Once again I've encountered a problem, which might not be difficult:
I'm given a sequence of random variables $ (X_n) $, each with density function $g_n(x) = nx^{n-1} \textbf{1}_{(0,1]} $.
I am to prove that this sequence converges almost surely. So to begin with, I think I am able to find the limit and prove convergence by probability.
We have:
$\mathbb{E} X_n = \int\limits_{0}^1 nx^n dx = \frac{n}{n+1} $
Applying Chebyshev inequality:
$ \lim\limits_{n \rightarrow \infty}\mathbb{P}(1-X_n < \epsilon) = \lim\limits_{n \rightarrow \infty} (1 - \mathbb{P}(1- X_n \geq \epsilon)) \geq \lim\limits_{n \rightarrow \infty}(1- \frac{\mathbb{E}(1-X_n)}{\epsilon}) = \lim\limits_{n \rightarrow \infty} (1- \frac{1}{\epsilon}(1-\frac{n}{n+1})) = 1$
Thus, $ X_n \rightarrow 1 $ by probability.
And there goes my question: how do I prove almost sure convergence? I've been trying to use the definition, but it doesn't seem to help.
Thanks in advance
$\endgroup$3 Answers
$\begingroup$$Pr(|X_n - 1| \geq \epsilon \mbox{ infintely often}) = \sum_n \int _{0}^{1 - \epsilon} nx^{n-1} dx = \sum _n (1 - \epsilon)^n < \infty $ for any $\epsilon > 0$.
Hence, by Borel Cantelli Lemma it converges to $1$ almost surely.
$\endgroup$ $\begingroup$Notice that for each integer $n$ and each $a_n>0$, $\mu\{1-X_n\gt a_n\}=(1-a_n)^n$.
Then choose $a_n$ such that $\sum_n (1-a_n)^n<\infty$ and $a_n\to 0$ and conclude by the Borel-Cantelli lemma.
$\endgroup$ $\begingroup$given $\varepsilon \in (0,1)$. Note that $P(|X_{n} -1|\geq \varepsilon) = (1-\varepsilon)^n$. $\sum_{n=1}^{\infty}(1-\varepsilon)^n < \infty$ thus $P(|X_{n} -1| \geq \varepsilon \ i.o.)=0$ by the Borel-Cantelli lemma. That is $P(|X_{n} -1| < \varepsilon \ evt.) = 1$. Hence $X_{n} \to 1$ almost surely.
$\endgroup$
