Expected Radial Distance on a Circular Disk
A frog starts at the center of a circular lily pad of radius 1 meter. The frog hops to a uniformly random point on the lily pad (uniform over the area of the disk).
Find the expected radial distance of the frog from the center after the hop.
Hints
- A uniform point on a disk is not uniform in its radial distance. How does the area enclosed within radius $r$ grow?
- The CDF of the radial distance is $F_R(r) = r^2$, giving PDF $f_R(r) = 2r$ on $[0, 1]$. Now compute $E[R]$.
- Alternatively, use the tail probability formula: $E[R] = \int_0^1 P(R \ge r) \, dr = \int_0^1 (1 - r^2) \, dr$.
Worked Solution
How to Think About It: A point chosen uniformly on a disk is not uniform in its radial distance -- points are more concentrated at larger radii because there is more area out there. The PDF of the radial distance $R$ is $f_R(r) = 2r$ on $[0, 1]$, not uniform. So $E[R]$ is not $1/2$ -- it should be larger, since the density is weighted toward the edge. Quick guess: somewhere around $2/3$.
Quick Estimate: The CDF of $R$ is $F_R(r) = r^2$ (ratio of areas: circle of radius $r$ over circle of radius 1). The median is at $r = 1/\sqrt{2} \approx 0.71$. The mean should be a bit less than the median since the density $2r$ is right-skewed on $[0,1]$... actually, $2r$ is increasing, so the mass is concentrated toward 1, meaning the mean is above $1/2$. Let us just compute: $E[R] = \int_0^1 r \cdot 2r \, dr = 2/3$.
Approach: Find the PDF of $R$ from the geometric probability, then compute $E[R]$ directly. Alternatively, use the survival function method.
Formal Solution:
Method 1: CDF approach.
The frog lands uniformly on the disk of radius 1. The probability of landing within radius $r$ is the ratio of areas:
$$F_R(r) = P(R \le r) = \frac{\pi r^2}{\pi \cdot 1^2} = r^2, \quad 0 \le r \le 1$$
Differentiating:
$$f_R(r) = 2r, \quad 0 \le r \le 1$$
This is a $\text{Beta}(2, 1)$ distribution. The expected value is:
$$E[R] = \int_0^1 r \cdot 2r \, dr = 2 \int_0^1 r^2 \, dr = 2 \cdot \frac{1}{3} = \frac{2}{3}$$
Method 2: Survival function (tail sum).
For a non-negative random variable:
$$E[R] = \int_0^{\infty} P(R \ge r) \, dr = \int_0^1 (1 - r^2) \, dr = \left[r - \frac{r^3}{3}\right]_0^1 = 1 - \frac{1}{3} = \frac{2}{3}$$
Both methods confirm the answer.
Answer: $E[R] = \dfrac{2}{3}$.
Intuition
The key insight is that "uniform on the disk" does not mean "uniform in radius." Area grows as $r^2$, so there is more room at larger radii -- the annulus between $r$ and $r + dr$ has area $2\pi r \, dr$, which grows linearly with $r$. This is why the radial PDF is $f(r) = 2r$, not $f(r) = 1$. The result $E[R] = 2/3$ is strictly greater than $1/2$ because the density is tilted toward the boundary.
This is a classic interview problem that tests whether you understand the Jacobian of polar coordinates. The same principle applies whenever you transform to polar or spherical coordinates: the "volume element" is not flat in the radial direction. In higher dimensions, the effect is even more dramatic -- a uniform point in a $d$-dimensional ball has $E[R] \to 1$ as $d \to \infty$, because essentially all the volume is near the surface.