laser acting as the arrow mounted on a pin at the center of a spinner. Suppose the center of the disk is one meter away from a wall of infinite extent marked with a linear scale, with zero at the point closest the center of the spinner and negative numbers to left, positive to the right.

The laser is spun and comes to a rest projecting for one of its ends at a point Y on the

scale (with probability zero the laser will stop parallel to the wall and miss it; we ignore

that possibility). Suppose that the angle X the laser makes to the perpendicular to the

wall is uniformly distributed over –π/2 to π/2. Find the probability density of Y.

## Homework Equations

## The Attempt at a Solution

The angle at which the laser lands must have PDF [itex]f_X(x)=1/{\pi}[/itex]

That is, it is the uniform distribution over pi radians.

Next, I set up a right triangle and find that the angle opposite the the 1 meter side is equal to [itex]{\pi}/2-X[/itex] and that the angle opposite the base (call it B) is simply X (given by the PDF above). Using the law of sines (and that the distance from the "infinite wall" to the spinner is 1 meter) to find:

[itex]B=sinx/sin({\pi}/2-X)[/itex]

It seems clear to me that Y is then the uniform distribution over [itex]lim B [/itex] as [itex] x->{\pi}/2 + - (lim B [/itex] as [itex] x->-{\pi}/2)[/itex]

This is the entire real number line and so the PDF of Y does not exist.

Is this right (even if my steps are less than rigorous)? Any help would be appreciated.