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Official PYQ · 2014Set 1 · Q2Probability: Random Variables and DistributionsNAT1 mark

GATE CS 2014 Set 1 Q2 — Probability: Random Variables and Distributions

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Suppose you break a stick of unit length at a point chosen uniformly at random. Then the expected length of the shorter stick is ________ .
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Answer: 0.24 : 0.27

Let the stick of unit length be represented by the interval [0,1][0, 1]. Let the break point be a random variable XX chosen uniformly at random from [0,1][0, 1].

The probability density function (PDF) of XX is:

f(x)=1for 0≤x≤1f(x) = 1 \quad \text{for } 0 \le x \le 1

Breaking the stick at XX divides it into two pieces of lengths XX and 1−X1-X. The length of the shorter stick, YY, is given by:

Y=min⁡(X,1−X)={Xif 0≤X<0.51−Xif 0.5≤X≤1Y = \min(X, 1-X) = \begin{cases} X & \text{if } 0 \le X < 0.5 \\ 1-X & \text{if } 0.5 \le X \le 1 \end{cases}

The expected length of the shorter stick, E[Y]E[Y], is calculated as:

E[Y]=∫01min⁡(x,1−x)f(x) dxE[Y] = \int_{0}^{1} \min(x, 1-x) f(x) \, dx E[Y]=∫00.5x dx+∫0.51(1−x) dxE[Y] = \int_{0}^{0.5} x \, dx + \int_{0.5}^{1} (1-x) \, dx

By symmetry, the two integrals are equal:

E[Y]=2∫00.5x dxE[Y] = 2 \int_{0}^{0.5} x \, dx E[Y]=2[x22]00.5=(0.5)2=0.25E[Y] = 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.

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