GATE CS 2014 Set 1 Q.12 (CS Q.2): Probability and Statistics

Suppose you break a stick of unit length at a point chosen uniformly at random. Then the expected length of the shorter stick is ________ .

Your answer

Official answer and solution

Accepted answers: 0.24 to 0.27
From: Official GATE 2014 answer key

Solution

Model the break point as a continuous uniform random variable on $(0, 1)$ and compute the expected value of the minimum of the two pieces.

Let the stick have length 1, and let the break point $X$ be chosen uniformly at random from $(0, 1)$ , so the probability density function is $f_X(x) = 1$ for $x \in (0, 1)$ .

The stick is broken into two pieces of lengths $X$ and $1 - X$ .
The length of the shorter piece is:

L = \min(X, 1 - X)

Notice that:

If $X \le 0.5$ , then $L = X$ .
If $X > 0.5$ , then $L = 1 - X$ .

By symmetry about $x = 0.5$ :

E[L] = \int_0^1 \min(x, 1 - x) dx = 2 \int_0^{0.5} x dx

E[L] = 2 \left[ \frac{x^2}{2} \right]_0^{0.5} = (0.5)^2 = 0.25

Thus, the expected length of the shorter stick is $0.25$ .

Final answer

0.25

Common trap

Integrating over the full interval $(0, 1)$ without accounting for the piecewise definition of the minimum.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2014 Computer Science and Information Technology (CS) Session 1
Question: 12 of 65, printed as CS Q.2
Section: Computer Science and Information Technology

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