Consider points uniformly distributed on the surface of a sphere with radius $R$ centered at the origin. I would like to find the marginal distribution of the $X$ coordinate of points on the surface of the sphere.

Why is it not correct that the marginal pdf is the ratio of the circumference of the circle in the y-z plane at a fixed $x$ over the surface area of the sphere?

The latter is $4 \pi R^2$. The former is $2 \pi \sqrt{R^2 - x^2}$. So we get $$ f_X(x) = \frac{\sqrt{R^2 - x^2}}{2R^2} $$

This marginal pdf is apparently wrong, but the approach makes sense.

Edit: Apparently the marginal pdf is constant (uniform) and independent of $x$. If this is true, then it's really going against my intuition.


Suppose $P = (X, Y, Z)$ is a randomly chosen point on the surface of such a sphere. Would you agree that the marginal cumulative distribution function of $X$ would be $$\Pr[X \le x] = \frac{\text{surface area of "cap" with } X \text{-values at most } x}{\text{total surface area}}$$ for some $x \in [-R, R]$? If so, then recall that the surface area of such a cap is given by $$A(x) = 2\pi R (x+R).$$ Hence $$F_X(x) = \Pr[X \le x] = \frac{2\pi R (x+R)}{4 \pi R} = \frac{x+R}{2R}.$$ Therefore $$f_X(x) = \frac{\mathbb 1(-R \le X \le R)}{2R};$$ that is to say, $X$ has uniform density on $[-R,R]$ as claimed.

Why does the calculation you proposed not work? The reason is because you are comparing an arclength (the circumference measure) against an area. The measures of these are not the same. In particular, by making such an argument, you are ignoring a factor that relates the particular $x$-value to the probability of being in a neighborhood of that $x$.

What I mean by this is that you need to be more careful about formalizing the argument. Consider instead the limiting behavior as $\epsilon \to 0$ of the probability $$\Pr[x - \epsilon \le X \le x + \epsilon].$$ Then instead of a circumference, we are interested in a thin "band" of the sphere's surface satisfying $|X - x| \le \epsilon$. The first thing you will notice is that for a fixed choice of $\epsilon > 0$, the lateral width of the band is not constant as a function of $x$: for instance, when $x$ is "close" to $R$, the band will appear thickest, compared to $x$ "close" to $0$, where the band appears thinnest, even though the band is sandwiched between planes equally spaced apart by $2\epsilon$. Therefore, you must adjust the calculation accordingly, in as much as computing the arclength of a circular sector does not give the same result as the projection of that arclength onto the $x$-axis.

  • $\begingroup$ I think this makes sense. However, the way I approached this problem was the same way I approached the question "What is the marginal pdf of $x$ for points uniformly distributed inside a 2D circle." In that case, if we take the ratio of the vertical length over the area, we get the correct answer. By vertical length, I mean if you fix $x$, the length would be $2y$, which is $2 \sqrt{R^2 - x^2}$. And hence $f_X(x) = \frac{2\sqrt{R^2 - x^2}}{\pi R^2}$, which is the correct answer. But why does this idea apply for that case? $\endgroup$ – David Aug 12 '20 at 19:32
  • $\begingroup$ In the above comment I am comparing a length to an area, which are also different measures. $\endgroup$ – David Aug 12 '20 at 19:38
  • $\begingroup$ In this case, you are looking at a volume in two dimensions, not a surface area, which is arc length. The analogy in three dimensions would be to say "what is the marginal PDF of $x$ for points uniformly distributed inside a sphere, not just on the surface of it. Then the same reasoning works because now you are looking at a volume integral. $\endgroup$ – heropup Aug 12 '20 at 21:02
  • $\begingroup$ Conversely, if you try to apply your argument to the two-dimensional case, you would obtain the incorrect result that the arclength of a semicircle equals its diameter. $\endgroup$ – heropup Aug 12 '20 at 21:03
  • $\begingroup$ Makes sense. I still find the uniform pdf of the marginal case to be not so intuitive. From a visual perspective, it just seems like the probability of $X$ being close to $-R$ or $R$ is smaller than the center because there's less surface area at the ends (assuming a constant $\Delta x$ $\endgroup$ – David Aug 12 '20 at 22:38

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