Uniform Probability over Real Line?

[gajohnson]

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 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.

 

[LCKurtz]

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.

 

[gajohnson]

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!

 

[gajohnson]

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]

Confirmed this elsewhere. It is indeed that simple. Appreciate your help!

Yes, it is. Sorry I couldn't get back to you yesterday. Too busy.