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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 ________ .

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

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.

But in expectation we consider both the scenarios, so why here only one part is being considered.
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

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

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:

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

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