The length of the third piece in the Stick breaking process

Question

I'm reading Nonparametric Bayesian Statistics Part I: some classical results.

In page 14 and section 3.2.1 I can't understand why the length of the stick after the second break is:

$$(1-Y_1)(1-Y_2)$$

and not

$$(1-Y_1-Y_2)$$

How is the extra term $$Y_1Y_2$$ part of the remaining piece of the original length 1 stick?

Answer 1

That's because the authors are considering the variables $Y_k$ to be in the standard range $[0,1]$.

So $Y_1$ is the length of the first cut and $1-Y_1$ the remaining part, out of which a part having a ratio $Y_2$ of $1-Y_1$ is cut, i.e. with length $Y_2(1-Y_1), and $(1-Y_1)(1-Y_2)$ is the length remaining ...

Answer 2

Thanks to G Cab I understand what's happening. First I noticed two key words: ratio and relative. The first pointed out by G Cab and the second one from the text I referenced in my question, i.e.,

We think of the remaining mass $1 − Y_1$ as a new stick, and break it into two pieces of relative lengths $Y_2$ and $1 − Y_2$...

The word relative, which in this context is just another way of saying ratio, basically means that the new absolute lengths, let's say $a$ and $b$ (i.e. after breaking the stick the first time) can be obtained from $$Y_2 = \frac{a}{1-Y_1}$$ and $$1-Y_2 = \frac{b}{1-Y_1}$$ thus $a=Y_2(1-Y_1)$ and $b=(1-Y_2)(1-Y_1)$.

So my confusion was because I didn't pay enough attention to the word relative in the text.