All depends on what you assume as known. For instance, if X  and  Y  are independent random variables that the pair (X,Y)  possesses pdf with circular symmetry, then the angle is uniformly distributed between  0   and  \pi  (with density  1/\pi, of course). A special explanation requires your term ,normal method' . In my opinion every method is normal if it starts from the assumption about the joint probability of  (X,Y). A normal method is only the simplest possible: Claim from nothing that the pdf of the angle is "such and such" giving here any description of the pdf concentrated on the interval [0,\pi].

Regards

In the absence of any other information in your question, the answer *has* to be uniform in [ 0, 2 pi] , since rotating the plane shouldn't affect the answer.  

Not that this makes the answer correct --- after all, how do you choose a uniform point in the plane?  

So, to be more precise, specify a distribution on the points of the plane (and also specify what you mean by the angle: I'm assuming it's the angle in a positive direction from the x-axis: was that what you meant?)  and then we can give you more concrete answers.  

But, if your distributions have rotational symmetry, so will the answer, and it will be uniform.

If you have some observed data,  you can plot its rose diagram (a kind of circular histogram that is availlable in MATLAB and perhaps sholuld be in R also.)  If it seems like a symmetric one about some point  better go for von Mises distribution.  Several analysis tools for vMD are availlable in the circ stat tool box in MATLAB.

A systematic way to get your desired result might be the transformation theorem outlined in Chapter 1 of Gut's "An intermediate course in probability". You can look at the relevant sections here https://books.google.de/books?id=ufxMwdtrmOAC

I assume that you have a joint pdf of your two points. First, you try to find a one-to-one mapping between your two points and the angle and some help variables. Then you apply the transformation theorem to get the pdf of the angle and the help variables. Finally, you integrate out the help variables to get the pdf of the angle only.

Mathematica might be helpful to get there. :)

Assume that both points P1 and P2 lie on a unit radius circle.

Note that the scalar product of the vectors (x1,y1) and (x2,y2) is equal Cos[\phi].

so, \phi= arc cos [x1 x2+y1 y2]. May be, this will help you.

Ilya

Vibekananda, 

To clarify: you are choosing two points in the plane independently from some distribution.  

a) I am assuming they are both from the *same* distribution.  Is this correct?

b) Are you choosing points (x,y) as cartesian coordinates, with x and y both 

uniformly and independently distributed between o and pi?  If so, my comments above are not of any use since the square doesn't have rotational symmetry

c) On the other hand are you choosing points (r, theta) as polar coordinates, with theta uniformly distributed?  I'm guessing that this is not the case, since you said 0,pi, not 0, 2 pi.

d) If the answer to b) is yes, is there a reason you are restricting the points to a square?  If you were to choose from a bivariate normal distribution centered 0 with rotational symmetry, say, everything would be easy:-)

Hi All Followers of Vibekananda's question

Seemigly it is now time for a more concrete example. Unfortunately I am giving them without the very detail. The following is just my guess about what you expect, based on the above comments.

1. Assume you have a robot equipped with sensors which supply at each instant the results of observation; first problem then is to write a code calculating the best suitted pdf of - say two vectors X an Y (points on the flat screen), representing the position of the "aim" and the "gun", respectively.

2. Assume you want to apply the joint probability distribution P_(X,Y) of the two vectors; then the problem is to choose the way of description of the (in our case 4-dimensional) probability distribution. The proper solution is a kind of art - sometimes one chooses a way convenient for further steps, sometimes this requires a few trials. Suggestion (comp. Neil's answers):

2.1. if the vectors lie on the unit circle - apply suitably the example you already know (for instance the wraped normal or some restricted polynomal)

2.2. if the vectors possess density of the probability distribution then start with two independent densities, e.g. take

2.2.1. normal two-dimensional densities; their parameters are (respectively)  mean vectors (mx1, mx2) and (my1, my2) covariance 2by2-matrices

[cxij; i,j =1,2] [cyij; i,j =1,2] 

2.2.2. some densities of the (r,\theta) variables (many choices are possible, just for instance take for rx and ry two different gamma pdf, and for the anle \thetax and thetay two different normal wrapped pdf on the uniy cirle

Junction od 1. AND 2. is a problem of statistics: how the data from sensors should be translated into the pdfs of the two vectors (follow the advices of the former answer, esp. by Michael and Hemangi)

3. The final step - the angle between the random vectors X and Y . (The formula by Ilya has an extention over arbitrary vectors after division by product of the lengths of the vectors; warning Ilya's x, y, 1, 2 are my 1, 2, x ,y - respectively). Calculation of the pdf of the angle is a result of what is called by Michael a transformation formula. This is a mathematical formula which should be applied with a special care. Everything depends on the original coordinates chosen for the description of the vectors AND ON THE CHOSEN PROBABILITY DISTRIBUTIONS modelling the randomness. For some choices the calculation is simple, for other might be impossible to calculate in writing (then use math packages).

