Expected Chord Length Through a Disk (Random Radius Method)
A chord is drawn through a unit disk as follows: pick a point uniformly at random inside the disk, then draw the chord that passes through that point and is perpendicular to the radius at that point.
What is the expected length of the chord?
Note: this is one of the three classical sampling methods in Bertrand's paradox -- each method gives a different answer. Be explicit about which method you are using.
Hints
- A chord perpendicular to the radius at distance $d$ from the center has length $2\sqrt{1-d^2}$ -- draw the right triangle formed by the center, the foot of the perpendicular, and one endpoint of the chord.
- For a point chosen uniformly inside a unit disk, the distance $D$ from the center has CDF $P(D \leq d) = d^2$ (area of inner disk divided by total area), so the PDF is $f_D(d) = 2d$. The factor of $2d$ reflects the fact that more area lies near the boundary.
- The expected chord length is $\int_0^1 2\sqrt{1-d^2} \cdot 2d \, dd$. Use the substitution $u = 1 - d^2$ to evaluate this integral.
Worked Solution
How to Think About It: A chord perpendicular to the radius at distance $d$ from the center has length $2\sqrt{1-d^2}$ (Pythagoras on the right triangle: center, foot of the perpendicular, endpoint). So the whole problem reduces to *what is the distribution of $d$* for a uniform point in the disk. The heuristic that makes this trivial in your head is a change of variable to the quantity that is actually uniform: for a uniform point, $d^2$ is Uniform$(0,1)$ (the fraction of area inside radius $d$ is $d^2$). Naively averaging over $d$ tempts you to plug in a 'typical' $d$ and stop — a Jensen trap, since $2\sqrt{1-d^2}$ is concave, so $E[\text{length}] < \text{length}(E[d])$.
Quick Estimate: Let $u = 1 - d^2$. Since $d^2$ is Uniform$(0,1)$, so is $u$, and the chord length is exactly $2\sqrt{u}$. Now the estimate is a one-liner mental integral: the average of $\sqrt{u}$ over $u \in (0,1)$ is $\int_0^1 u^{1/2}\,du = \tfrac{2}{3}$ (each intermediate: power rule bumps the exponent to $3/2$ and divides, $\tfrac{1}{3/2} = \tfrac23$). Hence $E[\text{length}] \approx 2 \cdot \tfrac23 = \tfrac43 \approx 1.33$. Sanity bracket: chords range from $2$ (center) down to $0$ (edge), and the area density $2d$ pushes the point toward the rim, so the mean chord should sit well below $1$ in $d$ but the length still averages above $1$ — $1.33$ fits, and the Jensen bound $2\sqrt{1-E[d^2]} = 2\sqrt{1/2} \approx 1.41$ correctly sits *above* it.
Approach: Find the PDF of $d$, then compute $E[2\sqrt{1-d^2}]$ — or use the $u = 1-d^2$ shortcut directly.
Formal Solution:
(a) Distribution of $d$. For a uniform point in the unit disk, $P(D \le d) = \dfrac{\pi d^2}{\pi} = d^2$ on $[0,1]$, so the PDF is $$f_D(d) = 2d, \qquad 0 \le d \le 1.$$ The factor $2d$ (the circumference-of-ring Jacobian) is why the point favors large $d$.
(b) Chord length. From the right triangle, $$\ell(d) = 2\sqrt{1 - d^2}.$$
(c) Expected length. $$E[\ell] = \int_0^1 2\sqrt{1-d^2}\cdot 2d\,dd = 4\int_0^1 d\sqrt{1-d^2}\,dd.$$ Substitute $u = 1 - d^2,\ du = -2d\,dd$: $$E[\ell] = 4\int_1^0 \sqrt{u}\left(-\tfrac{du}{2}\right) = 2\int_0^1 u^{1/2}\,du = 2\cdot\frac{u^{3/2}}{3/2}\Big|_0^1 = 2\cdot\frac23.$$ $$\boxed{E[\ell] = \frac{4}{3} \approx 1.333}$$ This matches the mental route: the substitution is exactly the statement that $u = 1-d^2 = 1-D^2$ is Uniform$(0,1)$, so $E[\ell] = 2\,E[\sqrt{u}] = 2\cdot\tfrac23$.
Context — Bertrand's paradox. This $4/3$ is the random-radius method. The other two classical samplings differ: random angle on a diameter gives $E[\ell] = \pi/2 \approx 1.571$; two independent uniform endpoints on the circle give $E[\ell] = 4/\pi \approx 1.273$. 'Random chord' is under-specified — the sampling mechanism must be stated.
Answer: $E[\ell] = \dfrac{4}{3} \approx 1.33$ (random-radius method).
Intuition
The key lesson here is the difference between the distribution of a random variable and the distribution of a function of that variable. The uniform distribution on the disk does NOT correspond to a uniform distribution on the radius $d$ -- the density of $d$ is $2d$, not $1$, because larger radii correspond to larger rings (more area). This is the Jacobian factor from converting Cartesian to polar coordinates, and forgetting it is the most common error on geometric probability problems.
The broader context -- Bertrand's paradox -- is genuinely important for anyone doing probabilistic modeling. The phrase 'choose a random X' is meaningless without specifying the sampling mechanism. In finance, the same ambiguity appears in 'random time horizon,' 'random price path,' and 'random portfolio weights.' The resolution is always the same: pin down your probability measure precisely before computing anything. Bertrand's paradox is a clean, beautiful illustration of why this matters.