break unit rod randomly somewhere, expected ratio between the two parts

Question

I have a stick of 1m length. I break it at a random point. Where I break it is described by a uniform distribution across the stick's length. The two parts are of length $x$ and $1-x$.

The question is: what is the expected ratio between the lengths of the two parts?

My initial solution is that the ratio is $\frac{x}{1-x}$, and to find the expected value, we multiply the PDF with $x$ and integrate across the relevant interval:

$$\int_0^1 \frac{x}{1-x} ~x ~\mathrm{d}x$$

But this integral does not converge. Switching $x$ with $1-x$ gives the same problem (independent of whether we integrate from 0 to 1 or 1 to 0):

$$\int_0^1 \frac{1-x}{x} ~(1-x) ~\mathrm{d} x $$ However,

$$\int_0^1 \frac{1-x}{x} ~x ~\mathrm{d}x$$

does converge, and even better, the two $x$ cancel out and we are left with

$$\int_0^1 1-x ~ \mathrm{d}x= 0.5$$

To be clear, I understand the math behind the above calculations and how to show that they do not converge. From the formulation of the problem and my common sense and intuition, and since $x$ and $1-x$ should be symmetrically distributed, I cannot decide why it should be relevant whether we take the ratio to be $\frac{x}{1-x}$ or $\frac{1-x}{x}$ (given that we integrate over $\mathrm{d}x$, not $\mathrm{d}(1-x)$, but we could also do the latter).

Is the problem not formulated precisely enough - is it necessary to define which is the numerator and denominator in the ratio of two numbers? Can I say now that the expected ratio of the two parts is 0.5, or do I need more information, or is it undefined?

How can the fact that the problem's logic should allow $x$ and $1-x$ to be interchangeable be related with these completely different results? What am I overseeing?

Answer 1

$\newcommand{\E}{\mathrm{E}}$ $\newcommand{\Var}{\mathrm{Var}}$ $\newcommand{\Cov}{\mathrm{Cov}}$ $\newcommand{\d}{\mathop{}\!\mathrm{d}}$ I identified the problem. What I want is the expected value of $\frac{x}{1-x}$, not of $x$, and its PDF is just $1$ on $[0,1]$ and $0$ elsewhere. I mistakenly took $\frac{x}{1-x}$ to be the PDF and multiplied with $x$. I therefore need to calculate $$\E\left[\frac{x}{1-x}\right]= \int_0^1 \frac{x}{1-x} ~\d x,$$ or alternatively $$\E\left[\frac{1-x}{x}\right]= \int_0^1 \frac{1-x}{x} ~\d x,$$ and these two are now indeed equivalent, as they should be (the two terms are symmetric around $\frac{1}{2}$).

Let's work on the second one: $$\E\left[\frac{1-x}{x}\right]= \int_0^1 \frac{1-x}{x} ~\d x$$ $$= \int_0^1 \frac{1}{x}-1 ~\d x$$ $$= \left.\ln ~x -x \right|_0^1$$ ...And clearly, this integral does not converge ($\mathrm{ln}~0$ undefined). So it really is necessary to specify that we need the ratio between the shorter and the longer part, otherwise the expected value is unbounded, as in this question: broken stick : average of the ratio of the shortest piece to the largest piece.