Does $X_n$ with $\mathbb{P}[X_n = \frac{1}{n}] = 1 - \frac{1}{n^2}$ and $\mathbb{P}[X_n = n] = \frac{1}{n^2}$ converges in probability? In $L^2$?

Aplying limit definition I see that it is convergent. But I don't understand what I need to do in case of $L^2$ because I think that $L^2$ is for vectors and I don't see vectors here.


We need to prove that there exists random variable $X$ such that for any $\varepsilon > 0$ and any $\delta > 0$ there exists number $N$ (which may depend on one or both) such that for all $n \ge N, \Pr\left(|X_n - X| > \varepsilon\right) \lt \delta$ (definition of limit).

Let $X = 0$, then if we choose $N$ such that $\frac{1}{N} \lt \varepsilon$ and $\frac{1}{N^2} \lt \delta$ we will have that either $X_n \lt \varepsilon,$ or probability that this is false is less than $\delta.$ Thus $X_n$ is convergent in probability.


Actually, $X_n \to 0$ almost surely. This requires Borel Cantelli Lemma.

$\sum P(X_n=n)=\sum \frac 1 {n^{2}} <\infty$. By Borel-Cantelli Lemma it follows that $X_n=n$ i.o with probability $0$. Hence, $X_n=\frac 1 n$ for all $n$ sufficiently large, with proability $1$. It follows that $X_n \to 0$ a.s.

$EX_n^{2}=n^{2}(\frac 1{n^{2}})+\frac 1{n^{2}}(1-\frac 1 {n^{2}})\to 1$. If at all $(X_n)$ converges in $L^{2}$, it can only converge to $0$ because a.s. limit is $0$. It follows that $X_n$ does not converge in $L^{2}$.

  • $\begingroup$ The confusion seems to be more about what $L^2$ means, based on the OP's post; not so much about what to do from there. $\endgroup$
    – Clement C.
    Mar 28 at 23:27
  • 1
    $\begingroup$ Thank you, +1. I updated the answer with my simple proof of convergence. I'm not sure why do you need this lemma. $\endgroup$
    – Yola
    Mar 29 at 1:17
  • $\begingroup$ @Yola I thought you were asking for a.s. convergence, which is also true. I have edites my answer accordingly. $\endgroup$ Mar 29 at 5:03

I think you are confusing $\ell_2$ norms for vectors and sequences: $$ \|x\|_2 = \sum_{n=1}^\infty |x_n|^2 $$ and $L^2$ norms for functions (or, here, random variables: it's the same): $$ \|X\|_2 = \sqrt{\mathbb{E}[|X|^2]} $$ There definitely are connections between the two, these are not really different concepts.

In your case, $$ \|X_n\|_2^2 = \mathbb{E}[|X_n|^2] = \frac{1}{n^2}\cdot \left(1-\frac{1}{n^2}\right) + n^2\cdot \frac{1}{n^2} > n^2\cdot \frac{1}{n^2} = 1 $$ so $\|X_n\|_2^2 \not\to 0$. $X_n$ does not converge to $0$ in $L^2$.


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