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Suppose that we know: as $n\to\infty$, $h\to 0$ and $nh\to\infty$. Why does it follow that

  1. $\frac{O(h)}{nh}=o[(nh)^{-1}]$,
  2. $O_p(h^2+(nh)^{-1/2})=o_p(1)$?

I'm learning kernel density estimation from Li and Racine (2007). The above appeared on pages 9--12 of the text without any explanation. Could someone please help me figure them out?

Community
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yurnero
  • 10,505

1 Answers1

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Recall that $$ f=\mathcal{o}(g) \quad \text{as} \quad x \rightarrow +\infty \quad \text{if} \quad \frac{f}{g} \rightarrow 0. $$ Here, from your hypotheses, you have

  1. $$ \frac{\frac{O(h)}{nh}}{(nh)^{-1}}=O(h)\rightarrow 0 \qquad \text{giving} \qquad \frac{O(h)}{nh}=o[(nh)^{-1}] $$

  2. $$ \frac{O_p(h^2+(nh)^{-1/2})}{1}=O_p\left(\max (h^2,(nh)^{-1/2})\right)\rightarrow 0 $$

giving $$ O_p(h^2+(nh)^{-1/2})=o_p(1). $$

Olivier Oloa
  • 120,989
  • Thank you! Could you please elaborate more on the steps $O(h)\to 0$ and $O_p(\max(h^2,(nh)^{-1/2})\to 0$. For the latter convergence, don't we need convergence in probability? – yurnero Dec 21 '14 at 17:07
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    @yurnero Thanks. Recall that $f=O(h)$ implies $|f|\leq Ch$ for some constant $C$, thus you have $O(h)\rightarrow 0$ as $h \rightarrow 0$ Please, what do you mean by writing $O_p(g)$ for some function $g$? – Olivier Oloa Dec 21 '14 at 17:20