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Arjun
asked
in Probability
Sep 26, 2014

14,266 views
55 votes

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

For proper solution please refer --> https://math.stackexchange.com/questions/350679/we-break-a-unit-length-rod-into-two-pieces-at-a-uniformly-chosen-point-find-the/350863#350863

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88 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 $L = \int_{0}^{0.5} L*p(L) dL = \int_{0}^{0.5} L*2 dL = 2*\left[\dfrac{L^2}{2}\right]^{0.5}_0 = 0.25$

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 $L = \int_{0}^{0.5} L*p(L) dL = \int_{0}^{0.5} L*2 dL = 2*\left[\dfrac{L^2}{2}\right]^{0.5}_0 = 0.25$

reshown
Oct 15, 2017
by Arjun

producing multiple pictures , But I go with 1/4.

http://math.stackexchange.com/questions/253780/expected-length-of-broken-stick

http://math.stackexchange.com/questions/13959/if-a-1-meter-rope-is-cut-at-two-uniformly-randomly-chosen-points-what-is-the-av

http://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-r

http://www.quora.com/Assume-a-stick-is-broken-at-random-into-three-pieces-What-is-the-probability-that-the-pieces-can-form-a-triangle

http://math.stackexchange.com/questions/253780/expected-length-of-broken-stick

http://math.stackexchange.com/questions/13959/if-a-1-meter-rope-is-cut-at-two-uniformly-randomly-chosen-points-what-is-the-av

http://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-r

http://www.quora.com/Assume-a-stick-is-broken-at-random-into-three-pieces-What-is-the-probability-that-the-pieces-can-form-a-triangle

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@Happy Mittal Can we also do this like this: First we calculate the expected value of shorter stick ( integration of x time 1 with limits 0 to 1. We now use this expected value as limits to a new expectation, which will the expected value of the shorter stick. I got the same answer using that. Just wanted to confirm.

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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 vote

**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 = $Mean$ = $E[X] = \frac{b+a}{2} = \frac{0+\frac{1}{2}}{2} = \frac{1}{4} = 0.25$