Stick Broken in Two: EV of Smaller Piece
A stick of length 1 is broken at a randomly chosen point (uniform on \([0,1]\)). What is the expected length of the smaller piece?
Answer
Let \(X\) be the break point.
- If \(0 \le X \le \tfrac12\), the smaller piece is \(X\).
- If \(\tfrac12 < X \le 1\), the smaller piece is \(1 - X\).
Hence, the expected length of the smaller piece is
\[
\int_{0}^{1/2} x \, dx \;+\; \int_{1/2}^{1} (1 - x)\, dx.
\]
Compute each part:
\[
\int_{0}^{1/2} x \,dx
= \left.\frac{x^2}{2}\right|_{0}^{1/2}
= \frac{(1/2)^2}{2}
= \frac{1}{8},
\]
\[
\int_{1/2}^{1} (1 - x)\,dx
= \left.\Bigl(x - \frac{x^2}{2}\Bigr)\right|_{1/2}^{1}
= \left(1 - \frac12\right) - \left(\tfrac12 - \tfrac{(1/2)^2}{2}\right)
= \frac12 - \left(\frac12 - \frac{1}{8}\right)
= \frac{1}{8}.
\]
Summing gives
\[
\mathbb{E}[\text{smaller piece}]
= \frac{1}{8} + \frac{1}{8} = \frac{1}{4}.
\]
Notes and comments
Comment 1: The area can also be thought of as $\frac{1}{4}$th of the area of a unit square, hence $\frac{1}{4}$.