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Let X and Y be uniformly distributed on a unit disk such that

$x^2 + y^2 \leq 1$

Let $R = \sqrt{X^2 + Y^2}$. What are the CDF and PDF of $R$?

I know that the area of the unit disk is

$A = \pi r^2 = \pi 1^2 = \pi$

Thus, I think that the joint PDF of $X$ and $Y$ is the following, but I am not sure about this:

$f_{X, Y}(x, y) = \frac{1}{\pi}, \ \ \ x^2 + y^2 \leq 1$

I know that

$P(R \leq r) = P(\sqrt{X^2 + Y^2} \leq r)$.

This is where I'm stuck.

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    $\begingroup$ $x^2 + y^2 + 2 \le 1$ does not define a unit disk. $\endgroup$
    – Moormanly
    Commented Nov 19, 2018 at 3:01
  • $\begingroup$ Reopened. For a uniform, probability is proportional to area. Consider what proportion of the area is within a distance $r$ of the center. (This relates immediately to where you're stuck.) $\endgroup$
    – Glen_b
    Commented Nov 27, 2018 at 3:48

1 Answer 1

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You are extremely close to the answer. You have already noted that the area of a circle with radius $r$ is $A(r) = \pi r^2$. Since you are selecting a uniform point on the circle, this means that you have the cumulative distribution function:

$$F_R(r) = \mathbb{P}(R \leqslant r) = \frac{A(r)}{A(1)} = \frac{\pi r^2}{\pi} = r^2 \quad \quad \quad 0 \leqslant r \leqslant 1.$$

The reasoning behind this is that with a uniform distribution we have a flat density over the circle. So the probability of falling within a given area is the relative size of that area compared to the size of the circle. This gives the corresponding density function:

$$f_R(r) = 2r \quad \quad \text{for all } 0 \leqslant r \leqslant 1.$$

As you can see from this result, although the chosen point is uniformly distributed on the circle, the resulting distance $R$ from the origin to the point does not have a uniform distribution; as expected, it is more likely to be nearer to the outer edge of the circle than its origin.

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  • $\begingroup$ Thanks, @Ben. I don't understand how you arrived at $\frac{A(r)}{A(1)}$. Could you please elaborate on that? $\endgroup$
    – MSE
    Commented Nov 27, 2018 at 14:51
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    $\begingroup$ Ah, I think that I understand it now. For each value of $r$, there is a circle with that radius, and it has the origin (0,0) at its centre. Thus, the CDF is the proportion of the coverage of this inner circle over the possible sample space, which is the circle of $r = 1$. (We can just divide the area of the inner circle by this outer circle, because the probability is uniform throughout the sample space.) @Ben: Is this correct? $\endgroup$
    – MSE
    Commented Nov 27, 2018 at 18:25
  • $\begingroup$ Yes, that is correct. $\endgroup$
    – Ben
    Commented Jun 28, 2019 at 2:09

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