Math Quiz 9

[Aren Siekmeier]

Originally Posted by Ether

[b]OK I'm going to drop a bombshell here.

The correct answer to the OP, as worded, is 0.33634

...

Sampling segments by choosing the segment endpoint coordinates from a random uniform distribution does not match the probability density of the segment lengths in the infinite set.

It's hard to disagree with your first figure.

If you can switch the order of integration and then integrate the pdf over theta, then you should have a pdf for L only. I'm trying to wrap my head around why that pdf for L should be different than the one effected by a uniform weighting of dx1dx2dy1dy2 differential elements. The concept in my head is that there's no invertible change of variables with which to set up that translation. The transformation from coordinates to length is not injective, so the Jacobian in the change of variables is zero: hard to divide by, conceivably giving the ratio between our answers in some sort of limit. Or from another view the four-dimensional differential element is of measure zero within the one-dimensional one, reflected by a pdf for the coordinates that is somehow undefined or not finite??

I'm very curious if there is an appropriate pdf for the coordinate approach that would give this result. Or is it just the wrong way to look at the problem?

This seems to boil down to the difference between:
[1] Average distance between all unordered pairs of points
[2] Average length of all line segments
But what's the difference?

[Ether]

Originally Posted by Aren Siekmeier

This seems to boil down to the difference between:
[1] Average distance between all unordered pairs of points
[2] Average length of all line segments
But what's the difference?

The difference is in how you choose samples.

If you want the average distance between all unordered pairs of points, then you choose coordinates of pairs of points from a uniform random distribution, and compute the corresponding distance.

If you want the average length of all segments, you randomly choose a segment length, a segment orientation, and the coordinates of the center of the segment. Then you discard any chosen segments which are not inside the square. If you run a Monte Carlo sim of that, you'll get 0.3363

...and you'll get a pdf of L only, if you make a histogram of the data and adjust it to have an area of 1.

[Aren Siekmeier]

Originally Posted by Ether

The difference is in how you choose samples.

If you want the average distance between all unordered pairs of points, then you choose coordinates of pairs of points from a uniform random distribution, and compute the corresponding distance.

If you want the average length of all segments, you randomly choose a segment length, a segment orientation, and the coordinates of the center of the segment. Then you discard any chosen segments which are not inside the square. If you run a Monte Carlo sim of that, you'll get 0.3363

...and you'll get a pdf of L only, if you make a histogram of the data and adjust it to have an area of 1.

How practical of you

I'd like to think it's possible to parametrize the choice of line segment by end points. Curious about the difference between that and the problem (simply choosing uniformly distributed end points) that we wrote 94 posts about. Obviously the end points of all line segments are not uniformly distributed in the square.

I've got an idea for finding just what their distribution is that I might get to if ever get unswamped this week.

[Ether]

FWIW: seglen pdf for L theta x y versus x1 y1 x2 y2 sampling

[Aren Siekmeier]

Still stuck in my old ways aren't I.

Here's the distribution of endpoints using your sampling procedure.
https://plot.ly/~compwiztobe/9/

from random import random
from math import pi,cos,sin,sqrt
import numpy

def randseg():
	hl=random()/sqrt(2)
	th=random()*pi
	cx=random()
	cy=random()
	return [cx-hl*cos(th),cx+hl*cos(th),cy-hl*sin(th),cy+hl*sin(th)]

def inbox(seg):
	return(all([x>=0 and x<=1 for x in seg]))

def go(n):
	l=[]
	for i in range(n):
		x=randseg()
		if inbox(x):
			l.append(x)
	return(l)

def pts(l):
	p=[]
	for x in l:
		p.append([x[0],x[2]])
		p.append([x[1],x[3]])
	return(p)

data=pts(go(1000000))

#then bin and plot 2d histrogram with plotly API

Curious about an analytical form, but this at least shows the idea. The way we solved the problem before assumed this plot was all one color.

[GeeTwo]

Originally Posted by Ether

The difference is in how you choose samples.

If you want the average distance between all unordered pairs of points, then you choose coordinates of pairs of points from a uniform random distribution, and compute the corresponding distance.

If you want the average length of all segments, you randomly choose a segment length, a segment orientation, and the coordinates of the center of the segment. Then you discard any chosen segments which are not inside the square. If you run a Monte Carlo sim of that, you'll get 0.3363

I believe this is fallacious, in particular that for a given center point and two different segment lengths, there are the same number of segments. For the center point at (0.5, 0.5) and length 0.2, the candidate segments are the diameters of a circle of radius 0.1. For the center point at (0.5, 0.5) and length 1.0, the candidate segments are the diameters of a circle of radius 0.5. As there are five times as many points on the large circle to serve as either end of a diameter, it logically has five times as many diameters.

