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?

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.
