# Is it mathematical correct proof? Need rigorous proof of uniformly tight sequence.

I need to prove that

$P(\|X_n\| > M_n) \to 0$ as $n \to \infty$ for every $M_n \to \infty$

$P(\|X_n\| > M) < \varepsilon$ for every $N \leq n$

I do the following:

I say that it is obvious that mostly for any $i$: $M_i<M_{i+1}$, where $i=1,\ldots,\infty$ That is why the probabilities for different values of $M$ obey this inequality: $$P(\|X_n\| > M_i) > P(\|X_n\| > M_{i+1}) > \cdots > 0$$

Hence, we can state that there exist some $\varepsilon$ for which there exist $M$ such that using it, $\varepsilon$ can be chosen such that $P(\|X_n\| > M) < \varepsilon$

But I think this Hence, we can state... is not really mathematical proof. Which steps mathematically are omitted by me for this proof to be mathematically correct? Really do not want to skip any step and want to make this proof as meticulous as possible.

• A statement that $\text{“}\varepsilon$ can be chosen such that $P(\cdots\cdots)<\varepsilon\text{''}$ is trivially true, since one could choose $\varepsilon=2$. ${}\qquad{}$ Oct 17, 2015 at 18:07
• that is nice remark! Oct 18, 2015 at 10:37
• The point of the remark is that you should edit your question to say 'leads to for all $\epsilon>0$ and for all $M>0$ there exists $N$ such that if $n \ge N$ then...'. Oct 19, 2015 at 2:37

Do you mean that "for all $\epsilon>0$" (instead of "there exists some $\epsilon>0$"), there exists some $M$ such that $\mathbb{P}(\Vert X_n\Vert >M)<\epsilon$ for large enough $n$? Because otherwise the proof is pretty simple, just pick $\epsilon=2$

A possible proof is by contradiction. Suppose that

Hence, we can state that there exist some $\epsilon>0$ for which there exist $M$ such that there is some $N$ such that $n\geq N\Rightarrow \mathbb{P}(\Vert X_n \Vert >M)<ϵ$

is false. Then, there is a certain $\epsilon>0$ such that for all $M$, for all $n$, there is $k\geq n$ such that $\mathbb{P}(\Vert X_k \Vert >M)\geq ϵ$

Then, we construct recursively the following sequence $M_n$. $M_1=1$, and, for $M_n$, let $$M_{n+1}=\begin{cases} M_{n} & \text{if \mathbb{P}(\Vert X_n\Vert>M_n)<\epsilon}\\ M_{n}+1 & \text{if \mathbb{P}(\Vert X_n\Vert>M_n)\geq\epsilon} \end{cases}$$ First, we prove that $M_n\rightarrow \infty$. For $n\in\mathbb{N}$, we already have that, for $M_n$, there is some $k\geq n$ such that $\mathbb{P}(\Vert X_k\Vert>M_n)>\epsilon$. Then we have $M_{k+1}=M_k+1$. This proves the sequence diverges to infinity.

Second, since for all $n$, there is some $k\geq n$ such that $\mathbb{P}(\Vert X_k\Vert>M_n)>\epsilon$ we note we can extract a subsequence $k_n$ such that $\mathbb{P}(\Vert X_{k_n} \Vert>M_{k_n})\geq\epsilon$. This contradicts that

$\mathbb{P}(\Vert X_n\Vert>M_n)\rightarrow 0$ as $n\rightarrow \infty$ for every $M_n\rightarrow \infty$

• Dear Nate River, sorry for confusing, you are right. The tasks sounds like this: for every $\epsilon>0$ there exist $M$ and $N$ such that $P(...) <\epsilon$ for every $n>=N$ Oct 18, 2015 at 10:40
• I think that there is a mistake in a line: is false. Then, there is a certain ..... $\mathbb{P}(\Vert X_k \Vert >M)<ϵ$ . Here must be $\mathbb{P}(\Vert X_k \Vert >M)>ϵ$ as you even call this when you prove first part. Yes? Oct 18, 2015 at 10:49
• Indeed. I just edited this. Oct 18, 2015 at 10:50