When you take this into account, the integral in the denominator becomes:

∫₀^π/4∫₀^1/cosθ (1- Lcosθ) (1- Lsinθ) L dL dθ =
∫₀^π/4∫₀^1/cosθ L - L²cosθ - L²sinθ + L³cosθsinθ dL dθ =
∫₀^π/4[ L²/2 - L³cosθ/3 - L³sinθ/3 + L⁴cosθsinθ/4 ]₀^1/cosθ dθ =
∫₀^π/4sec²θ/2 - sec²θ/3 - sec²θtanθ/3 + sec²θtanθ/4 dθ =
∫₀^π/4sec²θ/6 - sec²θtanθ/12 dθ =
[tanθ/6 - tan²θ/24 ]₀^π/4 =
[1/6 - 0 - 1/24 + 0 ] = 1/8

If you then integrate the adjusted calculation using 1/8 for the deominator, just substitute r for L and α for θ, and you will have the third equation of my most recent post in this thread.

Addition: Also, the idea of the average segment in a square of size 1 being within 1% of the average segment on any of its sides is .. counter-intuitive, to say the least.

[Ether]

For the center point at (0.1, 0.1) and length 0.2, the candidate segments are the diameters of a circle of radius 0.1. For the center point at (0.1, 0.1) and length 1.0, the number of candidate segments is zero.

[GeeTwo]

Originally Posted by Ether

For the center point at (0.1, 0.1) and length 0.2, the candidate segments are the diameters of a circle of radius 0.1. For the center point at (0.1, 0.1) and length 1.0, the number of candidate segments is zero.

Though it is not relevant, there are plenty of candidates, though none are selected.

Here's another case which more clearly shows the fallacy.

• What is the average distance from the origin to a point in the unit square?
• What is the average length of the radials from the origin to points in the first quadrant of a circle of radius √2?

Solving the first question in Cartesian coordinates, we find the answer to be (√2 + ln(1+√2))/3 = 0.765...

Solving the second question using the logic used by Ether to count radials, the answer comes out to √2/2 = 0.707..

As the quadrant of the large circle includes all the same points as the unit square plus others which area all farther away than either average, there is a fallacy here somewhere.

By weighting the density by r, the second question is properly answered 2√2//3 = .942...

Addition:
To more fully explore this point, I propose the following questions:

• What is the average length of a segment in a rectangle 1 unit wide and 0.5 units high?
• What is the average length of a segment in a rectangle 1 unit wide and 0.1 units high?
• What is the average length of a segment in a rectangle 1 unit wide and infinitesimally high? (that is, the limit as height goes to zero from above)
• How does that compare to the average length of a line segment within the line segment from 0 to 1?

[Ether]

Question for Gus:

What is the average length of all the vertical line segments in a unit circle?

[Hitchhiker 42]

Originally Posted by Ether

Question for Gus:

What is the average length of all the vertical line segments in a unit circle?

If I may,

Spoiler for Thoughts:

My approach would be to take the top semicircle (equation y=sqrt(1-x^2)) and find the average value of those vertical segments by integrating from -1 to 1 and dividing the area by the base (-1 to 1 is 2). Then, you could multiply by 2 for a full circle. Answer comes out to pi/2.

Work attached below.

EDIT: Gosh darn I only did those with both points touching the the edges. Lemme work on that.

[GeeTwo]

Originally Posted by Hitchhiker 42

EDIT: Gosh darn I only did those with both points touching the the edges. Lemme work on that.

Also, don't forget to weight by the number of segments - the x coordinates near the center contain more segments than those near the left and right edges.

In any case, to answer Ether's question, the short answer is "not the same as the average length of segments within a small angle (e.g. 1 second of arc) of vertical within the circle." We've got another build session this evening getting ready for Red Stick Rumble, so I don't know how much I'll get done on this tonight.

Addition:
Hitchhiker, thanks for answering a different question. While walking across the street to get lunch, I think I put some nails in this coffin. Consider these two questions:

• What is the average length of a vertical chord of the unit circle?
• What is the average length of a chord of the unit circle which passes through (-1,0)?

You answered the first - the area of the unit circle divided by its width: pi/2.
I answered the second in my paper a score or so posts back (first problem), in a form that is not incompatible with Ether's method, provided angle is checked before length: 4/pi.
If you answer the second using Ether's method but checking length first, then picking one of two possible angles, you will get an average value of 1.

Completing reducto ad absurdum is left to the reader - back to work!

Originally Posted by Ether

I just can't trip you up, can I

Well, not that easily.

[Ether]

Originally Posted by GeeTwo

In any case, to answer Ether's question, the short answer is "not the same as the average length of segments within a small angle (e.g. 1 second of arc) of vertical within the circle."

I just can't trip you up, can I