# How does $L^2$ differs for $X_n$ with $\mathbb{P}[X_n = \frac{1}{n}] = 1 - \frac{1}{n^2}$ and $\mathbb{P}[X_n = n] = \frac{1}{n^2}$?

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.

EDIT:

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.

## 2 Answers

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 $$L^{2}$$.

• 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. Mar 28 at 23:27
• Thank you, +1. I updated the answer with my simple proof of convergence. I'm not sure why do you need this lemma.
– Yola
Mar 29 at 1:17
• @Yola I thought you were asking for a.s. convergence, which is also true. I have edites my answer accordingly. 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 $$L^2$$.