6 ms·

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 dis

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?

Sharlin 9y ago
Mass is proportional to radius cubed. Cubing a normal distribution makes it not normal.
- tgb 9y ago
  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 https://en.wikipedia.org/wiki/Pushforward_measure
  - xyzzyz 9y ago
    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.
  - bonoboTP 9y ago
    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.
- leephillips 9y ago
  Yes, that's why.
- BeetleB 9y ago
  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.
curiousgal 9y ago
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).
- schoen 9y ago
  That factor should be 4/3 rather than 3/4. When you differentiate it, you need to get 4πr² for the surface area.
  - curiousgal 9y ago
    You are correct, my memory failed me and I didn't check. Thank you.
- user5994461 9y ago
  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.
untunedtheramin 9y ago
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.

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?

Sharlin 9y ago

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

tgb 9y ago
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 https://en.wikipedia.org/wiki/Pushforward_measure
- xyzzyz 9y ago
  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.
- bonoboTP 9y ago
  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.
leephillips 9y ago
Yes, that's why.
BeetleB 9y ago
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.

curiousgal 9y ago

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).

schoen 9y ago
That factor should be 4/3 rather than 3/4. When you differentiate it, you need to get 4πr² for the surface area.
- curiousgal 9y ago
  You are correct, my memory failed me and I didn't check. Thank you.
user5994461 9y ago
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.

untunedtheramin 9y ago

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.