The derivation could be found here:


or here https://www.youtube.com/watch?v=cTyPuZ9-JZ0

They assume that (1) the pdf f(x), f(y) are independent from each other and (2) the pdf of a point g(x,y) only depends on the distance of the point to the bull's eye.

There are definitely examples of pdfs that do not satisfy both (1) and (2). For example, imagine a discrete example where the dart landing at (2,1) or (1,2) both have a probability of 0.4 and the dart landing at (2,2) has a probability of 0.2. This satisfy the assumption (2) that the probability depends only on the distance but not (1). What kind of pdfs can satisfy both (1) and (2)?

  • $\begingroup$ You seem to be denying the analysis presented in your first link, which gives the answer at the line "so if I set $\tilde f\ldots$". Could you explain what might be unclear or missing in that post? $\endgroup$
    – whuber
    Commented Aug 25, 2018 at 13:25
  • $\begingroup$ This is broken down really well in this 3b1b video at 13:00 youtube.com/watch?v=cy8r7WSuT1I&t=2s $\endgroup$ Commented Apr 2, 2023 at 17:55

1 Answer 1


Joint pdfs $f_{X,Y}(x,y)$ that have circular symmetry about the origin (property 2 in the OP's description) exist in profusion, and they all have the property that if we think of $X$ and $Y$ in terms of $R=\sqrt{X^2+Y^2}$ and $\Theta = \arctan\left(\frac YX\right)$, then $R$ and $\Theta$ are independent random variables, and $\Theta \sim U[0,2\pi)$. But, properties 1 (independent $X$ and $Y$) and 2 (circular symmetry about the origin) are harder to satisfy.

It is shown in Sheldon Ross's A First Course in Probability, Prentice-Hall (see Example 2e in Chapter 6.2) that, under the slightly more onerous constraints that the marginal pdfs $f_X(x)$ and $f_Y(y)$ are differentiable functions of $x$ and $y$ respectively and that $f_{X,Y}(x,y)$ is differentiable with respect to $x$ and also with respect to $y$, the only pdfs $f_X$ and $f_Y$ that satisfy both properties 1 and 2 are zero-mean Gaussian pdfs with identical variances.

  • $\begingroup$ Wow, thank you very much!!! That's a great proof. Could you tell me what kind of approach would you take when solving such problems? Would you think more algebraically or geometrically? It was kind of hard to think of geometry that the functions must have to satisfy both properties. $\endgroup$
    – frt132
    Commented Aug 26, 2018 at 10:47

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