#cauchy distribution

[probaway]

The Physics of Wall Street by James Weatherall - review comments

The Physics of Wall Street by James Weatherall – review comments

A simple-minded person should pause before criticizing some of the most famous mathematicians of the last two hundred years. – Pause !!! - Okay, having paused, let me proceed. The book The Physics of Wall Street: A Brief History of Predicting the Unpredictable by James Owen Weatherallis about brilliant, highly motivated mathematicians making mistakes in an attempt to understand the fluctuations…

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[isomorphismes]

±∞

The Cauchy distribution (?dcauchy in R) nails a flashlight over the number line

and swings it at a constant speed from 9 o'clock down to 6 o'clock over to 3 o'clock. (Or the other direction, from 3→6→9.) Then counts how much light shone on each number.

In other words we want to map evenly from the circle (minus the top point) onto the line. Two of the most basic, yet topologically distinct shapes related together.

You've probably heard of a mapping that does something close enough to this: it's called tan.

Since tan is so familiar it's implemented in Excel, which means you can simulate draws from a Cauchy distribution in a spreadsheet. Make a column of =RAND()'s (say column A) and then pipe them through tan. For example B1=TAN(A1). You could even do =TAN(RAND()) as your only column. That's not quite it; you need to stretch and shift the [0,1] domain of =RAND() so it matches [−π,+π] like the circle. So really the long formula (if you didn't break it into separate columns) would be =TAN( PI() * (RAND()−.5) ). A stretch and a shift and you've matched the domains up. There's your Cauchy draw.

In R one could draw three Cauchy's with rcauchy(3) or with tan(2*(runif(3)−.5)).

What's happening at tan(−3π/2) and tan(π/2)? The tan function is putting out to ±∞.

I saw this in school and didn't know what to make of it--I don't think I had any further interest than finishing my problem set.

I saw as well the ±∞ in the output of flip[x]= 1/x.

1/−.0000...001 → −∞ whereas 1/.0000...0001 → +∞.

It's not immediately clear in the flip[x] example but in tan[x/2] what's definitely going on is that the angle is circling around the top of the circle (the hole in the top) and the flashlight of the Cauchy distribution could be pointing to the right or to the left at a parallel above the line.

Why not just call this ±∞ the same thing? "Projective infinity", or, the hole in the top of the circle.

[isomorphismes]

Stereographic projection is a way of mapping ℝ -- which is infinitely long -- onto a circle, which has a finite length (circumference).

In a sense, the circle is larger than ℝ. That is, you can map injectively every point from ℝ onto the circle, and then there's still a point left over at the top of the circle. That point gets associated to ∞.

This mapping also generates the Cauchy distribution from statistics.

An infinite thing is smaller than a finite thing. Weird, right? Well on the other hand, you could also map ℝ to the unit circle and exchange "x inches ∈ ℝ" for "x degrees ∈ circle", like wrapping an infinite string around the circle, and it would of course wrap around many times.

So either thing could be considered bigger. This is why the axiom of choice is messed up.

[isomorphismes]

I want to understand this connection better.

Although it sounds the most involved, arctan answers the simplest trigonometric question you could ask. "The full moon subtends seven degrees of my vision; so how big is the moon?" The arctan (with radius = distance from Earth to moon) is the answer.

But what does that simple operation have to do with:

a continuous sum of

an inversion of

one more than

some continuous range of numbers

times themselves?

That's confusing.

Four years later...

This comment shows how to get the answer. (Pair with this on the derivative of logarithms.) A consequence of the chain rule (or perhaps another way to state it!) is that the derivative of ƒ⁻¹ at a point p is flip( derivative( ƒ )) evaluated at ƒ⁻¹(p), presupposing that all of the maps fit together right.

arctan is defined, for simplicity, on a circle of unit radius. (This just means if you're looking at the moon then use units of "one moon-distance".) It takes as an argument a ratio of sides and returns an angle θ. Since derivative ( tan ) = derivative (sin⁄cos) = 1⁄cos² = sec × sec (reasoned with calc 101), by combining that derivative ( tan ) with JavaMan's perspective on the derivatives of inverse-functions we can argue that 

Java Man's idea gets us to look at the triangle which, since it's constrained to a circle by the equivalence-classing of triangles to be just the ones with a certain angle (see fibration), limits us to just one free parameter R (a ratio of opposite O to adjacent A side lengths of an equivalence-class of triangles). After following Java Man's logic to see why derivative( tan ) = sec² implies derivative( inverse of tan ) = cos² [at the angle which is implied by tan R], we're left knowing that A=the adjacent side implied by the cos R = cos O⁄A of the original O⁄A ratio we were given as the natural input space (ratios) which the tan function accepts. This isn't enough because the answer needs to be in O⁄A terms to match the input. We have to do some Pythagorean jiu jitsu involving A = A⁄1 and 1=1²=A²+O² to get the answer into an O⁄A form (since that was the information we were given). Using the 1=A²+O² is using the natural delimitation of the inscribing circle to make the two A and O move together the way they should on the circle, by the way. The algebraic jiu jitsu then yields A = flip( 1+R² ), now using the proper input R=O⁄A.

The point of all this mangling was merely to match up cosine's output with tangent's input. Sheesh with all the symbols!

But that's merely deriving the correct answer algebraically, with a bit of "why" from the comment's perspective on inverse functions generally. What about my original question? Why does this sequence of mappings, if iterated, subtend the moon?