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We break a unit length rod into two pieces at a uniformly chosen point. Find the expected length of the smaller piece

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    $\begingroup$ Someone said that I have to take the probability density function taking the min of either x or 1-x since those are the two values of the lengths of the pieces $\endgroup$
    – Alec Kartz
    Apr 4, 2013 at 1:40

4 Answers 4

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The length function is $f(x) = \min ( x,1-x )$ as you found. This is $x$ on the interval $[0,0.5]$ and $1-x$ on the interval $[0.5,1]$.

We are ultimately trying to find $E(f(X))$, the expected length. By the "law of the unconscious statistician," and since the distribution of the random variable $X$ is uniform:

$$E(f(X)) = \int_0^1 f(x) \cdot 1 \;dx = \int_0^{0.5} x \;dx + \int_{0.5}^1 (1-x) \;dx = \frac{1}{4}$$

Or you could just measure the area under the curve without integrating by noticing the graph forms a triangle with base $1$ and height $0.5$.

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With probability ${1\over2}$ each the break takes place in the left half, resp. in the right half of the rod. In both cases the average length of the smaller piece is ${1\over4}$. Therefore the overall expected length of the smaller piece is ${1\over4}$.

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Length of the shorter stick is uniformly distributed between 0 and 0.5, hence the answer is a mean of uniform distribution which is a midpoint, here it is 0.25. enter image description here

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There can be two events when choosing the break point, it can either lie in [0,1/2) or [1/2,1]. Assume x to be the shorter distance between an endpoint and the breakpoint. Since we are interested in the length of the smaller piece, the domain of x lies in [0,1/2]

  • in the first case, the expected length is x*1/2
  • in the second case the expected length is (1-x)*1/2

we add the expected lengths and integrate the sum between 0 and 1/2, and that would give us 1/4. Since these two events exhaust all possibilities of sample space, we have the answer 1/4.

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