Pdf of angle formed by two normal random variables

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Discussion Overview

The discussion revolves around finding the probability density function (pdf) of the angle formed by two normally distributed random variables represented as a vector in the x-y plane, specifically in relation to a fixed point. The context includes theoretical exploration and mathematical reasoning regarding the integration of these variables in polar coordinates.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant describes the initial approach using the joint pdf of the two normal random variables and expresses difficulty in integrating over angles to find the pdf of the angle.
  • Another participant suggests that due to symmetry, every angle may have the same probability, implying a uniform distribution of angles.
  • A different viewpoint proposes that if both random variables have a mean of one, the mean angle should be pi/4, while if one variable has a mean of one and the other negative one, the mean angle should be -pi/4.
  • Some participants question the need to integrate over angles, suggesting that the expected angle requires a clear definition due to the periodic nature of angles.
  • There is mention of the distribution of the ratio of y/x of two independent normal random variables, which involves a Cauchy distribution, and the angle can be derived using the arctangent function.
  • One participant proposes transforming the coordinates to center the means at zero before switching to polar coordinates, although they express uncertainty about whether this will simplify the problem.
  • Concerns are raised about integrating over angles to avoid redundancy and ensure the pdf integrates to one, but there is uncertainty about the correctness of this approach.

Areas of Agreement / Disagreement

Participants express differing views on the integration of angles and the implications of symmetry on the distribution of angles. There is no consensus on the best method to derive the pdf of the angle, and multiple competing perspectives remain.

Contextual Notes

Participants note limitations regarding the definition of "expected angle" due to the periodic nature of angles, and the integration approach may not yield a valid pdf without careful consideration of angle redundancy.

shazi
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Hi Everyone!
I have two normally distributed random variables. One on the x axis, the other on the y axis, like a complex normal random variable. I'm trying to find the pdf of the angle between a fixed point on the x-y plane(let's say point 1,0) and the vector formed by combining the two random variables.

I tried starting with the pdf of x and y:
p(x,y) =(1/(2πσ^2)) e^(-((x-μ_x)^2+(y-μ_y)^2)/(2σ^2))
where σ is the standard deviation for both random variables
μ_x is the mean for random variable x, and
μ_y is the mean for random variable y

Then, I converted to polar coordinates and integrated over all radii (negative infinity to infinity). Then, it seems like I should integrate from angles 0 to 2pi. However that last integration seems really hard. Does anyone know of another way to solve the pdf of the angle?

In case anyone is curious what the two input random variables actually represent, they are output from an accelerometer
 
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The angle between the vector (0,1) and a random (x,y)? I think you can just use symmetry - every angle has the same probability.
 
For example, if random variables x and y both have mean equal one then intuitively the mean of the random variable angle should be pi/4, because random variables x and y are going along the x and y-axis respectively. Also if x has mean equal one and y has mean equal negative one, the mean of angle should be -pi/4. Hope this makes the question clearer.
 
Oh sorry, I did not see the non-zero means of the variables.
Then, I converted to polar coordinates and integrated over all radii (negative infinity to infinity). Then, it seems like I should integrate from angles 0 to 2pi.
Why do you want to integrate over the angles? That is the distribution you want to get, I think.
Do you want the expected angle? That needs some definition of "expected angle" first, as the angles wrap around (2pi=0). I think you can use symmetry again there, and the "expected angle" will just be the angle between your vector and the vector (μ_x,μ_y).
 
The distribution of the ratio of y/x of two independent normal random variables involves a mixture that includes a Cauchy distribution. The angle the vector (x,y) makes with the x-axis would be the arctangent (ATAN2) of that ratio. A prominent statistician says there are practical ways of dealing with the ratio of normal random variables. http://www.google.com/url?sa=t&rct=...sg=AFQjCNEgO1dvktreWiL-rt-ZPcS3K1FmYQ&cad=rja

(I don't know how useful it is to transform his distributions by the inverse tangent.)
 
Stephen Tashi said:
The distribution of the ratio of y/x of two independent normal random variables involves a mixture that includes a Cauchy distribution. The angle the vector (x,y) makes with the x-axis would be the arctangent (ATAN2) of that ratio. A prominent statistician says there are practical ways of dealing with the ratio of normal random variables. http://www.google.com/url?sa=t&rct=...sg=AFQjCNEgO1dvktreWiL-rt-ZPcS3K1FmYQ&cad=rja
Thanks for the link - this is a pretty cool-looking paper! I recognize the name, too: Marsaglia invented the polar method for generating pairs of standard normal pseudorandom numbers.

It might be easier to represent the random point (X,Y) in polar coordinates (R,θ). I'd start by first choosing new coordinates so that the means of X and Y are both zero, then switching to polar coordinates. I'm not sure if that will help or not; sometimes new coordinates cause more problems than they solve.
 
mfb said:
Why do you want to integrate over the angles? That is the distribution you want to get, I think.
Do you want the expected angle? That needs some definition of "expected angle" first, as the angles wrap around (2pi=0). I think you can use symmetry again there, and the "expected angle" will just be the angle between your vector and the vector (μ_x,μ_y).

I considered integrating over 0 to 2pi to exclude redundant angles. Without that integration the PDF would be a sinusoid that continues forever, it wouldn't integrate to one, and therefore not a valid PDF. But I'm not sure this is a correct way to do it.

Yes, I'm trying to find the PDF of angle between a fixed vector and vector(mean_x,mean_y)
 
Thanks Everyone, I'll start with the method suggested by Stephen
 

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