Show that $\frac{X_1+\dots+X_n}{n}$ converges a.s. for $X_n \sim U([0,1-2^{-n}])$ independent

Question

Let $(X_{n})$ be a sequence of independent random variables and let $X_{n}$ have a uniform distribution on $[0, 1-2^{-n}]$. Prove that the sequence:

$$\frac{X_{1}+X_{2}+\dots+X_{n}}{n}$$

converges almost sure. Find its limit.

I've proved that this sequence converges a.s. using Kolmogorov's theorm. Because the series $\sum_{n=1}^{\infty} \frac{\text{Var} \, X_{n}}{n^{2}}=\sum_{n=1}^{\infty}\frac{(1-2^{-n})^{2}}{12n^{2}}<\infty$ and $(X_{n})$ are independent so our sequence converges a.s (satisfies LLN).

How can I compute its limit? Should it be equal to $\sum_{n=1}^{\infty}EX_{n}$?

Answer 1

Kolmogorov's theorem implies that $(X_n)_{n \in \mathbb{N}}$ satisfies the strong law of large numbers, i.e.

$$\frac{1}{n} \sum_{i=1}^n (X_i-\mathbb{E}X_i) \to 0 \tag{1}$$

almost surely. Note that this does not necessarily imply that

$$\frac{1}{n} \sum_{i=1}^n X_i$$

converges.

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Hints:

1. If the series $\frac{1}{n} \sum_{i=1}^n \mathbb{E}X_i$ converges, then, by $(1)$, $\frac{1}{n} \sum_{i=1}^n X_i$ converges almost surely to exactly this limit.
2. Show the following statement: If $(a_n)_{n \in \mathbb{N}}$ is a sequence such that $a_n \to a$, then $$\frac{1}{n} \sum_{i=1}^n a_i \to a.$$
3. Use step 2 to conclude that $$\frac{1}{n} \sum_{i=1}^n \mathbb{E}X_i \to \frac{1}{2}.$$

Answer 2

Hint:

Note that we can write $X_i=(1-2^{-i})U_i$ where $U_i$ are iid U[0,1] variables. Now, apply SLLN on $\overline{U}_n$ and compute the limit of the difference $\overline{U_n}-\overline{X_n}$.

To make your calculations easier; Guess a limit of $\overline{X_n}$. (compare $\overline{X_n}$ and $ \overline{U_n}$ for large n)

(this makes Kolmogorov unnecessary)

I will post a solution if needed.