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(I know this example here uses the $o_p$ which is a bit different than $o$, but I'm guessing the same rule applies to both).

In one proof my professor used the following:

$$\hat \theta_n =\theta + o_p (\frac{1}{\sqrt n}) $$ $$o_p(\sqrt n(\hat \theta_n-\theta))=o_p(1) $$ Technically, if I derive it, then $$o_p(\sqrt n(\hat \theta_n-\theta))= o_p(\sqrt n(\theta +o_p(\frac{1}{\sqrt n}) - \theta))=o_p(\sqrt n o_p(\frac{1}{\sqrt n}))=o_p(o_p(1))$$

Intuitively I understand we are talking magnitudes here, but I'm missing the formal/rigor/definition understanding why this is equal to $o_p(1)$

Maverick Meerkat
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  • $o_{t\to a}(1)$ is any function $t\mapsto g(t)$ defined locally near $t=a$ such that $g(t)\to0$ as $t\to a$. Thus $o_{t\to a}\left(o_{t\to a}(1)\right)$ is any function $t\mapsto f(g(t))$ defined locally near $t=a$ such that $f(t)\to0$ as well as $g(t)\to 0$ as $t\to a$. So $f(g(t))\to 0$ as $t\to a$ and thus $o_{t\to a}\left(o_{t\to a}(1)\right)$ is also $o_{t\to a}(1)$. – Nicolas Jan 11 '21 at 12:54

1 Answers1

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By definition of $o(g)$, we have that $$ f(n) = o(g(n)) \iff \lim_{n \to \infty} \frac{f(n)}{g(n)} = 0.$$ Read it as '$f$ is eventually much smaller than $g$'. What your professor is essentially claiming, is that if $f = o(g)$ and $g = o(h)$, then $f = o(h)$. This is saying that you can compose little-oh with itself. Indeed, $f = o(h)$ because $$ \lim_{n \to \infty} \frac{f(n)}{h(n)} = \lim_{n \to \infty} \frac{f(n)}{g(n)}\frac{g(n)}{h(n)} = 0.$$ In fact, more is true: if we weaken $g = o(h)$ to $g = O(h)$, it still works.

Thomas Bakx
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