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Comment by leephillips | Hacker News Reader

An example from a book I used to own, called, I think, _Treatment of Experimental Data_: imagine a factory manufacturing ball bearings. They want them all to have the same radius, but because of random errors, the radii will be normally distributed about some mean. If this is true, other random variables, such as the mass, can not be normally distributed.

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Technically true, but the smaller the standard deviation of the radius (relative to the mean), the more closely the distribution of masses will approximate a normal distribution.

This. Technically the radii can't even have a normal distribution, because the support of the normal is (-Inf, Inf) but the radii are bounded below by 0.

So you might say that the radii = r + e, where r is constant and e is normal around 0 with variance way, way smaller than r. But then the volume of the ball bearings = K * (r+e)^3, and because r >> e, the largest source of randomness term is going to be 3Kr^2e, which makes the whole thing look pretty normal.

Sorry, could you explain why mass can't be normally distributed? If, say, the mass of the bearing is related to its radius, then shouldn't it follow similar distribution?

Mass is proportional to radius cubed. Cubing a normal distribution makes it not normal.

Though it should be noted that "cubing" doesn't refer to just taking the cube, i.e if f(x) = e^{-x^2/2} is proportional to the standard normal distribution then the cube of f(x), namely f * f * f is not the distribution you're considering here. Rather it's the "pushforward measure" [1], in this case is 1/3 y^{-2/3} e^{-y^{2/3}/2} where y=x^3 is the mass and x is the radius. (Up to constants.)

[1] https://en.wikipedia.org/wiki/Pushforward_measure

Cubing does refer to just taking the cube, but the cube of random variable, and not its probability density function -- the density after cubing is indeed the pushforward.

This is indeed one confusion that I ran into very often when first learning stats.

Transforming the random variable is a very different thing than transforming the probability density function directly.

The distinction between a random variable and a distribution (or density) is not made clear enough in classes I think.

    R = r + e

    R^{3} = (r+e)^{3} = r^{3} + 3r^{2}e + O(e^{2})

Unless you expect the error to be significant with respect to r, you can ignore the higher order terms. And voila, you get an approximately normal distribution.

Yes, that's why.

mass = 4/3 pi radius cubed

The cube of a normal distribution is indeterminate.

Had the relationship between the mass and the radius been linear (M = a * R + b) then yes the mass would have followed a normal distribution as well (with different parameters of course).

That factor should be 4/3 rather than 3/4. When you differentiate it, you need to get 4πr² for the surface area.

You are correct, my memory failed me and I didn't check. Thank you.

That distribution still has the same median, with half the samples on one side and the other half on the other side.

That looks a lot like a normal distribution even if it's not one.

Suppose the radius is normally distributed, and is a random varable X. The mass will be proportional to the volume, which increases with the cube of the radius, so the mass will be a random variable kX^3 (k = 4/3 x pi x density here). X and kX^3 won't have the same distribution.

Depends on whether the radius errors are generated from a separate process than the mass errors. (1) Get a glob of molten metal, mass determined by how much metal is there, some error there; (2) shape it into a ball, and suppose the radius is mainly determined by air that gets trapped, then some nearly independent error there.

That's true. But the idea that distributions related non-linearly cannot both be normal holds.

That's kind of an odd metric to begin with. Radius would probably be derived from the imperfections in the consistency of matter. Whether that distribution is gaussian probably depends on the fabrication process and the materials...

This kinda blows my mind.

What happens if they do quality control on both mass and radius?

I imagine the reason why, in your example it is the radius rather than the mass that is normal, is due to what QA is focused on.

The reason that QA is done on radius is because the purpose of a ball bearing is to sit inside its housing to within a fairly tight tolerance. If the balls are out of spec, then potentially the thing the bearing is coupling will be out of alignment/won't rotate freely. The mass is less important.

If you did QA on mass, you could easily calculate what mass would be required for a particular radius, but that would let misshapen bearings through.

You could do QA on both mass and radius, but unless you have potential contaminants, the radius gives you the mass for free so there isn't much point.

But its still log-normal

Hmm, why is that?

with constant density, p, and spherical particles:

$ m = pi * r^3 * p $

Have some Unicode symbols!

m = πr³ρ

(with no disrespect to your use of TeX)

But also it should be m = (4/3)πr³ρ because the volume of the sphere is (4/3)πr³, not πr³.

Alternative: have a mathjax bookmarklet :) https://www.checkmyworking.com/misc/mathjax-bookmarklet/

Wow, nice!

Not enough Unicode. Let's make it:

m = (¾)⁻¹πr³ρ

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