Two vectors and two perpendicular lines

[LCDF]

TL;DR Summary

This post is an extension of the post: https://www.physicsforums.com/threads/probability-that-two-points-are-on-opposite-sides-of-a-line.994977/#post-6406442

In $\mathbb{R}^2$, there are two lines passing through the origin that are perpendicular to each other. The orientation of one of the lines with respect to $x$-axis is $\psi \in [0, \pi]$, where $\psi$ is uniformly distributed in $[0, \pi]$. Also, there are two vectors in $\mathbb{R}^2$ from the origin corresponding to points $(x_1, y_1)$ and $(x_2, y_2)$. The angle between these vectors is $\theta \in (0, 2\pi)$ measured in anticlockwise direction, and has the probability density function $f_{\theta}$. I want to calculate the probability that none of the perpendicular lines lies between the two vectors.

My attempt: I notice that this is equivalent to calculating the probability that the two points $(x_1, y_1)$ and $(x_2, y_2)$ lie on the same side of each of the perpendicular lines. In other words, a line does not lie between the two vectors iff $(a_1x_1 + b_1y_1 +c_1)(a_1x_2 + b_1y_2 +c_2) > 0$ AND $(a_2x_1 + b_2y_1 +c_1)(a_2x_2 + b_2y_2 +c_2) > 0$, where $a_1x + b_1y +c = 0$ and $a_2x + b_2y +c = 0$ are the equations of the lines. But I am unable to obtain an expression for the probability.

For the case of only one line passing through the origin (ignore the second line), the probability that the line does not lie between two vectors is
$p = \frac{1}{\pi}\int_{0}^{\pi} (\pi -\rho) f_{\theta}(\rho)\mathrm{d}\rho + \frac{1}{\pi}\int_{\pi}^{2\pi} (\rho-\pi) f_{\theta}(\rho)\mathrm{d}\rho.$

 

[PeroK]

How are you selecting the points $(x_1, y_1)$ and $(x_2, y_2)$?

Introducing $\psi$ may be unnecessary. Why not take the two lines to be the $x$ and $y$ axes?

In any case,the problem reduces to the two points being in the same quadrant, with respect to your axes.

 

[LCDF]

How are you selecting the points $(x_1, y_1)$ and $(x_2, y_2)$?

Introducing $\psi$ may be unnecessary. Why not take the two lines to be the $x$ and $y$ axes?

In any case,the problem reduces to the two points being in the same quadrant, with respect to your axes.

You are right that the problem reduces to the points being in the same quadrant. But I am unsure how to calculate the probability of it. The points $(x_1, y_1)$ and $(x_2, y_2)$ are the points of the Poisson point process (so they are random), and I know the distribution of the angle between them.

 

[PeroK]

You are right that the problem reduces to the points being in the same quadrant. But I am unsure how to calculate the probability of it. The points $(x_1, y_1)$ and $(x_2, y_2)$ are the points of the Poisson point process, and I know the distribution of the angle between them.

The angle between them doesn't specify the relationship between, say, $(x_1, y_1)$ and your axes. You need to specify how $(x_1, y_1)$ is chosen and then the distribution for $\theta$.

Are you assuming that $(x_1, y_1)$ is uniformly distributed in the plane?

 

[LCDF]

The angle between them doesn't specify the relationship between, say, $(x_1, y_1)$ and your axes. You need to specify how $(x_1, y_1)$ is chosen and then the distribution for $\theta$.

Are you assuming that $(x_1, y_1)$ is uniformly distributed in the plane?

Yes, $(x_1, y_1)$ is uniformly distributed in the plane.

 

[LCDF]

The angle between them doesn't specify the relationship between, say, $(x_1, y_1)$ and your axes. You need to specify how $(x_1, y_1)$ is chosen and then the distribution for $\theta$.

Are you assuming that $(x_1, y_1)$ is uniformly distributed in the plane?

But I am not sure why one needs to specify the location of $(x_1, y_1)$. For instance, when there is only one line (or, only one axis), the probability that the axis will not lie within the angle $\theta$ is simply $(\pi - \theta)/\pi$ as the axis is oriented uniformly between $[0, \pi]$. Then one averages over the distribution of $\theta$. Am I missing something here?

 

[PeroK]

Yes, $(x_1, y_1)$ is uniformly distributed in the plane.

Then $\psi$ is irrelevant. Take the $x$ and $y$ axes. Assume, without loss of generality, that $(x_1, y_1)$ is in the first quadrant, with an angle $\phi$ uniform between $0$ and $\pi/2$.

For each $\phi$ you need $\theta$ in the ranges $[0, \frac \pi 2 - \phi)$ or $(2\pi - \phi, 2\pi)$.

Draw a diagram to see this.

 

[LCDF]

Then $\psi$ is irrelevant. Take the $x$ and $y$ axes. Assume, without loss of generality, that $(x_1, x_2)$ is in the first quadrant, with an angle $\phi$ uniform between $0$ and $\pi/2$.

For each $\phi$ you need $\theta$ in the ranges $[0, \frac \pi 2 - \phi)$ or $(2\pi - \phi, 2\pi)$.

Draw a diagram to see this.

I get it in the case when $(x_1, y_1)$ is uniformly distributed. But I am not sure why one needs to specify the location of $(x_1, y_1)$. So, please let me repeat my previous comment in the case you missed it. For instance, when there is only one line (or, only one axis), the probability that the axis will not lie within the angle $\theta$ is simply $(\pi - \theta)/\pi$ for $\theta \in (0, \pi)$ as the axis is oriented uniformly between $[0, \pi]$. Then one averages over the distribution of $\theta$. One need not say anything else about $(x_1, y_1)$. Am I missing something here?

