$X_n$ is uniformly distributed on $[0,n]$

Question

Let $(X_1 ,X_2, \ldots)$ be an independent sequence of random variables, where for each $n$, $X_n$ is uniformly distributed on $[0,n]$.

How can I calculate the $$ P(\{w: X_n(w)\rightarrow \infty \ \ as \ n \rightarrow \infty\}).$$

I have been thinking about calculating the $P(\{w: X_n(w) \geq \frac{1}{n}\}$ and with the theorem of continuity we have that this probability is zero but if I calculate $P(\{w: X_n(w) \geq n-\frac{1}{n}\}$ applying again the theorem of continuity we have that this probability is 1.

I have obviously a mistake but I really try to understand why I am wrong but I can't solve this, even with $Borel-Cantelli$ the serie $\sum \frac{1}{n}$ diverges and then the probability is $1$ and the serie of $\sum \frac{1}{n^2}$ converges so the probability is $0$.

Could someone help me to solve this pls?

Thanks for your time and help.

Answer 1

Note that $\displaystyle P(X_n\to \infty)= P\left(\bigcap_N\bigcup_n \bigcap_{k\geq n}(X_k \geq N)\right)$.

Since the sequence of sets (indexed by $N$) $\bigcup_n \bigcap_{k\geq n}(X_k \geq N)$ is decreasing, $$P\left(\bigcap_N\bigcup_n \bigcap_{k\geq n}(X_k \geq N)\right)=\lim_N P\left(\bigcup_n \bigcap_{k\geq n}(X_k \geq N)\right)$$

so it suffices to compute $P\left(\bigcup_n \bigcap_{k\geq n}(X_k \geq N)\right)$ for each $N$.

But $P\left(\bigcup_n \bigcap_{k\geq n}(X_k \geq N)\right)$ is nothing other than $P(\liminf_n X_n\geq N) = 1-P(\limsup_n X_n<N)$.

Since $\sum_{n=1}^\infty P(X_n< N)=\sum_{n=1}^N 1 + \sum_{n=N+1}^\infty \frac{N}{n}=N+\infty = \infty$ and the $X_n<N$ are independent, Borel-Cantelli lemma yields $P(\limsup_n X_n<N) = 1$, hence $P(\liminf_n X_n\geq N)=0$ for all $N$.

Therefore, $P(X_n\to \infty)=0$.

Answer 2

For every positive number $x$, the probability that $X_n <x$ is $x/n$ for sufficiently large $n.$ Since the sum of these diverges, by Borel Cantelli, the probability that $X_n < x$ infinitely often is one. So the probability that $X_n\to\infty$ is zero.

(In fact with a small amount of additional argument, we can say something stronger, that the $\liminf$ of the sequence is zero almost surely.)