# Distribution of final piece in stick breaking process $x_{k+1} = (x_{k} -1) \cdot u_{k+1}$

Consider a stick breaking process where we start with a large stick of length $$x_1 \gg 1$$.

As long as the stick is larger than one, $$x_k > 1$$, we break the stick by, removing a piece of length 1 and in addition remove another piece by break at a random spot (chosen by a uniform distribution),

$$x_{k+1} = (x_{k} -1) \cdot u_{k+1} \quad \text{where} \quad u_{k+1} \sim U(0,1)$$

We stop this process when the length is below 1.

What is the probability distribution for the final piece?

A motivation/background for this question is a short discussion from the tenfold chat.

• A motivation/background for this question is a short discussion from the tenfold chat. Commented Feb 3, 2023 at 18:08

### Simulation

We can simulate the distribution with a computer code

### function to perform a stick breaking process
sample = function(x=10) {
while(x>1) {
x = (x-1) * runif(1)
}
return(x)
}

# create an Empirical sample
n = 10^5
z = replicate(n, sample())
z = z[order(z)]
p = c(1:n)/n

# plot empirical distribution along with simple estimate
plot(z,p, type = "l", main = "Empirical distribution (black) \n compared to \n estimated CDF (red)", xlab = "x", ylab = "P(stick length < x)")
lines(z, z*(1-0.5*log(z)), col = 2)


### First level estimate

The distribution function can be estimated with a function $$F(x) = x(1-0.5 \log(x))$$

That function is a mixture distribution of a uniform variable and the product of two uniform variables.

That mixture is based on the idea that you get

• A uniform component for the cases that you end end up below one, $$x_{k+1}<1$$, while the previous value was above two, $$x_k > 2$$.
• You get a product of two uniform variables component for the cases that you end end up below one and while $$1 < x_k \leq 2$$ and also $$x_{k-1}>3$$. (That latter condition means that $$x_k$$ is uniformly distributed between one and two)

There are some more components, if $$1 < x_k \leq 2$$ and also $$x_{k-1}<3$$, but those are more complicated to compute.

• I noticed the spelling typo but the graph was already done. Nevertheless, an insightful post (specially how the distribution was conceived) given the discussion in the chat. Commented Feb 3, 2023 at 18:09