Frederick Mosteller , 'Fifty challenging problems of probability', Q.42
If a stick is broken in 2 at random, what is the average ratio of the smaller length to the larger ?
the answer is given as:
Breaking 'at random' means that all points of the stick are equally likely as a breaking point (uniform distribution).
We might suppose that the point fell in the right half. Then $\frac{1-x}{x}$ is the fraction of the stick if the stick is of unit length. Since $x$ is evenly distributed from $1 \over 2$ to $1$ the average value – instead of the intuitive $1 \over 3$ – is:
$$ 2 \int_{1 \over 2}^{1} \frac {1-x}{x}dx = 2 \ln(2)-1 \approx 0.386$$
I don't understand this. Why don't I need to somehow derive the probability density of the ratio of the 2 uniform random variables involved, and then integrate:
$$ 2 \int_{1 \over 2}^{1} \frac {1-x}{x}\ \text{pdf_of_the_ratio}(x) \ dx =?$$
Best Answer
You need to take the expected value of the random variable the ratio $R$, which is given in terms of the uniform variable $X$ (which is defined as "the length of the long half of the stick") as $$ R = \frac{1-X}{X}.$$
Thus you should integrate against the PDF of $X$ which is $2$ since it's uniform on $[1/2,1]$