Calculate convergence of random variable sum

Question

I have a problem with following task. I need to show that this series convergence almost surely and find its limit. $$\frac{X_1+\ldots+X_n}{n}$$ where $X_n$ has distribution $U(\frac{1}{n},1)$ and all of this random variables are independent. First thought modify $X_n$ to make it $U(0,1)$ distributed and then use law of large numbers. Let $X_n^{\star}+\frac{1}{n}=X_n$ then $X_n^{\star}$ has $U(0,1-\frac{1}{n})$ distribution. $$\frac{X_1+\ldots+X_n}{n}=\frac{X_1^{\star}+\frac{1}{1}\ldots+X_n^{\star}+\frac{1}{n}}{n}=\frac{X_1^{\star}+\ldots+X_n^{\star}}{n}+\frac{1+\ldots + \frac{1}{n}}{n}$$ So second factor will be asymptotically $\frac{\pi^2}{6}$ so lets leave that one for a while. Now lets $X_n^{\star}=X_n^{\circ}(1-\frac{1}{n})$ Then $X_n^{\circ}$ will be $U(0,1)$ $$\frac{X_1^{\star}+\ldots+X_n^{\star}}{n}=\frac{X_n^{\circ}(1-\frac{1}{1})+\ldots+X_n^{\circ}(1-\frac{1}{n})}{n}=\frac{X_1^{\circ}+\ldots+X_n^{\circ}}{n}-\frac{\frac{1}{1}X_1^{\circ}+\ldots+\frac{1}{n}X_n^{\circ}}{n}$$ Now the problem is second part. I guess I can't estimate it a s i need exact value(unless it is zero but it doesnt seem like is). I will be very glad for help, regards!

Answer 1

Your second part will be zero

$u_n=\frac{1}{n}$ converges towards zero

According to Cesaro :

$U_n=\frac{\sum_{k=1}^{n}u_k}{n}$ converges towards zero as well

Since $U_n$ is greater than your second term, it converges towards zero.

Answer 2

Here is an argument, via Kolmogorov's two series theorem and Kronecker's lemma.

Lemma (Kronecker) Suppose $a_n \nearrow \infty$, $\displaystyle \sum_{n\geq 1}\frac{x_n}{a_n}$ converges. Then, $\displaystyle \frac{1}{a_n}\sum_{k=1}^n x_k \to 0$.

We will show that $\displaystyle\frac{X_1+X_2+\dots+X_n}{n}\to \frac{1}{2}$. Let $Y_ k = X_k-\mathbb{E}[X_k]$. Notice that $\mathbb{E}[X_k]=\displaystyle\frac{k+1}{2k}\to \frac{1}{2}$ as $k\to\infty$. Hence, it is equivalent to show that $\displaystyle \frac{1}{n}\sum_{k=1}^n Y_k\to 0$, almost surely; since that would imply: $\displaystyle\frac{1}{n}\sum_{k=1}^n X_k - \displaystyle\underbrace{\frac{1}{n}\sum_{k=1}^n \mathbb{E}{X_k}}_{\to \frac{1}{2}, \text{ Cesaro mean}}\to 0 \implies \frac{1}{n}\sum_{k=1}^n X_k \to \frac{1}{2}$ almost surely.

To prove the result, it suffices to prove (due to Kronecker's lemma) that $\displaystyle \sum_{n=1}^\infty\frac{Y_n}{n}<\infty$ almost surely; which is implied (due to Kolmogorov's two series theorem) if $\displaystyle\sum_{n=1}^{\infty}\mathbb{VAR}\left(\frac{Y_n}{n}\right)=\sum_{n=1}^{\infty}\frac{1}{12}\frac{(n-1)^2}{n^4}\leq C\displaystyle\sum_n \frac{1}{n^2}<\infty$. We are done.