A problem about a rod breaking into $n$ pieces and probability theory

Question

A rod of length $1$ is broken at random, which means that the remaining part has an even distribution at $(0,1)$. The remainder is broken in a similar way, and so on.
$1$. Let $X_{n}$ be the length of the part left over after the rod has been broken $n$ times. Describe $X_{n}$ as a product.
$2$. Show that the sequence $\{\log(X_{n})/ n \}_{n\geq 1}$ almost sure converges, answering what the limit is.
$3.$ Obtain an approximation for the probability that $X_{36}\leq e^{-24}$

My attempt:

1. I think $\displaystyle X_{n}=\prod_{i=1}^{n}X_{i}$, where $i\in [0,n]$ with $x_{i}$ choosen uniformly from $(0,1)$. is it correct?
2. We need to prove that $$U_{n}=\frac{\log (X_{n})}{n} \overset{a.s}{\to} U$$ but $$\frac{\log(X_{n})}{n}=\frac{\log\left(\prod_{i=1}^{n}X_{i} \right)}{n}=\frac{\sum_{1\leq i\leq n}\log(X_{i})}{n}$$ but since that $X_{i}$ are i.i.d (I don't sure about it. Can you correct me if I'm wrong?) and since that $$\mathbb{E}[\log(X)]=\int_{0}^{1}1\cdot \log(x)dx=-1$$ where $f(x)=\frac{1}{1-0}$ is the density function of $X\sim U(0,1)$. So, by Strong Law of Large Numbers we have $$U_{n} \overset{a.s}{\to}-1$$
3. I don´t know how to solve this option.
   Can you teach me how can I solve this problem?

Answer 1

1. $X_i$ are already defined in the problem, so it would be better to define a different random variable for your uniform random variables, e.g. $X_n = \prod_{i=1}^n Y_i$ for $Y_i \overset{\text{i.i.d}}{\sim} \text{Unif}(0,1)$.
2. You are correct that $\frac{1}{n} \log (X_n) = \frac{1}{n} \sum_{i=1}^n \log(Y_i) \overset{\text{a.s.}}{\to} E[\log(Y_1)]$, but $E[\log(Y_1)] = \int_0^1 \log x \, dx = -1$.
3. $P(X_{36} \le e^{-24}) = P(\log X_{36} \le -24) = P(\frac{1}{36}\sum_{i=1}^{36} \log(Y_i) \le -2/3)$. By a normal approximation (central limit theorem), you can approximate $\frac{1}{36}\sum_{i=1}^{36} \log (Y_i)$ with a $N(E[\log(Y_1)], \text{Var}(\log(Y_1))/36)$ normal random variable to estimate the above probability.