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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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3 Answers

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85 votes
Best answer
The length of the shorter stick can be from $0$ to $0.5$ (because if it is greater than $0.5,$ it is no longer a shorter stick).

 This random variable $L$ (length of shorter stick) follows a uniform distribution, and hence probability density function of $L$ is $\dfrac{1}{0.5-0}= 2$ for all lengths in range $0$ to $0.5$

Now expected value of $L = \int_{0}^{0.5} L*p(L) dL = \int_{0}^{0.5} L*2 dL = 2*\left[\dfrac{L^2}{2}\right]^{0.5}_0 = 0.25$
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4 Comments

Length of the shorter stick can be from 0.5 to 1 OR  0 to 0.5. so even if you take range 0.5 to 1 you will get same answer.

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But in expectation we consider both the scenarios, so why here only one part is being considered.
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Both scenarios are identical so expectation will not change.

if you consider scenario 1 then you get 0.25, if you consider scenario 2 then you get 0.25 .

expecation = (scenario1 + scenario 2 )/2 = (0.25 +0.25) /2 = 0.25
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3 votes
3 votes

Expectation in case of a Uniform Random Variable = $\frac{b-a}{2}$

https://www.ucd.ie/msc/t4media/Uniform%20Distribution.pdf

As shorter length stick is specified in question, take a=0 and b=0.5

$\frac{0.5}{2}$ = ¼ = 0.25

1 comment

@ashley they have given it wrong, right one is a+b/2 not b-a/2 you can try to calculate what they have they did a mistake in calcuation so be careful
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0 votes
0 votes

Answer = $0.25$

Let $X$ = Length of Shorter Stick

$\implies 0<X<\frac{1}{2}\implies X$ follows Uniform Distribution over $(0,\frac{1}{2})$

We know, In uniform distribution, the mean (first moment) of the distribution is:

${\displaystyle E(X)={\frac {1}{2}}(b+a).}$   Link : Continuous uniform distribution - Wikipedia

Hence, the expected length of the shorter stick is = $Mean$ = $E[X] = \frac{b+a}{2} = \frac{0+\frac{1}{2}}{2} = \frac{1}{4} = 0.25$