1
$\begingroup$

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:

$$ 2 \int_{1 \over 2}^{1} \frac {1-x}{x}\ \text{pdf_of_the_ratio}(x) \ dx =?$$

  • 0
    I still believe in the intuitive solution. There are several answers on MSE which show it – 2017-01-29
  • 0
    Have you had a look to other answer on MSE ? – 2017-01-29

2 Answers 2

1

It is a well known fact that if $y = f(x)$, $$\mathbb{E}(Y) = \int f(x) p_x(x)dx.$$ Another way is $$\mathbb{E}(Y) = \int y p_y(y)dy.$$

The solution is using the first one, and what you're suggesting is none of them.

  • 0
    yes of course thank you - i got confused by mistakenly thinking that this was a ratio of independent uniform rvs – 2017-01-29
2

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 $[1/2,1]$