# Uniform Probability over Real Line?

## 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.

## The Attempt at a Solution

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

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 ${\pi}/2-X$ 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:

$B=sinx/sin({\pi}/2-X)$

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

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.

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LCKurtz
Homework Helper
Gold Member
Let the origin of the Y axis along the wall be the foot of the 1 meter perpendicular side. Then for a point Y on that axis you have ##\frac Y 1 =\tan X##. Try starting with that.

• 1 person
So much simpler! My trig is too rusty.

Ok, so then I find that g(Y)=arctan(Y) and, after taking the derivative, I make the transformation assuming the uniform distribution described before for X.

I get $f_Y(y)=1/{\pi}(1+y^2)$

Is it that simple? Thanks!

Last edited:
So much simpler! My trig is too rusty.

Ok, so then I find that g(Y)=arctan(Y) and, after taking the derivative, I make the transformation assuming the uniform distribution described before for X.

I get $f_Y(y)=1/{\pi}(1+y^2)$

Is it that simple? Thanks!
Confirmed this elsewhere. It is indeed that simple. Appreciate your help!

LCKurtz