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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 $\sum_{n=1}^{\infty}EX_{n}$?

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  • $\begingroup$ Hint or solution? $\endgroup$ – Conformal Jun 13 '15 at 14:45
  • $\begingroup$ Give me a hint :D $\endgroup$ – mrnobody Jun 13 '15 at 14:47
  • $\begingroup$ Which "Kolmgorov theorem" do you use? (There are several ones with this name.) The one I'm thinking of actually requires that the series of the expectations converges and not only the series of the variances. $\endgroup$ – saz Jun 13 '15 at 14:56
  • $\begingroup$ Woops. Made a calculation error, so i deleted my answer. sorry. Ill get back to you :) $\endgroup$ – Conformal Jun 13 '15 at 15:02
  • $\begingroup$ @saz: I use this one: "If $(X_{n})$ is a sequence of independent random variables with $Var X_{n}<\infty$ for n=1,2,.... and $\sum _{n=1}^{\infty}\frac{Var X_{n}}{n^{2}}$, then $(X_{n})$ satisfies LLN." $\endgroup$ – mrnobody Jun 13 '15 at 15:11
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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.


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 $$\frac{1}{n} \sum_{i=1}^n \mathbb{E}X_i \to \frac{1}{2}.$$
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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.

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