 

[PeroK]

I get it in the case when $(x_1, y_1)$ is uniformly distributed. But I am not sure why one needs to specify the location of $(x_1, y_1)$. So, please let me repeat my previous comment in the case you missed it. For instance, when there is only one line (or, only one axis), the probability that the axis will not lie within the angle $\theta$ is simply $(\pi - \theta)/\pi$ for $\theta \in (0, \pi)$ as the axis is oriented uniformly between $[0, \pi]$. Then one averages over the distribution of $\theta$. One need not say anything else about $(x_1, y_1)$. Am I missing something here?

That's a different problem.

 

[PeroK]

But I am not sure why one needs to specify the location of $(x_1, y_1)$.

You don't. It's the same in every quadrant. Which is why we can take it to be the first, for simplicity.

It's a form of decluttering the problem.

 

[PeroK]

$p = \frac{1}{\pi}\int_{0}^{\pi} (\pi -\rho) f_{\theta}(\rho)\mathrm{d}\rho + \frac{1}{\pi}\int_{\pi}^{2\pi} (\rho-\pi) f_{\theta}(\rho)\mathrm{d}\rho.$

You should check this when $f_\theta$ is uniformly distributed. It doesn't look right.

If I understand the problem, you have a single line amd a pair of lines separated by an angle $\theta$. As above, we take the line to be the x-axis, $(x_1, y_1)$ to be above the x-axis at an angle $\phi$. Then the probability that the second point is also above the x-axis is:
$$p = \int_0^{\pi - \phi} f(\theta) d\theta + \int_{2\pi - \phi}^{2\pi} f(\theta) d\theta$$
And if you check with a uniform $f(\theta)$ you get $p = \frac 1 2$, as expected.

 

[LCDF]

Then $\psi$ is irrelevant. Take the $x$ and $y$ axes. Assume, without loss of generality, that $(x_1, x_2)$ is in the first quadrant, with an angle $\phi$ uniform between $0$ and $\pi/2$.

For each $\phi$ you need $\theta$ in the ranges $[0, \frac \pi 2 - \phi)$ or $(2\pi - \phi, 2\pi)$.

Draw a diagram to see this.

You mean $(x_1,y_1)$ instead of $(x_1, x_2)$?

 

[PeroK]

You mean $(x_1,y_1)$ instead of $(x_1, x_2)$?

Yes, fixed.

 

[jbriggs444]

Yes, $(x_1, y_1)$ is uniformly distributed in the plane.

There is no such thing as a uniform distribution in the plane.

 

[PeroK]

There is no such thing as a uniform distribution in the plane.

That's a good point. I guess we could limit it to points at a fixed distance; or, have an arbitrary radial distribution.

 

[LCDF]

There is no such thing as a uniform distribution in the plane.

Right. I think saying that the angle is uniform would be accurate.

 

[LCDF]

Then $\psi$ is irrelevant. Take the $x$ and $y$ axes. Assume, without loss of generality, that $(x_1, y_1)$ is in the first quadrant, with an angle $\phi$ uniform between $0$ and $\pi/2$.

For each $\phi$ you need $\theta$ in the ranges $[0, \frac \pi 2 - \phi)$ or $(2\pi - \phi, 2\pi)$.

Draw a diagram to see this.

I do not see why $\psi$ is irrelevant. A uniformly distributed $\psi$ in $[0, \pi]$ is equivalent to fixing the line as the x-axis and one of the points makes an angle $\psi$ that is uniformly distributed in $[0, \pi]$. But this is possible only if the orientation of the line is uniformly distributed $\psi$ in $[0, \pi]$.

 

[PeroK]

I do not see why $\psi$ is irrelevant. A uniformly distributed $\psi$ in $[0, \pi]$ is equivalent to fixing the line as the x-axis and one of the points makes an angle $\psi$ that is uniformly distributed in $[0, \pi]$. But this is possible only if the orientation of the line is uniformly distributed $\psi$ in $[0, \pi]$.

If $\psi$ is relevant, then $p$ will be a function of $\psi$.

If you take $\psi = 0$ and $\psi = \frac \pi 4$, say, and do the calculations, you would get a different answer. Do you expect that? If so, why?

If we haven't understood the problem and you get a different scenario for different $\psi$ then it is relevant. From what I understand the probability should be independent of $\psi$.

 

[LCDF]

If $\psi$ is relevant, then $p$ will be a function of $\psi$.

If you take $\psi = 0$ and $\psi = \frac \pi 4$, say, and do the calculations, you would get a different answer. Do you expect that? If so, why?

If we haven't understood the problem and you get a different scenario for different $\psi$ then it is relevant. From what I understand the probability should be independent of $\psi$.

Your suggestion was very useful, and I obtained the answer based on it. And I agree that the probability should not be dependent on $\psi$. What I was thinking is that the case of a random line with orientation $\psi$ that is uniformly distributed in $[0, \pi]$ and a fixed point at $(x_1, y_1)$ is equivalent to the case of considering a fixed line as the x-axis and a point whose angle with the x-axis is uniformly distributed in $[0, \pi]$. Isn't it?

If I am not misunderstood you earlier suggestion, you were exactly implying the second interpretation, while I the first. And I get the desired answer by putting things in your interpretation that matches my simulation while considering a fixed point and a randomly oriented line!

 

[PeroK]

If I am not misunderstood you earlier suggestion, you were exactly implying the second interpretation, while I the first. And I get the desired answer by putting things in your interpretation that matches my simulation while considering a fixed point and a randomly oriented line!

I don't understand that. The calculations should be independent of $\psi$.

 

[LCDF]

I don't understand that. The calculations should be independent of $\psi$.

Certainly, the calculations should be independent of $\psi$ and my answer reflects that. But I now see that it is nature of the randomness in points than the randomness in the orientation. Thanks!