4. And now the really final step: what should be the decision about the movement of the "gun" to get it afterwards at most probably close to the "aim". This is the core of problems of automatic control, which is relatively simple for stationary "aim" (however, no particular code is known to me) and much much more complex for moving "aims".

5. Appendix - An example of calculation the pdf of the angle between two random unit vectors. Let  FRAC{ a }   denote fraction of the real number  a. If the angles of  X  and  Y  (on the unit circle) are given by

  thetax = 2 \pi FRAC{ U }, thetay = 2 \pi FRAC{ V },

with independent U and V distributed according to the normal distributions N(mx, vx) and N( my, vy) (thus, if both angles have (different) wrapped normal distributions), then the angle between X and  Y  (counted counter-clockwise from  x  to  y )  is distributed according to the wrapped normal:

  thetay - thetax mod  (2 \pi)  =  2 \pi FRAC{ W },

where  W possesses N( my - mx, vx + vy).

Regards

@ Vibekananda. Would you please chose the place of your problem within the above structure (or explain it as a problem outside this scheme). JoD

@ Vibekananda

Simply:  the first vector is the  aim position. Stationary means  it does not move during the observation by sensors. Otherwise, " moving aim"  means  we admit it moves and the robot takes picture at one instant analysis it, turns the gun toward, say the most probable direction, makes a shut; if failed, in the next instant doing the same taking into account that the position could be changed. this can be decided with multisteps backward analysis (syatem with memry), etc.

And what do you mean about between stationary and moving aim

Regards

This is just a guess. Denote by \phi_1 the angle between the vector (x1,y1) and the

X-axis. The angles are measured counterclockwise. Seems that \phi_1 is uniformly distributed on [0,2\pi]. The same with the angle \phi_2 for the second point (x2,y2).

what we are looking for is the angle \phi=\phi_1-\phi_2. So with have a random variable

which is the difference Y of two uniformly distributed independent random variables.

Seems that Y will have a triangular distribution (?) with the support [-2\pi,+2\pi].

Sorry if I made wrong assumptions.

Ilya Gertsbakh.

@ Ilya

Indeed, under the assumption that \phi _1 and \phi_2 are independent and uniformly distributed on [0; 2 \pi] , the difference \phi_1 - phi_2  possesses the triangular pd on [-2\pi,+2\pi], but we observe its value       modulo  2 \pi,     i.e.

                     \phi_{between} = \phi _1 - \phi_2  mod (2 \pi)

(comp. my last but one answer). And it again possesses the uniform pd on  [0, 2 \pi].

Regards

I second Ilya Gertsbach's response.

Let me continue with a more detailed  explanation that the resulting observed difference   of  angles  defined  by the last Ilya question  is  uniformly distributed  within the interval [0; 2 \pi ).

First, let us assume, that the resulting angle is a number which shoud have value in a range of lenth 2\pi. Without loss of generality, let this be the angle counted from vector 2  to  vector  1  counterclockwise. This means, that we are not looking for the distribution of the difference  \phi = \phi_1-\phi_2 but for the reduced  angle

\phi_o := \phi  mod  q  = \phi_1-\phi_2  mod  q,       where q:=2 \pi

obtained with application of the operation modulo  q  defined as follows:

 x mod q :=  x - max{ n \cdot q : n \in Z, n \cdot q \le x}, for  x \in R,  q>o,

(R and  Z  stand for reals  and integers, respectively).

Next, as Ilya said, the difference \phi has a symmetric triangle distribution on  [-2\pi;+2\pi], which leads to the following probabilities for  a \in [0;2\pi):

F(a) := P{ \phi_o \le a } =  P {  \phi \in [0; a } + P{ \phi \in [-2\pi; a-2\pi] = p1 + p2,

where the integrals

p1 = \int_{[ 0; a ]   f(x)  dx     and   p2 = \int_{ [-2\pi; a-2\pi] } f(x) dx 

use the  density f(x)= (1/ (2\pi) )  -  |x| / (2\pi)^2,  After simple calculus one gets:

F(a) = a/ (2\pi), for   a \in [0;2\pi).

Regards