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For a scalar random sequence $X_n$, we write $X_n=O_p(a_n)$ if for every $\epsilon$ there exists $M_{\epsilon}$ such that $\limsup_{n}\Pr(|X_n/a_n|>M_{\epsilon})<\epsilon$.

What's the extension of this definition to the vector case? For example, when we write random vector $Z_n=O_{p}(a_n)$, do we mean that for every $\epsilon>0$ there exists $M_{\epsilon}$ such that $\limsup_{n}\Pr(||Z_n/a_n||>M_{\epsilon})<\epsilon$, where $||\cdot||$ is the Euclidean norm?

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    $\begingroup$ I think there is a problem with your scalar definition of $O_p(\cdot)$. If $Z_n \equiv Z \sim N(0,1)$ then of course $Z_n = O_p(1)$ but it is not the case that there exists $M$ such that $\Pr(|Z_n| > M) \to 0$. Instead, you want that for every $\epsilon$ there exists an $M$ such that $\limsup_n \Pr(|Z_n| > M) \le \epsilon$. $\endgroup$ – guy Jul 16 '20 at 22:27
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    $\begingroup$ I get the feeling that it might not matter which ($L_p$) norm you choose; there will still be such an $M$. It might be a different $M$ for a different norm, but the requirement is that $M$ exists, not that $M$ is a particular value. $\endgroup$ – Dave Jul 16 '20 at 22:29
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    $\begingroup$ I don’t know that I’m right. I just started wondering what would happen if you used an $L_1$ norm, and I think we still get an $M$. $\endgroup$ – Dave Jul 16 '20 at 22:47
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    $\begingroup$ In Wooldridge, "Econometric analysis and panel data" is said, that the definition of $O_p(\cdot)$ in scalar case is applied element by element to vectors, what is equivalent to say, that Euclidean norm of vector is bounded in probability. Analogously, definition could be extended to random matrices. $\endgroup$ – Mentossinho Jul 16 '20 at 22:55
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    $\begingroup$ @JTS365 To answer you question about the wikipedia link, your translation is not correct. The correct translation is that for every $\epsilon$ there exists an $M$ such that $\limsup_n \Pr(|Z_n| > M) \le \epsilon$, which is quite different from what you have. $\endgroup$ – guy Jul 17 '20 at 19:54
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With vectors of fixed finite dimension, say $m$, you can do this in various equivalent ways. Write $X_{ni}$ for the components of $X_n$.

  1. $X_n=O_p(1)$ when the components $X_{ni}$ of $X_n$ are $O_p(1)$ as scalar sequences.
  2. $X_n=O_p(1)$ when the Euclidean norm $\|X_n\|_2=\sqrt{\sum_{i=1}^m X_{ni}^2}$ is $O_p(1)$ (note: this is the random quantity $\|X_n\|_2$, not the possibly non-existent $E[\|X_n\|_2])$
  3. Like 2, but for any $p$-norm $\|X_n\|_p=\sqrt[p]{\sum_{i=1}^m |X_{ni}|^p}$, for $0<p\leq \infty$

These are all equivalent to each other, and they are equivalent to tightness of the sequence, which is a topological condition: for any $\epsilon$ there exists a compact set $K_\epsilon$ such that $P(X_n\in K_\epsilon)>1-\epsilon$ for all $n$.

These are all different if the dimension is not finite or is allowed to grow with $n$; in that context you would usually be explicit and say $\|X_n\|_2=O_p(1)$ or $X_{ni}=O_p(1)$ componentwise. In particular, definition 1 is a very weak condition for infinite-dimensional vectors.

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    $\begingroup$ Thanks! This is very helpful! $\endgroup$ – T34driver Jul 19 '20 at 18:21

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