Stick Break Problem Setup

A stick of length $L$ is broken at a random point. Let the break point be $X$, which is a random variable uniformly distributed on the interval $[0, L]$. The probability density function (PDF) is $f(x) = 1/L$ for $0 \le x \le L$.

This random break results in two pieces with lengths $X$ and $L - X$.

Smaller Piece Length Definition

We need to find the average length of the smaller piece. Let $S$ denote the length of the smaller piece. Mathematically, this is defined as:

$S = \min(X, L - X)$

The value of $S$ depends on where the break occurs:

• If the break point $X$ is in the first half of the stick ($0 \le X \le L/2$), then the first piece $X$ is the smaller one, so $S = X$.
• If the break point $X$ is in the second half of the stick ($L/2 < X \le L$), then the second piece $L - X$ is the smaller one, so $S = L - X$.

Average Length Calculation E[S]

The average length of the smaller piece is its expected value, denoted by $E[S]$. We calculate this using the definition of expected value for a continuous random variable:

$E[S] = \int_{0}^{L} \min(x, L - x) f(x) dx$

Substitute the PDF $f(x) = 1/L$ and split the integral into two parts based on the definition of $S$:

$E[S] = \int_{0}^{L/2} x \cdot \frac{1}{L} dx + \int_{L/2}^{L} (L - x) \cdot \frac{1}{L} dx$

Integral Evaluation Steps

We evaluate each integral separately:

1. First Integral (for $0 \le X \le L/2$): Here $S = x$. $\frac{1}{L} \int_{0}^{L/2} x dx = \frac{1}{L} \left[ \frac{x^2}{2} \right]_{0}^{L/2} = \frac{1}{L} \left( \frac{(L/2)^2}{2} - 0 \right) = \frac{1}{L} \cdot \frac{L^2}{8} = \frac{L}{8}$
2. Second Integral (for $L/2 < X \le L$): Here $S = L - x$. $\frac{1}{L} \int_{L/2}^{L} (L - x) dx = \frac{1}{L} \left[ Lx - \frac{x^2}{2} \right]_{L/2}^{L} = \frac{1}{L} \left[ \left( L^2 - \frac{L^2}{2} \right) - \left( \frac{L^2}{2} - \frac{L^2}{8} \right) \right] = \frac{1}{L} \left[ \frac{L^2}{8} \right] = \frac{L}{8}$

Final Average Length Result

Add the results from the two integrals to find the total expected value:

$E[S] = \frac{L}{8} + \frac{L}{8} = \frac{2L}{8} = \frac{L}{4}$

Thus, the average length of the smaller piece is $L/4$.