• Support PF! Buy your school textbooks, materials and every day products Here!

Uniform Probability over Real Line?

  • Thread starter gajohnson
  • Start date
  • #1
73
0

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.
 

Answers and Replies

  • #2
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,535
751
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.
 
  • Like
Likes 1 person
  • #3
73
0
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 [itex]f_Y(y)=1/{\pi}(1+y^2)[/itex]

Is it that simple? Thanks!
 
Last edited:
  • #4
73
0
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 [itex]f_Y(y)=1/{\pi}(1+y^2)[/itex]

Is it that simple? Thanks!
Confirmed this elsewhere. It is indeed that simple. Appreciate your help!
 
  • #5
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,535
751
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.
 

Related Threads on Uniform Probability over Real Line?

Replies
2
Views
890
Replies
5
Views
941
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
7
Views
592
  • Last Post
Replies
5
Views
9K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
1
Views
3K
Replies
5
Views
732
Replies
5
Views
6K
Top