Diagonalisation Argument for an Array of Random Variables

Question

Let $(A_{mn})_{m\in \mathcal{I},n\in\mathbb{N}}$, where $\mathcal{I}$ is some index set (for example, $\mathcal{I} = \mathbb{R}^+$ or $\mathcal{I} = \mathbb{N}$). Suppose further that for each fixed $m \in \mathcal{I}$, $A_{mn}\xrightarrow{p} 0$ as $n\rightarrow \infty$.

Question: I wish to show that there exists a sufficiently slowly non-decreasing $m \equiv m(n)$ such that $A_{m(n) n} \xrightarrow{p} 0$ as $n\rightarrow \infty$. Is this possible using some diagonalisation argument? I can't seem to formalise this right now.

EDIT: It is required that $m(n)\rightarrow \infty$ as $n\rightarrow \infty$.

You could consider $\mathcal{I}=\mathbb{N}$ and $A_{mn}=\frac{m}{n}$ (constant random variables). Then, for each fixed $m\in\mathbb{N}$, $A_{mn}\rightarrow0$ as $n\rightarrow\infty$ in the real number sense, so in probability. Let $m:\mathbb{N}\rightarrow\mathbb{N}$ be an increasing sequence. It is easy to prove by induction that $m(n)\geq n$ (indeed, for $n=1$, we have $m(1)\geq1$, and if we assume by induction hypothesis that $m(n-1)\geq n-1$, then $m(n)\geq m(n-1)+1\geq n-1+1=n$). Therefore $A_{m(n),n}=m(n)/n\geq 1$, so it cannot converge to $0$. — user39756, Jan 01 '17 at 10:34
Why can't I choose my sequence $m(n) = \lfloor\sqrt{n}\rfloor$ for example, then wouldnt $A_{m(n),n}$ would behave like $(n)^{-1/2} \rightarrow{0}$, hence there exists a sufficiently slowly increasing sequence $m(n)$ such that $A_{m(n),n} \rightarrow 0$? — user61038, Jan 01 '17 at 10:48
I understood by increasing the definition from here. So you are refering to $m(n+1)\geq m(n)$, and not $m(n+1)>m(n)$ as I thought, right? — user39756, Jan 01 '17 at 11:37
But if you allow having $m(n+1)\geq m(n)$, why don't you take $m(n)=1$ constant? Then $A_{m(n),n}=A_{1,n}\rightarrow 0$ by hypothesis. — user39756, Jan 01 '17 at 11:40
The user39756 observations are good. If you require $m(n)$ to be nondecreasing and $\lim_{n\rightarrow\infty} m(n)=\infty$, you can just hold $m(n)$ constant for a while and then increase it only when we want. This question seems similar to the fact "if a sequence of random variables converges in probability, then there is a subsequence that converges with prob 1." — Michael, Jan 01 '17 at 20:54
OFcourse that I understand, but the point is I want $m(n)$ to diverge to infinity. I DO NOT need a sequence that is STRICTLY increasing, I suppose I could have said nondecreasing instead. — user61038, Jan 01 '17 at 23:57
Thanks @Michael that's the kind of intuitive argument that I'm looking for, do you have any references where I might be able to find a logical step by step proof of it? — user61038, Jan 02 '17 at 00:16
A proof of the fact I mentioned above can be found here: http://math.stackexchange.com/questions/222264/another-question-on-almost-sure-and-convergence-in-probability The current question is different, but you can answer it with similar argumentation. — Michael, Jan 02 '17 at 00:27

user61038 · Accepted Answer · 2017-01-03T05:15:59.143

Define a sequence $m_k \rightarrow \infty$ as $k\rightarrow\infty$. Then for each fixed $k >0$, we have that $P(|A_{m_k n}| > 2^{-k})\rightarrow 0$ as $n\rightarrow \infty$. In particular, there exists an $N_k$ such that

$$ P(|A_{m_k n}| > 2^{-k}) \leq k^{-1}$$

for all $n\geq N_k$. WLOG suppose that the sequence $(N_k)_{k\in\mathbb N}$ is strictly increasing. Define $m(n) = m_k$ , $r(n) = 2^{-k}$ whenever $N_k\leq n < N_{k+1}$. Then observe that $m(n)\rightarrow \infty$, and $r(n)\rightarrow 0$ as $n\rightarrow \infty$.

Now fix $\epsilon > 0$. Then there exists an $k^*$ such that $\epsilon > 2^{-k^*}$, and in particular

$ \begin{align*} P(|A_{m(n)n}| > \epsilon) \leq P(|A_{m(n)n}|> 2^{-k^*}) \leq k^{*^{-1}}, \end{align*} $

for $N_{k^*} \leq n < N_{k^*+1}$. Whenever $n> N_{k^*+1}$, observe that $r(n) \leq 2^{-k^*}$ and $P(|A_{m(n)n}| > r(n)) \leq (k^*+1)^{-1} \leq k^{*^{-1}}$ and therefore we have that

$$ P(|A_{m(n)n}| > \epsilon) \leq k^{*^{-1}} $$

for all $n \geq N_{k^*}$. In fact, this argument can be applied to any $k>k^*$ and so we have that $P(|A_{m(n)n}| > \epsilon) \leq k^{-1}$ for all $n > N_k$. Therefore, we have constructed a sequence $m(n)$ diverging sufficiently slowly enough such that $|A_{m(n)n}| = o_p(1)$.

Very nice. You can simplify by just using $m_k\equiv k$. Also, a minor typo: It should say "there exists a $k^$ such that $\epsilon>2^{-k^}$" (rather than $\epsilon > 2^{-k}$). — Michael, Jan 03 '17 at 04:21

Diagonalisation Argument for an Array of Random Variables

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