# Connection between derivation of normal distribution and the central limit theorem

I was under the impression that the reason that the normal distribution occurs naturally could be explained by the central limit theorem (CLT). I recently watched a video that described a derivation of the normal probability density function using the case of darts clustered around the center of a dartboard. As far as I can tell, the example is distinct from the CLT.

The assumptions about the distribution of darts are as follows:

1. Darts are clustered around a center and are less dense further from the center.
2. The distribution of darts in the X direction is statistically independent of the distribution of darts in the Y direction.
3. The distribution of darts is equivalent for any point with a given radius $$r = \sqrt{x^2+y^2}$$.

If you told me that dart throws are normally distributed, I would be unphased (something something CLT, presumably). But, it is shocking to me that these very natural assumptions for a two-dimensional distribution are met uniquely by the normal distribution. Is it a coincidence that the normal distribution appears in this context when there doesn't appear to be any connection to the central limit theorem? Or is there a deeper connection?

• You say that the conditions are met uniquely by a bivariate Gaussian, but why wouldn’t independent $t$ distributions meet the conditions? That certainly meets the first two conditions ($t$s taper off, even if slower than normal distributions, and they’d be independent of one another) and I think the third.
– Dave
Commented Jul 5, 2020 at 1:25
• Independent $t$ don't satisfy the third condition: they have outliers preferentially near the horizontal and vertical. Commented Jul 5, 2020 at 1:32
• @Dave I only say it because the normal pdf can be mathematically derived from these assumptions alone (unless I missed something?) as demonstrated in the linked video. I had the same thought as you about the t distribution. Commented Jul 5, 2020 at 1:52

Why aren't there others? The issue is that condition 2 says $$X$$ and $$Y$$ are independent, and condition 3 imposes strict conditions on the relationship between $$X$$ and $$Y$$, and these don't go together.
Suppose $$X$$ and $$Y$$ are independent (condition 2). We also want $$r$$ and $$\theta$$ in polar coordinates to be independent (condition 3). So the joint density has to be given both by $$f(x)f(y)$$ and by $$g(r)$$ (not depending on $$\theta$$).
Transforming the density $$g(r)$$ to rectangular coordinates gives $$g(\sqrt{x^2+y^2})(1/r)$$ so we will only satisfy condition 2 if $$g(\sqrt{x^2+y^2})/r$$ happens to factor into $$f(x)f(y)$$, which is obviously not going to happen very often, but does happen for the Normal pdf (which has a square to undo the square root, then an exponential to turn addition into multiplication).